CHEMICAL KINETICS
1. IIT–JEE Syllabus
Rates of Chemical reactions, Order of reactions, Rate constant, Effect of concentration and Temperature, Kinetics of first–order reactions, Arrhenius Equation, Kinetics of radioactive disintegrations.
- Introduction
Chemical Kinetics is the branch of science that deals with rate of reaction, the factors affecting the rate of reaction mechanism Chemical reactions need varying length of time for completion depending upon nature of reactants and products and conditions under which reaction is run. Many reactions such as explosive reaction take place in a fraction of a second while the rusting of a wrecked ship may take centuries to complete. Similarly ionic reactions in solution such as that between HCl (aq) and AgNO3 (aq) giving a white precipitate of silver chloride within twinkling of eye. In chemical kinetics we actually study the rates of only those reactions which are in between the above two extremes, i.e. neither so fast nor so slow.
A chemical reaction involves breaking of bonds in reactant molecules and making of bonds in product molecules. A reaction that involves breaking of weak bond(s) is faster than one involving breaking of strong bond at a given temperature. Ionic compounds remain completely ionised in solution and in any ionic reaction no bond is to be broken. This is why an ionic reaction is so fast. Different reactions differ in respect of the strength of the bonds to be broken and hence they occur at different rates. Reactions which involve less bond rearrangements are generally faster than those which involve considerable bond rearrangements, at a given temperature.
- Rate of a Chemical Reaction
The change in the number of moles / litre of any reactant or product per unit time is known as rate of reaction. Thus,
Rate =
Where ΔC is change in the concentration of reactant or product during time interval Δt.
– = rate at which concentration of reactant decreases
+ = rate at which concentration of product increases
Rate of reaction is not uniform. It goes on decreasing from moment to moment due to decrease in the concentration(s) of reactant(s) with the progress of reaction i.e. with time as shown below by rate vs time curve. Thus, the rate defined above is actually the average rate of reaction during the time interval considered. |
The rate of reaction at any time t also called instantaneous rate is as defined below.
rinst =
Where dc is the infinitesimal change in the concentration of the reactant or product during the infinitesimal small time interval dt after time t i.e. between t and t + dt.
dt being infinitesimal small, the rate of reaction may be assumed to be constant (uniform) during this time interval.
For the reaction:
2N2O5 ⎯→ 4NO2 + O2
the rate can be expressed as one of the following
, + and +
These individual rates may be made equal to one another by dividing them by the respective stoichiometric coefficients in the chemical equation for the reaction. Thus,
Rate of reaction = – = +
Unit of rate of reaction =
= mole litre–1time–1
The rate of reaction at any time t may be determined by plotting concentration (of reactant or product) vs. time and drawing a tangent at the point P of the curve corresponding to time t at which rate is to be determined. The slope of the tangent gives rate of reaction at the required time.
([R] = concentration of reactant and [P] = concentration of product)
Rate at the time t = slope of tangent = tanθ
- Order and Molecularity of Reaction
Chemical reactions may be classified as “elementary” (or simple) and “complicated”. Reactions which occur in one and only one step are actually called elementary reactions and reactions occurring in the sequence of two or more steps are called complicated reactions. The sequence of two or more steps in which a complicated reaction occurs is called its mechanism. Thus, complicated reaction has mechanism while elementary reaction has not. Further, each step of a complicated reaction is an elementary reaction.
The molecularity of an elementary reaction is defined as the minimum number of molecule(s), atom(s) or ion(s) of reactant(s) required for the reaction to occur i.e. which are actually needed for transition state formation of the reaction, and is equal to the sum of the stoichiometric coefficients of the reactants as appearing in the chemical equation for the reaction. Thus, for the reaction represented by the chemical equation.
aA + bB ⎯→ Products
molecularity = a + b
Examples:
Reactions Molecularity
PCl5 PCl3 + Cl2 1
H2 + I2 2HI 2
Reactions are classified as unimolecular, bimolecular, ter molecular etc according to their molecularity equal to one, two, three etc., respectively. A complicated reaction has, in fact, no molecularity of its own but molecularity of each of the steps involved in its mechanism. For example, the reaction: 2NO + 2H2 ⎯→ N2 + 2H2O, is complicated and takes place in this sequence of following three steps.
I NO + NO N2O2 (fast and reversible)
II N2O2 + H2 ⎯→ N2O + H2O (slow)
III N2O + H2 ⎯→ N2 + H2O (fast)
The molecularity of each step is 2 and hence we say that the reaction takes place in the sequence of three bimolecular steps.
The rates of various elementary reactions in any reaction mechanism generally differ from one another. The rate of overall reaction and the rate of product formation evidently cannot be faster than the rate of the slowest reaction. To take analogy, suppose in a bicycle factory, the different components are manufactured at the rate of 200, 150, 50, 250 and 100 pieces per day. The rate of bicycle production cannot be more than 50 bicycles per day rather it must be exactly 50 bicycles per day. Thus, the slowest step is always the rate determining step (r.d.s.). Based on this concept there is another view regarding molecularity of a complicated reaction. According to which molecularity of a complicated reaction or better to say the overall molecularity of the reaction mechanism is given by the number of molecules atoms or ions of reactant(s) and/or intermediates coming into contact and colliding simultaneously in the slowest step of the mechanism i.e. the r.d.s. of the reaction. To understand this view let us consider the following nucleophilic substitution reaction.
(CH3)3C – OH + HCl (CH3)3C – Cl + H2O
The reaction occurring in the sequence of following three steps is said to follow SN1 i.e. substitution nucleophilic unimolecular mechanism since r.d.s. is unimolecular.
I (CH3)3 C –+ (CH3)3 C – +Cl– (fast)
II (CH3)3 C –– (CH3)3C⊕ + (slow r.d.s.)
III (CH3)3 + – ⎯→ R – (fast)
The mathematical equation expressing concentration dependence of rate is commonly known as rate law or rate expression of the reaction and sum of the indices of the concentration terms appearing in the experimentally observed rate law of a reaction is known as order of the reaction.
Thus, if for a reaction: aA + bB ⎯→ products, rate law will be as follows:
Rate ∝ [A]m [B]n
Then
Order w.r.t. A = m
Order w.r.t. B = n and
Overall order = m + n
Note that m and n are experimental quantities which may or may not be equal to the respective stoichiometric coefficients (a and b) i.e. molecularity, so, order of reaction cannot be predicted by just seeing the chemical equation.
Let us consider the reaction:
2NO + 2H2 ⎯→ N2 + 2H2O
Kinetic experiment carried on the reaction has revealed following facts.
- i) Rate increases four fold when concentration (or partial pressure) of nitric oxide (NO) is doubled keeping that of H2
- ii) Rate gets doubled with doubling of concentration (or partial pressure) of hydrogen keeping that of NO constant.
iii) Rate increases 8-fold when concentrations (or partial pressure) of both NO and H2 are doubled simultaneously.
From these facts it is evident that the reaction obeys following kinetics (rate law).
Rate ∝ [NO]2 [H2]
Thus, order w.r.t. NO = 2
Order w.r.t. H2 = 1
Overall order = 2 + 1 = 3
Reactions are classified as zero order, 1st order, 2nd order etc.
4.1 Distinction between order and molecularity
- i) Molecularity is a theoretical property and it concerns with mechanism while order is an experimental property and it concerns with kinetics.
- ii) Molecularity is always an integer excepting zero while order may be any number integral or fractional including zero.
iii) Order may change with change in experimental conditions while molecularity is a fixed property of a reaction and is independent of experimental conditions.
Example: The isomerisation of cyclopropane into propene is a 1st order at high cyclopropane gas pressure but 2nd order at low cyclopropane gas pressure.
Rate ∝ [Cyclo-propane] … at high pressure
Rate ∝ [Cyclo-propane]2 … at low pressure
Order may differ from the molecularity if reaction is complicated or if concentrations of any of the reactants be taken in large excess as compared to other.
For example, let us consider the reaction “the acid catalysed hydrolysis of an ester say ethyl acetate” represented by the following chemical equation.
CH3COOC2H5 + H2OCH3COOH + C2H5OH
As it is evident from the chemical equation, the reaction is bimolecular. But the order of reaction is “one”. It obeys the following rate law as observed experimentally.
Rate ∝ [CH3COOC2H5]
i.e. order w.r.t. ester = 1
order w.r.t. water = 0
overall order = 1 + 0 = 1
The reaction is actually carried out taking dilute aqueous solution of ester and dilute acid (HCl). In the reaction mixture so employed, the concentration of water remains in large excess as compared to the concentration of ester. Though, ester and water both are consumed in equimolecular amounts during the reaction, the decrease in the concentration of water is negligibly small in comparison to its large initial concentration. In other words, the concentration of water remains practically constant inspite of change in its concentration during the reactions.
As per Law of Mass Action, the rate of reaction may be given as
Rate ∝ [ester] [water]
or Rate = k [ester] [water]
or Rate = K[ester]
Where K = k [water] = another constant.
The reaction is called “pseudo unimolecular” and K is called “pseudo rate constant” or apparent or observed rate constant. In the light of the above example, the order of reaction may be defined as the number of molecules of the reactant(s) whose concentration(s) alters during the chemical charge.
Illustration- 1: The rate of hydrolysis of an ester (0.01 M) in aqueous solution is carried out. How will you explain the fact that rate of hydrolysis with respect to ethyl acetate can be determined but with respect to water can not be determined.
Solution: When reaction is complete 0.01 moles of ester will completely react with 0.01 moles of water according to following equation-
As water is present in large excess its concentration remains constant throughout the reaction –
i.e. Rate = K [ester]m [H2O]n [where m and n are orders with respect to ester and water respectively]
∴Rate = K[ester]m [K’]n [K’ = constant]
∴Rate
∴Rate = K″ [ester]m
rate only depends upon concentration of ester order with respect to water can not be determined.
4.2 Rate Law and Mechanism
The main important job of Chemical Kinetics is to propose reaction mechanism. The reaction mechanism is proposed keeping the rate law in mind. If the proposed reaction mechanism explains the rate law, it goes in terms of its acceptance. From reaction mechanism rate law is derived using the following facts.
- i) The slowest step is the rate-determining step i.e. the rate of overall reaction is the rate of the slowest step.
- ii) Rate of any elementary step as according to Law of Mass Action varies as the product of the molar concentration of the reactants, each concentration term being raised to the power equal to stoichiometric coefficient of the species concerned in the chemical equation.
iii) The concentration of any intermediate must be inserted in terms of the concentrations of the reactants for writing finally the rate law of the overall reaction.
For example, the reaction
2NO + 2H2 ⎯→ N2 + 2H2O is found to obey the following third order reaction.
Rate ∝ [NO]2 [H2]
The rate is not consistent with chemical equation of the reaction indicating thereby that the reaction is not elementary but complicated. Following mechanism has been proposed to explain the observed rate law.
I NO + NO N2O2 (fast and reversible)
II N2O2 + H2 ⎯→ N2O + H2O (slow)
III N2O + H2 ⎯→ N2 + H2O (fast)
Derivation of the Rate Law from Mechanism:
Step II being the slowest is the rate determining step. So,
Rate of overall reaction = Rate of Step II ∝ [N2O2] [H2]
or Rate of overall reaction = KII [N2O2] [H2] …(1)
Where kII is the rate constant of step II
The step I being reversible, the equilibrium constant of the reaction is given by
Keqlbm =
or [N2O2] = Kequl [NO]2
Putting this in equation 1 mentioned above
Rate = k[NO]2 [H2]
Or Rate ∝ [NO]2 [H2]
Where k = KII ⋅ Keql = rate constant of the overall reaction
This is the same rate law as observed experimentally. Hence, the above mechanism is acceptable.
There may be more than one mechanism leading to the observed rate law. In such a case most probable mechanism is decided by detecting the intermediate and also taking into account of the fact that a bimolecular collision is more probable than a trimolecular collision which in turn is more probable than a tetra molecular collision and so on. For example, the following two mechanisms of the reaction: reduction of nitric oxide by hydrogen as considered above also confirm to the experimentally observed rate law. But they are less probable than one considered above, as which of these two mechanisms involves ter molecular rate determining step, the mechanism mentioned above involves bimolecular rate determining step.
I 2NO + H2 ⎯→ N2O + H2O (slow)
N2O + H2 ⎯→ N2 + H2O (fast)
II 2NO + H2 ⎯→ H2O2 + N2 (slow)
H2O2 + H2 ⎯→ 2 H2O (fast)
4.3 Examples of reactions of various orders
- i) First order reactions
- a) N2O5 ⎯→ 2NO2 + O2
- b) Br2 ⎯→ 2Br
- c) 2HNO3 ⎯→ 2NO + H2O
- d) H2O2(aq) ⎯→ H2O + O2
These reactions obey following rate law.
Rate ∝ [R]
Where R = N2O5, Br2, HNO3 and H2O2 for the reactions (a), (b), (c) and (d), respectively.
- ii) Second – Order Reactions
- a) H2 + I2 2HI, Rate ∝ [H2] [I2]
- b) CH3COOH + C2H5OH CH3COOC2H5 + H2O
Rate ∝ [CH3COOH] [C2H5OH]
- c) CH3COOEt + NaOH ⎯→ CH3COOH + EtOH
Rate ∝ [CH3COOEt] [NaOH]
- d) 2NO2 ⎯→ N2O2 + O2
Rate ∝ [NO2]2
iii) Fractional – order reactions
CH3CHO ⎯→ CH4 + CO
Rate ∝ [CH3CHO]3/2
- iv) Pseudo first – order Reactions
- a) CH3COOEt + H2O CH3COOH + EtOH
- b) + H2O +
These reactions follow the following 1st order kinetics.
Rate ∝ [R], R = CH3COOEt and C12H22O11
Pseudo first – order reaction also called pseudo unimolecular reaction is a bimolecular reaction conforming to the first-order kinetics due to large excess concentration of one of the reactants.
- v) Zero-order Reactions
2NH3 N2 + 3H2 [when catalyst surface is saturated]
Reaction between chlorine and hydrogen in sunlight over water in inverted tube
- Integrated Rate Law for a First Order Reaction
The differential rate law only tells us how the rate of a reaction depends on the concentration. In order to get some information on how the concentration of a reactant depends on time, we need to integrate the differential rate law. The integrated version in turn is called the integrated rate law. For example
A ⎯→ Product
Let us assume that the reaction is of first order
= k [A] ⇒ = kdt
The left hand side of the equation is a function of only concentration, and right hand side apart from a constant is only the differential of time. Therefore we can integrate both sides taking limits as [A]o which is the concentration at t = 0 and [A]t which is the concentration left at time t.
Thus
⇒ ⇒ ⇒
or k=
This is the integrated rate law for a first order reaction.
Illustration-2: Calculate the average rate of decomposition of N2O5 of the reaction
2N2O5(g) → 4NO2(g) + O2(g)
During the time interval t = 600s to t = 1200s. Use the following data:
Time [N2O5]
600s 1.24 × 10–2 M
1200s 0.93 × 10–2 M
Solution: We know that the decomposition of N2O5 is a first order reaction. Therefore,
or k =
Now rate of decomposition at 600s
= (4.8 × 10–4s–1) (1.24 × 10–2M)
= 5.95 × 10–6 M s–1
Similarly rate of decomposition at 1200s
= (4.8 × 10–4 s–1) (0.93 × 10–2 M) = 4.46 × 10–6 M s–1
Average rate of decomposition = Ms–1
= 5.205 × 10–6 Ms–1
Exercise- 1: The first order reaction has k = 1.5 × 10-6 per second at 200°C. If the reaction is allowed to run for 10 hours, what percentage of the initial concentration would have changed in the product? What is the half-life period of this reaction?
5.1 Half Life
It is customary to use a quantity called the half life of a reaction. It is the time required for the concentration of a reactant to decrease by half of its initial value. It should be understood that it is not defined as the time required for the concentration of a reactant to decrease to half of its initial value. One may wonder that when both of them mean the same, so why should we be so rigorous about the definition? The answer lies in the fact that when we define, say, three fourth life time, we define it as the time required for the reactant to decrease by ¾th of the initial amount and not to decrease to ¾th of the initial amount.
For a first order reaction
= kt½
∴ t½ =
Therefore the time required for a reactant to decrease by half of its initial value is independent of the initial concentration of the reactant for a first order reaction.
- Half – Life of a nth Order Reaction
Let us venture on to find out the t½ for a nth order reaction where n ≠1.
∴
⇒ ⇒
⇒ ⇒
⇒ ⇒ ( order n≠1)
Therefore for a nth order reaction, the half life is inversely related to the initial concentration raised to the power of (n–1).
Note : It can be noted that for a zero order reaction t1/2 = .
Illustration-3: Find out the order of reaction under following conditions:
- a) When t1/2 of the reaction is halved as initial concentration of reactant is doubled.
- b) When t1/2 of the reaction is doubled as the initial concentration of reactant is doubled.
- c) When t1/2 of the reaction remains unchanged as initial concentration of reactant is doubled.
Solution: a) In present case
t1/2 α [a0 = initial concentration] …(1)
…(2)
∴ On dividing equation (1) by equation (2)
2 = 2n–1
n – 1 = 1
or n = 2
- b) In this case
2 × t1/2 α …(3)
On dividing equation (1) by equation (2)
= 2n–1
0.5 = 2n–1
log 0.5 = (n –1) log2
∴ n – 1 =
n –1 = –1
∴ n = 0
- c) In this case
t1/2 α …(4)
On dividing equation (1) by equation (4)
1 = 2n–1
log1 = (n –1)log2
∴ n –1 = = 0
or n = 1
Exercise- 2: Find out the unit of rate constant for an nth order reaction
- Some Simple First Order Reactions
As we have to generally deal with only first order reactions, we examine some of these reactions from the point of calculating the rate constant based on different experimental data. Now we present several problems in which we shall learn how to calculate the rate constant of reactions based on the variety of data given.
Illustration 4: We are given a first order reaction
A → B+C where we assume that A,B and C are gases.The data given
to us is
Time | 0 | t |
Partial pressure of A | P1 | P2 |
And we have to find the rate constant of the reaction.
Solution: Since A is a gas and assuming it to be ideal, we can state that PA = [A]RT
[From PV = nRT]. ∴ at t = 0, P1 = [A]o RT and at t = t, P2 = [A]tRT. Thus the ratio of the concentration of A at two different time intervals is equal to the ratio of its partial pressure at those same time intervals.
∴
∴
Illustration 5: Now we consider the same reaction with different set of data
A ⎯→ B +C
Time | 0 | t |
Total pressure of A +B+C | P1 | P2 |
Find k.
Solution: In this we are given total pressure of the system at these time intervals. The total pressure obviously includes the pressure of A, B and C. At t = 0, the system would only have A. Therefore the total pressure at t = 0 would be the initial pressure of A ∴ P1 is the initial pressure of A. At time t let us assume that same moles of A decomposed to give B and C because of which its pressure is reduced by an amount x while that of B and C is increased by x each. That is
A ⎯→ B + C
Initial P1 0 0
At time t P1–x x x
∴ total pressure at time t = P1 + x = P2 ⇒ x = P2–P1
Now the pressure of A at time t would be P1– x = P1 – (P2–P1) = 2P1–P2
∴ k = ln
Illustration 6: In this case we have
A ⎯→ B +C
Time | t | ∞ |
Total pressure of A+B+C | P2 | P3 |
Find k.
Solution: Here ∞ means that the reaction is complete. Now we have
A ⎯→ B + C
At t=0 P1 0 0
At t =t (P1– x) x x
At t =∞ 0 P1 P1
∴ 2P1 = P3
⇒ P1 =
At time t,
P1 + x = P2
⇒ = P2
⇒ x = P2–
⇒P1– x = –(P2–) = P3–P2
∴
Illustration 7: A ⎯→ B +C
Time | T | ∞ |
Total pressure of (B+C) | P2 | P3 |
Find k.
Solution: A ⎯→ B + C
At t=0 P1 0 0
At t =t P1– x x x
At t = ∞ 0 P1 P1
∴ 2P1=P3
⇒ P1=
2 x = P2
⇒ x =
∴ P1– x =
∴ k=
Exercise 3: A⎯→ B + C
Time | t | ∞ |
Partial pressure of B | P2 | P3 |
Find k.
Illustration 8: Now let us assume that A,B and C are substances present in a solution. From a solution of A, a certain amount of the solution (small amount) is taken and titrated with a suitable reagent that reacts with A. The volume of the reagent used is V1 at t = 0 and V2 at t = t
Time | 0 | T |
Volume of reagent | V1 | V2 |
The reagent reacts only with A. Find k.
Solution: It can be understood that the volume of the reagent consumed is directly proportional to the concentration of A. Therefore the ratio of volume of the reagent consumed against A at t=0 and t=t is equal to [A]0 /[A]t
∴ k =
Illustration 9: A ⎯→ B + C
Time | 0 | t |
Volume of reagent | V1 | V2 |
The reagent reacts with A,B and C. Find k.
Solution: As we had seen in the previous case that the result of illustration 5 matches that of illustration of 1, we can use a similar logic for solving this one. That is
A ⎯→ B + C
At t = 0 V1 0 0
At t = t V1– x x x
At t =∞ 0 V1 V1
where x is the volume of the reagent for those moles of A which have been converted into B and C
∴ V1 + x = V2 ⇒ x = V2 –V1 ⇒V1– x = 2V1 –V2
∴
This result looks perfectly OK. The problem is that this is true only if we make an assumption in the beginning of this problem (which you may have thought about). The assumption is that the ‘n’ factor of A,B and C with the reagent is same.
This can be made clear as follows. Let y moles of A be converted to B and C. If y moles of A require x ml of reagent and if the ‘n’ factor of A is n1 with the reagent then the equivalents of A converted are n1y and in
x ml of reagent n1y equivalents of the reagent are present. Now since the same y moles of B and C are formed and we find the same volume of reagent used for B and C, it can be calculated that y moles of B and C also contain n1y equivalents, each, which implies that ‘n’ factor of B and C are also ‘n1’.
Exercise 4: A ⎯→ B +C
Time | 0 | t |
Volume of reagent | V1 | V2 |
Reagent reacts with all of them (A, B & C). A,B and C have ‘n’ factors in the ratio of 1:2:3 with the reagent ( assuming its n factor remains same with all of them). Find k.
Illustration 10: A ⎯→B+C
Time |
T |
∞ |
Volume of reagent | V2 | V3 |
Reagent reacts with all A,B and C and have ‘n’ factors in the ratio of 1:2:3 with the reagent. Find k.
Solution: A ⎯→ B + C
At t = 0 V1 0 0
At t = t V1– x 2 x 3x
At t =∞ 0 2V1 3V1
∴ 5V1=V3
V1=V3/5
V1+4 x = V2
x =
∴V1–x =
∴ k = ln
k =
Exercise 5: Now, we will consider a reaction A B+C which is catalysed by D. We will assume in this problem that the concentration of the catalyst remains constant throughout. The data given to us is
Time | 0 | t | ∞ |
Volume of reagent | V1 | V2 | V3 |
The reagent reacts with all (A,B,C and D). Assume that ‘n’ factor of A,B and C are in the ratio of 1:1:1 and that D is not known. Find k.
Illustration 11: A B+C
Time | 0 | t | ∞ |
Volume of reagent | V1 | V2 | V3 |
The reagent reacts with only B,C and D. Find k.
Solution: V1=VD
V2 = 2x + VD
V3 = 2VA + VD
∴ = VA
∴
Illustration 12: A B+C
Time |
0 | t | ∞ |
Volume of reagent | V1 | V2 | V3 |
The reagent reacts with only C and D. Find k.
Solution: V1 = VD
V2= x + VD
V3 = VA + VD
V3–V1 = VA
V3–V2 = VA –x
∴
Exercise 6: A B+C
Time | 0 | t | ∞ |
Volume of reagent | V1 | V2 | V3 |
The reagent reacts with all of them. The ‘n’ factor of A,B and C are in the ratio of 1:2:3 with the reagent. Find k.
Illustration 13: Now we shall see how to find the rate constant of a reaction using a very different set of data. There are some organic compounds which have a property of rotating a plane polarized light in a particular direction by a particular value. The compounds are called optically active compounds. One reaction in which an optically active substance converts to some other optically active substance is, Sucrose Glucose + Fructose
Sucrose, Glucose and Fructose are all optically active and while the first two compounds are dextro rotatory (rotating the plane polarised light in the right hand direction and the last is laevo rotatory (rotating the plane polarized light in the left hand direction). All the three compounds rotate the plane polarised light by different angles and their rotation is directly proportional to concentration.
Now the problem is S⎯→ G + F and the data is
Time | 0 | t |
Rotation of sucrose | r0 | rt |
Find k.
Solution: Let the rotation of Sucrose be r1° per mole and the initial moles of Sucrose be a. ∴ r0 =
Let the moles of Sucrose that is converted to Glucose and Fructose be x.
∴ rt = (a–x) ∴ =
∴ k =
Illustration 14: S ⎯→ G + F
Time | t | ∞ |
Rotation of Glucose & Fructose | rt | r∞ |
Find k.
Solution: Let the rotation of Glucose be and that of fructose be –(since it is laevo–rotatory) per mole.
∴ S ⎯→ G + F
At t = 0 a 0 0
At t = t a– x x x
At t = ∞ 0 a a
∴ rt = x
r∞ = a
, x = ∴ a–x =
∴
Exercise 7: S⎯⎯→ G + F
Time |
0 | t | ∞ |
Rotation | ro | rt | r∞ |
The rotation is due to all (total rotation). Find k.
Illustration 15: Now let us assume that the first order reaction is A ⎯→ B+C and this solution is taken in a tube. We radiate this solution with a monochromatic light of wavelength λ and we will assume that this radiation is absorbed only by A. Let the intensity of incident light be I0 and that of the transmitted light be It. log is called transmittance and it is inversely proportional to the concentration of A.
That is log = transmittance ∝
The data given is
Time |
0 | t |
log It /I0 | x | y |
Find k.
Solution: x ∝ and y ∝
∴
- Some Complex First Order Reactions
8.1 Parallel Reactions
In such reactions (mostly organic) a single reactant gives two products B and C with different rate constants. If we assume that both of them are first order, we get. |
…(1)
…(2)
and …(3)
Let us assume that in a time interval, dt, x moles / lit of B was produced and y moles / lit of C was produced.
∴ and
∴ = . We can also see that from (2) and (3),
∴ . This means that irrespective of how much time is elapsed, the ratio of concentration of B to that of C from the start (assuming no B and C in the beginning ) is a constant equal to k1/k2.
Illustration 16:
Let k1:k2 = 1 :10. Calculate the ratio, at the end of one hour assuming that k1 = x hr–1 |
Solution: =
∴ = (k1+k2) dt
Integrating with in the required limits, we get
= (k1 + k2 )t
∴ ln = (k1+k2) t
Since = =
∴ ln = 11x
8.2 Sequential Reactions
ABC. In this A decomposes to B which in turn decomposes to C.
∴ = k1[A] … (1)
= k1 [A] –k2[B] …(2)
= k2 [B] … (3)
Integrating equation (1), we get
Now we shall integrate equation (2) and find the concentration of B related
to time t.
= k1 [A] – k2 [B] ⇒
substituting [A] as ⇒ = k1 … (4)
Integration of the above equation is not possible as we are not able to separate the two variables, [B] and t. Therefore we multiply equation (4) by an integrating factor , on both the sides of the equation.
= k1[A]0
We can see that the left hand side of the equation is a differential of .
∴ =
= dt
Integrating with in the limits 0 to t.
⇒ = k1[A]0
⇒ = ⇒
[B] = … (5)
Now in order to find [C], substitute equation (5) in equation (3), we get
=
∴ d[C] =
On integrating
=
⇒[C] =
⇒[C]=
⇒[C]=
[C]=
Bmax and tmax: We can also attempt to find the time when [B] becomes maximum. For this we differentiate equation (5) and find & equate it to zero.
∴
, taking log of both the sides
∴tmax = … (6)
Substituting equation (6) in equation (5)
Bmax =
9. Arrhenius Equation
In 1889 Arrhenius recognised the temperature dependence of rates or rate constant. He has given an emperical relation which can be written as
k = … (1)
k is the rate constant ( of any order other than zero order), A is the pre-exponential factor, Ea is the activation energy, R is the universal gas constant and T is the absolute temperature. Activation energy (Ea) is the minimum energy required by a reactant at a certain temperature to undergo transformation into product. Arrhenius clearly assumed that reactions occur because of collisions between atoms and molecules of the reactant. He |
assumed the activation energy to be the least value of energy which the colliding molecules must possess for the collision to yield a product. If we plot graph between activation energy and progress of a reaction (expressed as reaction co-ordinate),we get a graph as shown in figure.
The difference between the energies of the reactant and the transition state (TS) is called.
ΔH =
A, the pre-exponential factor or the frequency factor or the Arrhenius constant is the number of effective collisions occuring per unit time. Effective collision is the number of collisions occuring per unit time in which orientation of the colliding molecules is proper. gives the fraction of the effective collision that have the sufficient activation energy. Therefore the product of gives the number of collisions per unit time that forms the product and is called its number of productive collisions which is the rate constant, k.
For all practical calculations, we shall assume that Ea and A are temperature independent. Both A and Ea are characteristics of the reaction.
Determination of A and Ea:
First Method:
Taking log of both sides of equation (1)
ln k = ln A –
Converting natural log to common log, log k = log A –
If log k is plotted against 1/T, a straight line is obtained which is shown as:
The slope of this line is given by
slope =
Thus, knowing the slope, the Ea can be easily calculated . The intercept of the line will give the value of log A.
Second Method : The logarithmic form of Arrhenius equation is rearranged as ln k = – + ln A Differentiating with respect to temperature, we get |
=
Integrating with in the limits of temperature T1 and T2 , we get
or log … (2)
Where k1 and k2 are rate constants at temperatures T1 & T2 respectively. Thus, knowing these values Ea can be calculated. When the value of Ea is known, the value of A can be calculated by substituting its value in equation k = . In equation (2), the value of R has to be inserted in the same unit in which Ea is desired.
Temperature Coefficient : “Temperature coefficient of a chemical reaction is defined as the ratio of rate constants of a reaction at two temperatures differing by 10°C”.
∴ Temperature coefficient =
where kT is the rate constant at temperature T K and kT+10 is the rate constant at temperature (T+10) K. This ratio generally falls between 2 and 3 which means for most of the chemical reactions, the rate becomes two or three folds for every 10°C rise in temperature.
Exercise 8. The specific reaction rate of a reaction is 1 × 10–3 min–1 at 27°C and
2 × 10–3 min–1 at 37°C. Calculate the energy of activation of this reaction and its specific reaction rate at 57°C.
Exercise 9. Calculate Ea for a reaction whose rate constant at room temperature
- a) Doubles by a 10°C increase in temperature
- b) Triples by 10°C increase in temperature
Exercise 10. The number of chirps per minute of a snowy tree cricket (occanthus fultoni) at several temperatures is 178 at 25.0°C, 126 at 20.3°C and 100 at 17.3°C.
- a) Find the activation energy for chirping process
- b) The chirping rate to be expected at 14°C
Exercise 11. The activation energy of a reaction is 167.4 kJ/mol for decomposition of a gas. What percent of the molecules has sufficient energy to cross over the energy barrier at 400°K.
- Role of a Catalyst
A catalyst is a substance that takes part in a reaction system by providing a new path way with a lower activation energy between the two fixed points of reactant and product.
It is very important to understand the role of catalyst and their significance. We shall explain this with the following examples.
There are several reactions which are catalysed by H+ or OH–. An example of this is the hydrolysis of an ester in the presence of an acid. Ester + H2O Acid + Alcohol The rate of the reaction is given as Rate = k [Ester] [H+] |
In this reaction H+ plays the role of the catalyst. Its concentration in the beginning of the reaction and at the end is same. In fact the rate can be written as
Rate = k’ [Ester]
Where k′ = k [H+]
These reactions are called pseudo first order reactions, and k’ is the pseudo or effective rate constant.
Now if we go by the well known definition of a catalyst that ‘catalysts increase the rate of reaction by their presence but their presence is not required for the reaction to occur (of course now the reaction will occur slowly).
If we take this definition on its face value then the absence of H+ should not let the reaction to happen, therefore the rate of the reaction in the absence of H+ should be,
Rate = k [Ester]. This puts us in a spot because if concentration of H+ is taken less than 1 molar then the rate of the reaction would become less than the rate which we had in its absence.
This dilemma can be resolved by understanding what catalysts do. There are two types of catalyst (a) Reactant catalyst and (b) Non reactant or surface catalyst. The former increases the rate of the reaction by existing in the rate law. Therefore, this implies that in their absence the reaction rate will be zero. In fact the reactant catalysts are just like any other reactant. We call them as catalysts only because they are regenerated back at the end of the reaction. For example, during dehydration of alcohols to alkene H+ acts like a catalyst which is in first step participates in the reaction and then in second step is regenerated.
CH3-CH2OH + H+ ⎯→ CH3+ H2O
CH3-CH2+ ⎯→ H+ + CH2=CH2
The non-reactant catalysts, on the other hand, do not exist in the rate expression. For example, when alkenes are hydrogenated, Pt is used as a catalyst. Now, the rate of the reaction depends on the concentration of H2 and that of alkenes only. Changing the concentration of Pt has no effect on the rate. These catalysts increase the rate of reaction by either decreasing the activation energy or increasing the value of Arrhenius constant or by doing both.
11. Radioactivity
All radioactive decay follow 1st order kinetics and this is where the similarity ends. This will be explained in a short while.
We measured the rate of reaction in chemical kinetics based on the rate of change of concentration of reactants or products. But this procedure will not work for calculating the rate of radioactive reaction. This is because most of the time the radioactive substance is a solid. Therefore its concentration would be a constant with time (assuming it to be pure and that the product does not remain with the reactants). Therefore the rate of radioactive reactions are measured by calculating the rate of change of nuclei of the radioactive substance. For a radioactive decay A →B, the rate of reaction is calculated as
=λ NA
Where λ = decay constant of reaction
NA = number of nuclei of the radioactive substance at the time when rate is calculated.
As you can see the above rate law is very much similar to the rate law of a first order chemical reaction, but all other similarities ceases here. For example unlike a chemical reaction the decay constant (λ) does not depend on temperature. Arrhenius equation is not valid for radioactive decay.
Integrating the differential rate law we get
= λ t Where No = number of nuclei of A at t = 0
Nt = number of nuclei of A at t = t
λ = decay constant
The expression can be rearranged to give
Nt = Noe-λt … (1)
This suggests that the number of nuclei of radioactive substance A at any instant of time can be calculated, by knowing the number of nuclei at t = 0, its decay constant
and the time.
11.1 Half – Life
Just like a 1st order reaction the half life of radioactive decay is given by
Note : Let us start with 10 nuclei. If the half life is 5 minutes, then at the end of first 5 minutes, number of nuclei would be 5. Now what would be the number of nuclei after next 5 minutes? Will it be 2.5 or 2 or 3? We can clearly see that it cannot be 2.5 and if it is 2 or 3 then it cannot be called as half life. This dilemma can be overcome by understanding that all formula relating to kinetics are only valid when the sample size is very large and in such a large sample size ,a small difference of 0.5 will be insignificant.
The fact that radioactive decay follows the exponential law implies that this phenomenon is statistical in nature. Every nucleus in a sample of a radionuclide has a certain probability of decaying, but there is no way to know in advance which nuclei will actually decay in a particular time span. If the sample is large enough – that is, if many nuclei are present – the actual fraction of it that decays in a certain time span will be very close to the probability for any individual nucleus to decay. To say that, a certain radioisotope has a half –life of 5 hr., then, signifies that every nucleus of this isotope has a 50 percent chance of decaying in every 5 hr. period. This does not mean a 100 percent probability of decaying in 10 hr. A nucleus does not have a memory, and its decay probability per unit time is constant until it actually does decay. A half life of 5 hr. implies a 75% probability of decay in 10 hr., which increases to 87.5% in 15 hr, to 93.75% in 20 hr, and so on, because in every 5 hr. interval the probability of decay is 50 percent.
11.2 Average – Life Time
Average life time is defined as the life time of a single isolated nucleus. Let us imagine a single nucleus which decays in 1 second. Assuming 1 second time interval to be very small the rate of change of nuclei would be 1/1 (because –dN = 1 and dt =1). We can also see that since = λN, for a single isolated nucleus N = 1, = λ. Therefore in this present case λ=1.
Now let us assume the same nucleus decays in 2 seconds, we can see that i.e., λ is equal to ½. You will also notice that in the 1st case the nucleus survived for 1 second and in the second case it survived for 2 second. Therefore the life time of a single isolated nucleus is .
∴ tav =
11.3 Activity
Activity by definition is the rate of decay of a radioactive element. It is represented as ‘A’ and is equal to λN. By no means should activity be confused with rate of change of radioactive nuclei represented by . This is because talks about the overall change in the number of nuclei in a given instant of time while activity only talks about that change which is decay. For example if you go out to a market with Rs. 50 in your pocket and you spend Rs.20 in 5 minutes then your rate of change of money in the wallet is Rs. 4 / min and in fact the rate of spending the money is also Rs.4 / min. Here you can see both are same. But if while spending Rs. 20 in 5 minutes, somebody keeps Rs.10 in your wallet, then the rate of change of money in your wallet would became Rs. 2.5 /min, while the rate of spending the money is Rs. 4 / min. This implies that as long as the radioactive substance is only decaying the rate of change of nuclei and activity are same and equation (1) in terms of activity of a radioactive substance can be written
as At =Ao e–λt. But if the radioactive substance is also being produced this = rate of production – activity, (Of course it’s a different matter, rate of production may or may not be a constant).
11.4 Specific Activity
This is activity per unit mass of the sample. Let radioactive sample weighing w gms have a decay constant λ. The number of nuclei in the w gms would be , where
M = molecular weight of the radioactive substance and Av= Avogadro’s number.
∴ specific activity = =
It should be remembered that if a radioactive sample is pure and the product does not remain with reactant then specific activity is a constant.
Units of Activity:
The unit of radioactivity of a substance is measured as the rate at which it changes into daughter nucleus. It has been derived on the scale of disintegration of Radium.
Let us consider 1g of radium (atomic mass = 226 and t½=1600 yrs) undergoes decay, then
Rate of decay of radium = λ × Number of nuclei of Ra in 1 g
= = 3.7×1010 dps = 3.7 ×1010 Becquerrel.
= 1 curie ( 1 Ci = 3.7 ×1010 dps) = 3.7 ×104 Rutherford ( 1 rd = 106 dps)
The SI unit of activity is dps or Becquerrel.
- Solutions to Exercises
Exercise- 1: For first order reaction
k =
Putting the different values,
1.5 × 10–6 s–1 =
or = 0.0234
or
or x = 0.52 = 52%
Hence initial concentration changed into product = 52%
We know,
t1/2 = = 4.62 × 105 s = 128.33 hours
Exercise- 2: Rate = K[Reactant]n
Mol L–1S–1 = K[mol L–1]n
∴ K = = mol1–nLn-1S–1
Exercise- 3: Taking clue from the previous illustrations
P1=P3 and x = P2
∴P1– x = P3–P2
∴ k =
Exercise- 4: A ⎯→ B + C
At t = 0 V1 0 0
At t = t V1–x 2x 3x
∴ V1+4x = V2
⇒ x =
⇒ V1–x = V1– + =
∴ k =
Exercise- 5: Let VA be the volume of the reagent required by A initially and VD be the volume required by D.
∴V1 = VA + VD
V2 = (VA – x) + x + x +VD (x is the volume of the reagent required for those moles of A that have reacted to give B and C).
V3 =
We can see that, V3–V1 = VA
And V3–V2 = VA – x
∴
Exercise- 6: V1 = VA + VD
V2 = (VA – x) + 2 x + 3 x + VD
V3 = 2VA + 3VA + VD
We can see that
and
∴
Exercise- 7: S ⎯⎯→ G + F
At t = 0 a 0 0
At t = t a– x x x
At t = ∞ 0 a a
∴ r0 = a …(1)
rt = (a– x) + x …(2)
r∞ = a …(3)
Simplifying equation (2)
rt = a – x
Since a , is r0
∴ x =
Now, we can write ‘a’ as or . But the problem in writing either of them is that the constants of a and x are different. This implies that the constants of a–x would also be different from that of ‘a’.
So we write, ro –r∞ = a
∴ a = ⇒ a– x =
∴
Exercise- 8: We know that
log
or Ea =
Substituting different values in the equation, we get
Ea =
log = (42835.8 cal mol–1) (log 2)
= (12893.6 cal mol–1) = 12900 cal mol–1
To calculate specific reaction rate at 57°C, we have
log
or log k2 = log k1 +
or log k2 = log 1 × 10–3 min–1
+ = –3 + –2.7575
or k2 = 2.656 × 10–3 min–1
Exercise-9: a)
Ea = 1.987 cal mol–1 K–1 × 298°K × 308°K × 10K–1 ln2
= 13 kcal/mol
- b) Ea = 1.987 cal mol1 K–1 × 298°K × 308°K × 10 K–1 ln3
= 20 kcal/mol
Exercise 10: a)
log1.4127 =
∴Ea =
= = 53.41 kJ mol–1
b)
∴ = 2.7894 ×
∴ = 0.3588
∴ = 100.3588
∴ K1 = min–1
= min–1
= 77.91 min–1
Exercise- 11: Fraction of molecules (x) having sufficient energy to react is given by
e–Ea/RT
x = e–Ea/RT
x =
x = e–50.336
∴ Inx = – 50.336
∴ 2.303logx = – 50.336
∴ logx =
∴logx = – 21.857
∴ x = 10–21.857
= 1.389 × 10–22
- Solved problems
13.1 Subjective
Problem-1: Rate of reaction: aA + bB ⎯→ Product, is given as a function of different initial concentration of A and B.
[A] (mole dm–3) |
[B] (mole dm–3) |
Initial rate (mol dm–3 min–1) | |
i)
ii) iii) |
0.01
0.02 0.01 |
0.01
0.01 0.02 |
0.005
0.010 0.005 |
Determine the order of reaction w.r.t. A and w.r.t. B. What is the half life of A in the reaction?
Solution: From the kinetic data given in question it is evident that (i) rate is doubled by doubling the concentration of A when concentration of B is constant i.e.
Rate ∝ [A] when [B] is constant
So, order w.r.t. A = 1, and
- ii) rate remains unchanged when concentration of B is changed keeping that of A constant so
Rate ∝ [B]0 when [A] is constant
So, order w.r.t. B = 0
The rate law of the reaction may, therefore, be put as
Rate = k[A] [B]0 = k[A]
or k = = 0.5 min–1
t1/2 = = 1.386 min
Alternatively:
Let the rate r be given as
r = k[A]m [B]n
where m and n are individual order w.r.t. A and B, respectively.
Putting the data as given in (i) and (ii) in the above rate equation one by one and then dividing one by another, we get
or ∴ m = 1
Again putting the data as given in (i) and (iii) in the rate law as mentioned above in turn and the dividing one by another, we get
or 1 = 1m ⋅ = ∴ n = 0
Problem- 2: From the following data for the reaction between A and B.
[A] (mole dm–3) |
[B] (mole dm–3) |
Initial rate (mol dm–3 min–1) | ||
300 k | 320 k | |||
i)
ii) iii) |
2.5 × 10–4
5.0 × 10–4 1.0 × 10–3 |
3.0 × 10–5
6.0 × 10–5 6.0 × 10–5 |
5.0 × 10–4
4.0 × 10–3 1.6 × 10–2 |
2.0 × 10–3 |
Calculate the following
- i) The order w.r.t. A and w.r.t. B
- ii) The rate constant at 300 K
iii) The energy of activation
- iv) The pre-exponential factor
Solution: i) Let the rate law be
Rate = k[A]m [B]n
Where m and n are the order w.r.t. A and B respectively, and k = rate constant of the reaction, a constant at constant temperature.
From (a): 5.0 × 10–4 = k(2.5 × 10–4)m (3.0 × 10–5)n
From (b): 4.0 × 10–3 = k(5.0 × 10–4)m (6.0 × 10–5)n
Dividing one by another
or 2m⋅2n = 8
From (c): 1.6 × 10–2 = k(1 × 10–3)m (6 × 10–5)n
Dividing this by the rate law from (b), we get
or 2m = 4 ∴ m = 2
Putting the value of m in the above equation
22 ⋅ 2n = 8, or n = 1
- ii) Rate = k[A]2 [B]
∴ k =
= 2.67 4 × 108 L2 mol–2 s–1
iii)
=
upon solving:Ea = 55333 J mol–1 = 55.333 kJ mol–1
- iv) k =
or 2.67 × 108 = A exp
or A = 1.145 × 1018
Problem- 3: The half-life of first order decomposition of nitramide is 2.1 hours at 15°C.
NH2NO2 (aq) ⎯→ N2O(g) + H2O(l)
If 6.2 g of NH2NO2 is allowed to decompose calculate (I) time taken for NH2NO2 to decompose 99% and (ii) the volume of dry N2O produced at this point, measured at STP.
Solution: i) K = = 0.33 h–1
Let t be the time for 99% decomposition. This means at the expiry of time t; x = 99 when a = 100. So,
K =
0.33 =
∴ t = 13.96 h
- ii) of mole of NH2NO2 taken = = 0.1
No. of mole of NH2NO2 decomposed = 0.99 × 0.1 = 0.099
From stoichiometry: NH2NO2 ≡ N2O
∴ No. of mole of N2O evolved = 0.099
Volume of N2O evolved at STP = 0.099 × 22.4
= 2.2176 L
Problem- 4: The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is a first order reaction. (i) After 55 seconds at 400 K, the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate the rate constant. (ii) Calculate the pressure of Cl2O7 after 100 seconds of decomposition at this temperature.
Solution: i) k =
= = 6.2 × 10–3 s–1
- ii) Again t = 100 s, Pt = ?
6.2 × 10–3 = log or Pt = 0.033 atm
Problem- 5: Show that for 1st order reaction the time required for completion of 75% reaction is twice that for 50% reaction.
Solution:
∴
∴ t3/4/t1/2 = 2
Problem-6: The time required for 10% completion of first order reaction at 298 K is equal to that required for its 25% completion at 308K. If the preexponential factor for the reaction is 3.56 × 109 s–1, calculate its rate constant at 318 K and also the energy of activation.
Solution: For first order reactions,
t =
At 298 k; t =
At 309 K; t =
Since time is the same hence,
or
or = 2.73
According to Arrhenius equation
2.303 log
or 2.303 log 2.73 = = 76.65 kJ
and K318 = 9.036 × 10–4 s–1
Problem 7: Show that following reaction is of first order with the help of data given below
2H2O2 ⎯→ 2H2O + O2
Time (min) | 0 | 10 | 20 |
Volume of KMnO4 (ml) | 22.9 | 13.9 | 8.4 |
Solution: Here volume of KMnO4 used is proportional to the concentration of H2O2 present
or V0 α a
Vt α a – x
∴ K = log
When t = 10 min
K =
= = = 0.04998 min–1
When t = 20
K =
= = × 0.4355 = 0.0502 min–1
Thus K, comes out to be nearly constant, we can say that reaction is of first order.
Problem 8: 5 ml of ethylacetate was added to a flask containing 100 ml of 0.1 N HCl placed in a thermostat maintained at 30°C. 5 ml of the reaction mixture was withdrawn at different intervals of time and after chilling, titrated against a standard alkali. The following data were obtained :
Time (minutes) | 0 | 75 | 119 | 183 | ∞ |
Volume of alkali used in ml | 9.62 | 12.10 | 1.10 | 14.75 | 21.05 |
Show that hydrolysis of ethyl acetate is a first order reaction.
Solution: The hydrolysis of ethyl acetate will be a first order reaction if the above data confirm to the equation.
Where V0, Vt and V∞ represent the volumes of alkali used at the commencement of the reaction, after time t and at the end of the reaction respectively, Hence
V∞ – V0 = 21.05 – 9.62 = 11.43
Time V∞ – Vt k1
75 min 21.05 – 12.10 = 8.95
119 min 21.05 – 13.10 = 7.95
183 min 21.05 – 14.75 = 6.30
A constant value of k shows that hydrolysis of ethyl acetate is a
first order reaction
Problem 9: The optical rotations of sucrose in 0.5 N HCl at 35°C at various time intervals are given below. Show that the reaction is of first order :
Time (minutes) | 0 | 10 | 20 | 30 | 40 | ∞ |
Rotation (degrees) | +32.4 | +28.8 | +25.5 | +22.4 | +19.6 | -11.1 |
Solution: The inversion of sucrose will be first order reaction if the above data confirm to the equation ,
Where r0, rt and r∞ represent optical rotations initially, at the commencement of the reaction after time t and at the completion of the reaction respectively
In the case a0 = r0 – r∞ = +32.4 – (-11.1) = +43.5
The value of k at different times is calculated as follows :
Time rt rt – r∞ k
10 min +28.8 39.9 = 0.00864 min-1
20 min +25.5 36.6 = 0.008637 min-1
30 min +22.4 33.5 = 0.008702 min-1
40 min +19.6 30.7 = 0.008714 min1
The constancy of k1 indicates that the inversion of sucrose is a
first order reaction.
Problem 10: The reaction given below, involving the gases is observed to be first order with rate constant 7.48 × 10−3 sec−1.Calculate the time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm and also find the total pressure after 100 sec.
2A(g) → 4B(g) + C(g)
Solution: 2A(g) → 4B(g) + C(g)
initial Po 0 0
at time t Po − P′ 2P′ P′/2
Ptotal = Po − P′ + 2P′ + P′/2 = Po +
P′ = (0.145 − 0.1) = 0.03 atm
k =
t =
t = 47.7 sec
Also, k =
7.48 × 10−3 =
0.1 − P′ = 0.047
P′ = 0.053
Ptotal = 0.1 + (0.053)
≈ 0.180 atm.
Problem 11: What will be the ratio of rate constants for two reactions that have same A value but Ea values at room temperature differ by
- a) 1 kcal/mol
- b) 10 kcal/mol
Solution: a)
= exp = 5.4
- b) = 2 × 107
Problem 12: 227Ac has a half-life of 22.0 years with respect to radioactive decay. The decay follows two parallel paths, one leading to 222Th and the other to 223Fr. The percentage yields of these two daughter nuclides are 2.0 and 98.0 respectively. What are the decay constants (λ) for each of the separate paths?
Solution: The rate constant of the decay is
k=
If k1 and k2 are the rate constants of the reactions leading to 222Th and 223Fr, respectively we have
k1 + k2 =
On solving for k1 and k2, we get
k2 = 0.03087 y−1
k1 = 0.00063 y−1
Problem 13: At 278°C the half life period for the first order thermal decomposition of ethylene oxide is 363 min and the energy of activation of the reaction in 52,00 cal/mole. From these data estimate the time required for ethylene oxide to be 75% decomposed at 450°C.
Solution: = 1.122
t1/2 (at 450°C) = 118.24 min.
Now t0.75 =
∴ t0.75 = = 236.48 min
Problem 14: The nucleidic ratio of to in a sample of water is 8.0 × 10–18:1. Tritium undergoes decay with half life period of 12.3 years. How many tritium atoms would 10.0 g of such a sample contains 40 years after the original sample in collected.
Solution: 18 g H2O has 2H atom in it = 6.023 × 1023× 2
18 g H2O has atoms = 8 ×10–18 × 6.023 ×1023 ×2
∴ 10 g H2O has atoms =
i.e No of = 5.354 × 106 atoms
Now t =
40 = = 5.624 ×105 atoms
Problem 15: On analysis a sample of Uranium was found to contain 0.277 g of 82Pb206 and 1.667 g of 92U238. The half life period of 92U238 is 4.51×109 years. If all the lead were assumed to have come from decay of 92U238, what is the age of the earth?
Solution: 92U238 = 1.667 g = mole
82Pb206 = 0.227g = mole
All the lead have come from decay of U.
∴ Moles of Pb formed =
∴ moles of U decayed =
∴ Total moles of Uranium = + , ie N0
Also N for U238 =
for U238 t =
= log
t= 1.143 × 109 yrs .
13.2 Objective
Problem 1: The rate constant for a reaction is 2 × 10-4 mol L-1 S-1. The reaction is –
(A) First order (B) Second order
(C) Third order (D) Zero order
Solution: Unit of rate constant of nth order reaction is given by – mol1-n Ln-1 S-1
When n = 0 unit of K will be mol L-1 S-1
∴ (D)
Problem 2: For an endothermic reaction, where ΔH represents the enthalpy of the reaction in kJ/mol, the minimum value for the energy of activation will be
(A) less than ΔH (B) Zero
(C) more than ΔH (D) Equal to ΔH
Solution: For an endothermic reaction, the activation energy that never be less than ΔH as indicated in the diagram.
∴ (C)
Problem 3: The rate constant, the activation energy and Arrhenius parameter of a chemical reaction at 25°C are 3.0 × 10–4 s–1, 1044 kJ mol–1 and 6.0 × 1014 s–1 respectively. The value of the rate constant as T → ∞ is
(A) 2.0 × 1018 s–1 (B) 6.0 × 1014 s–1
(C) Infinity (D) 3.6 × 1030 s–1
Solution: The Arrhenius equation is
K = Ae–Ea/RT
As T → ∞ the value → 0
& e–Ea/RT → e–0 ⇒ 1
Hence k → A
Hence k = 6.0 × 1014s–1
∴ (B)
Problem 4: For a first order reaction
(A) the degree of dissociation is equal to (1 – e–-kt)
(B) a plot of reciprocal concentration of the reactant versus time gives a straight line
(C) the time taken for the completion of 75% reaction is thrice the t1/2 of the reaction
(D) the pre-exponential factor in the Arrhenius equation has the dimension of time t.
Solution: (A) 1 – e–kt represents amount of molecules dissociated into products starting from unit quantity of reactants and hence represents degree of dissociation.
(B) The pre-exponential factor A = = k(e+Ea/RT)
Here e-Ea/RT is the fraction of molecules having sufficient activation energy to react and hence being a fraction it has no dimensions. Therefore A will have same dimensions as that of k i.e. (time)-1
Problem 5: Rate law for a gaseous reaction
A + B ⎯→ C + D is given by –
Rate = K[A]2 [B]0
The volumes of reaction vessel containing these gases is suddenly reduced to one fourth the initial volume. The rate of reaction relative to original rate would be –
(A) (B)
(C) (D)
Solution: On reducing the volume to ¼th increases the concentration of each reacting species increases four times. As rate is proportional to square of concentration of A and is independent of concentration of B rate increases sixteen times.
∴(B)
Problem 6: On introducing a catalyst at 500 K the rate of a first order reaction increases by 1.718 times. The activation energy in presence of catalyst is
4.15 kJ mol–1. The slope of the plot of ln k (in sec–1) vs (T in Kelvin) in absence of catalyst is (R = 8.3 J mol-1K-1)
(A) + 1 (B) –1
(C) + 1000 (D) – 1000
Solution: k = A e-Ea/RT
k’ = A e-Ea’/RT
(where k = rate constant for non catalysed reaction and k’ = rate constant for catalysed reaction. Ea = activation energy for non -catalysed reaction and Ea’ = activation energy for catalysed reaction)
Also given k’ = k + 1.718 k = 2.718 k
∴2.718 =
loge 2.718 =
Ea – Ea’ = 4.11
Ea’ = 4.15
∴Ea = 8.3 KJ/mol-1
k = A e-Ea/RT
∴loge k = logeA –
This is an equation for straight line with slope = =
= -1000 ∴ (D)
Problem 7: If a second order reaction involving a single reactant in 25% complete in 20 minutes then what is its half life if initial concentration of the reactant is being made double at the same temperature.
(A) 50 min (B) 30 min
(C) 60 min (D) 15 min
Solution: (for 2nd order reaction)
= k × 20
= k × t1/2 ⇒ ⇒ t1/2 = 30 minutes
Problem 8: In the following graph if r implies rate of reaction, c implies rate of reaction, c implies concentration of reactant. The slope of the straight line tells about –
(A) rate constant (B) order
(C) activation energy (D) frequency factor
Solution: r = K Cn [K = rate constant]
[n = order of reaction] or log r = log k + nlogC this is an equation for straight line, whose slope = n i.e. order of reaction |
Problem 9: For a reaction 2 A → B variation of concentration of A with time is shown below
The order of reaction is
(A) First (B) Second (C) Third (D) Zero |
Solution: Rate = K [A]n n = order of reaction
Rate remains constant with change in concentration of reactant [A], n must be = 0
as Rate = K [A]o
Rate = K = constant
∴ (D)
Problem 10: A reaction between A and B is second order. The rate law expression might possibly to the reaction is –
(A) Rate = K[A]1 [B]1 (B) Rate = K [A]2
(C) Rate = K [B]2 (D) All of the above
Solution We can rewrite the rate law-
- A) Rate = Rate = K[A]1 [B]1
Order 1+1 = 2
- B) Rate = K [A]2 [B]0
order = 2+0 = 2
- C) Rate = K[B]2 [Ao]
order = 2+0=2
∴ (D)
Problem 11. An endothermic reaction A →B has an activation of 15 KJ/mole and the heat of reaction is 5 KJ/mole. The activation energy of reaction B → A will be –
(A) 15 KJ/mole (B) 20 KJ/mole
(C) 0 (D) 10 KJ/mole
Solution: ∴ EA′ (Activation energy for product) = 15 – 5 = 10 KJ/mol |
Problem 12: The rate constant for a reaction is 2 × 10-4 mol L-1 S-1. The reaction is –
(A) First order (B) Second order
(C) Third order (D) Zero order
Solution: Unit of rate constant of nth order reaction is given by – mol1-n Ln-1 S-1
When n = 0 unit of K will be mol L-1 S-1
∴ (D)
Problem 13: The Arhenius equation for is-trans isomerisation of 2-buteneand 2-butene nitrile are given as follows:
- a) For 2-butene CH3CH=CHCH3 K(S-1) = 108
- b) For 2-butene nitrile CH3CH=CH.CN K’(S-1) = 1011
The temperature at which K=K’
(A) 913.87 oK (B) 533.43 oK
(C) 1000.02 oK (D) 407.05 oK
Solution:
= 1011
or 1013.8-11 =
or 630.957 =
∴2.303 log 630.957 =
∴T = = 913.87 oK
∴ (A)
Problem 14: Thermal decomposition of a compound is of first order. If 50% of a sample of the compound is decomposed in 120 minutes how long will it take for 90% of the compound to decompose?
(A) 399 min (B) 410 min
(C) 250 min (D) 120 min
Solution: First calculate the values of k
K = = = 5.77 × 10–3 min–1
Now we know that for a first order reaction
K =
Here the initial concentration a = 100 and n = 90
∴ 5.77 × 10–3 =
t = log
Solving t = 399 minute ∴ (A)
Problem 15: The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is a first order reaction.
After 55 seconds at 400 K the pressure of Cl2O7 falls from 0.062 to 0.044 atm. The rate constant and pressure of Cl2O7 after 100 sec of decomposition at this temperature are
(A) 5.2 × 10–4 sec–1; 0.05 atm (B) 6.2 × 10–3 sec–1; 0.033 atm
(C) 5.8 × 10–3 sec–1, 0.44 atm (D) 4.6 × 10–3 sec–1; 0.005 atm
Solution: The equation for first order reaction is
K =
- i) Here a = 0.062 a – x = 0.044
= 6.2 × 10–3 sec–1
- ii) Here we are to find the value of a – x
6.20 × 10–3 =
a – x = = 0.033 atm
∴ (B)
Problem 16: A substance having a half-life period of 30 minutes decomposes according to the first order rate law. The fraction decomposed, the balance remaining after 1.5 hours and time for 60% decomposition on its doubling the initial concentration will be.
(A) 87.4; 0.126; 39.7 mts (B) 80.6; 0.135; 40.8 mts
(C) 90.5; 0.144; 2829 mts (D) 802; 0.135; 26.6 mts
Solution: We know that
= 0.023 / min–1
Let the initial concentration of the substance = 100 and the substance decomposed in 1.5 hours (90 min) = x then, a – x = 100 – x
Substituting these values in
K = ⇒ 0.0231 = log
log ⇒ x = 87.4
So the fraction decomposed in 1.5 hours = = 0.874
Fraction remaining behind after 1.5 hours = 1 – 0.874 = 0.126
The time required for 60% of decomposition
K = ⇒ 0.023 =
= 39.7 minutes
Since the reaction is of first order the time required to complete specific fraction is independent of initial concentration (or pressure). Hence 60% of the reaction will decompose in 39.7 mins in this case also.
∴(A)
Problem 17: In the Arrhenius equation for a certain reaction the value of A and Ea (activation energy) are 4 × 1013 sec–1 and 98.6 kJ mol–1 respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is
(A) 280 K (B) 290 K
(C) 311.35 K (D) 418.26 K
Solution: Calculation of K
We know that k = (t1/2 = 10 × 60 sec) = 1.1558 × 10–3
According to Arrhenius equation
logk = logA –
Substituting the various values in the above equation
log1.155 × 10–3 = log4 × 1013 –
On usual calculation, T = 311.35 K
Problem 18: A first order reaction is 50% complete in 30 minutes at 27°C and in 10 minutes at 47°C. The rate constant at 47 oC and energy of activation of the reaction in kJ/mole will be
(A) 0.0693; 43.848 kJ mol–1 (B) 0.0560; 45.621 kJ mol–1
(C) 0.0625; 42.926 kJ mol–1 (D) 0.0660; 46.189 kJ mol–1
Solution: We know that k =
Substituting the values at the two given conditions
We also know that
=
or Ea =
=
= 43.848kJ mol–1
∴ (A)
Problem 19: If for a reaction 1 and 2, activation energies are EA and and rate constants are K1 and K2 at temperature T and at temperature T’ where T’ is higher than T then which among the following is true if
(A) (B)
(C) (D) None
Solution: ….(1)
….(2)
Dividing equation (1) by (2)
∴
∴ (A)
Problem 20: The value of maximum rate constant is equal to
(A) Frequency factor (A) (B) Activation energy (EA)
(C) Order of the reaction (n) (D) infinite
Solution: K = Ae-Ea/RT
When T = α , then
Or e-Ea/RT = 1
Kmax = A
∴ (A)
- Assignments (Subjective Problems)
LEVEL I
- Rate of reaction A + B → P is given below as a function of different initial concentration of A and B.
[A] moles lt–1 | [B] moles lt–1 | Initial rate (moles lt–1 min–1) |
0.01 | 0.01 | 0.002 |
0.02 | 0.01 | 0.004 |
0.01 | 0.02 | 0.002 |
- i) Determine the order of reaction with respect to A and with respect to B.
- ii) Calculate the rate constant
iii) Half life of A in the reaction
- iv) Calculate the rate constant when conc. of A and B are 0.3 and 0.4 M respectively.
- Arsine produces hydrogen gas on heating. The pressure of produced hydrogen was measured, at constant volume and temperature, as follows:
Time (hours) 0 5.5 6.5 8
Pressure (atm) 0.965 1.060 1.076 1.100
Calculate the specific rate and half-life period of reaction assuming the reaction to be of first order
- Find out the order of reaction with the help of graph given below:
- When marble chips react with hydrochloric acid, CO2 is released. The reaction has been followed by measuring the volume of CO2 produced. The results are given in following table.
Time (min) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ∞ |
Volume of CO2 in cm3 | 0 | 26 | 49 | 68 | 80 | 91 | 97 | 102 | 106 | 108 | 111 | 120 |
Find out the rate order and rate constant of the reaction
- While studying the decomposition of gaseous N2O5 it is observed that a plot of logarithms of its partial pressure versus time is linear. What kinetic parameters can be obtained from this observation?
- The first order reaction 2A → 2B + C is 35% complete after 325 seconds
- a) Find out the value of rate constant
- b) How long will it take for the reaction to be 70% complete
- c) 90% complete
- What will be the initial rate of reaction if its rate constant is 10–3 min–1 and the concentration of the reactant is 0.2 mol dm–3? How much of the reactant will be converted into the products in 200 minutes?
- Show that time required for 99.9% completion of a first order reaction is about 10 times that required for 50% completion of the same reaction.
9. | The progress of the reaction A nB, with time is presented in the figure. Determine.
i) the value of n ii) the equilibrium constant, K, and iii) the initial rate of conversion of A. |
- For the reaction :
C2H5I + OH− → C2H5OH + I−
the rate constant was found to have a value of 5.03 × 10−2 mol−1 dm3 s−1 at 289 K and 6.71 mol−1 dm3 s−1 at 333 K. What is the rate constant at 305 K.
- For the reaction 2 NO + Cl2 ⎯→ 2 NOCl it was found that on doubling concentration of both reactants the rate increases 8 fold. But on doubling the concentration of Cl2 alone rate only doubles. Calculate overall order.
- One of the hazards of nuclear explosion is the generation of Sr90 and its subsequent incorporation in bones. This nuclide has a half life of 28.1 years. Suppose one microgram was absorbed by a new born baby, how much Sr90 will remain in his bones after 20 yrs. [IIT–JEE ‘95]
- A painting claimed to be an original of Raphael (1483–1520) is analysed. A small piece of canvas gives C–14 content to be 0.94 of that in the living plants. Is the painting a forgery ( t1/2 of C14 = 5500 years).
- Benzene diazonium chloride decomposes in aqueous solution unless kept below about 5°C. The chemical equation is
C6H5N2+Cl– (aq.) + H2O(l) ⎯→ C6H5OH(aq) + N2(g) + HCl(aq)
The data in the following table refer to [HCl] at different times.
Time (min) | 0 | 1 | 2 | 4 | 6 | 8 | 12 | 14 | ∞ |
[HCl]10–2mol/lit | 0 | 0.83 | 1.45 | 2.90 | 3.88 | 4.60 | 5.56 | 6.05 | 7.50 |
Find out the order and rate constant for the reaction considering the other reactant water as solvent which is present in large excess.
- A sample of 238U (t1/2 = 4.5 ×109 years ) ore is found to contain 23.8 g of 238U and 20.6 g of 206Pb. Calculate the age of the rock.
LEVEL – II
- For a reaction, the energy of activation is zero. What is the value of rate constant at 300 K if k = 1.6 × 106 s–1 at 280 K? (R = 8.31 J K–1 mol–1)
- Find out the values of rate constants for the following first order reactions
- a) M → N, given that concentration of M decreases to one half of its initial value in 500 seconds.
- b) M → N, given that concentration of M decreases from 0.45 miol L–1 to 0.15 mol
L–1 in 50 sec. - c) 2M →L + N, given that M0 = 0.05 mol L–1 and after 100 seconds concentration of L rises to 0.02 mol L–1.
- Two reactions (I) A → products (ii) B → products, follow first order kinetics. The rate of the reaction (I) is doubled when the temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minutes. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (i) is half that of reaction (ii), calculate the rate constant of the reaction (ii) at 300 K.
- The time required for 10% completion of first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the preexponential factor for the reaction is 3.56 × 10-9 s-1, calculate its rate constant at 318 K and also the energy of activation.
- If Ea = 30 kcal/mol find the effect of 10°C temperature increase on room temperature rate constant.
- At 380°C, the half-life period for the first order decomposition of H2O2 is 360 min. the energy of activation of the reaction is 200 kJ mol–1, Calculate the time required for 75% decomposition at 450°C.
- The rate constant for the first order decomposition of a certain reaction is described by the equation:
logK(s–1) = 14.34 –
- i) What is the energy of activation for this reaction
- ii) At what temperature will its half-life period be 256 minutes?
- If reactions 1st and 2nd have A1 = 5A2 and K1 = 100 K2 at room temperature, find out Ea1 – Ea2.
- The EA of a non-catalysed reaction at 27°C is 300 kcal/mol and EA of the same reaction catalyzed by an enzyme is 10 kcal mol–1. Calculate the ratio of rate constants for catalysed and non-catalysed reaction.
- For the first order decomposition reaction
(CH3)3COOC(CH3)3 ⎯→ 2CH3COCH3 + C2H6
in the gaseous phase, the pressure of the system at t = 0 and t = 15 min were found to be 169.3 torr and 256.0 torr, respectively. Calculate (a) the rate constant of the reaction, (b) half-life period and (c) the pressure of the system at 9 min.
- For a first order reaction A → 3B when [A]0 = 0.04 mol L–1, the concentration of B increases to 0.03 in 3 minutes.
- a) Determine the rate constant for the reaction
- b) How much more time would be needed for the concentration of B to increase to 0.06 mol L–1
- The gas phase decomposition of dimethyl ether follows first order kinetics
CH3–O–CH3(g) ⎯→ CH4(g) + H2(g) + CO(g)
The reaction is carried out in a constant volume container at 500°C and has a half life of 14.5 minutes. Initially only dimethyl ether is present at a pressure of 0.4 atm. What is the total pressure of the system after 12 mins. Assume ideal gas behaviour. [IIT–JEE ‘93]
- A sample of pitch blende is found to contain 50% Uranium and 2.425% Lead. Of this Lead only 93% was Pb206 isotope. If the disintegration constant is
1.52 × 10–10 yr–1 , how old could be the pitch blende deposit.
- At 30°C a sample of milk sours in 36 hours but when stored in refrigerator at 5°C it takes one week to sour. Calculate EA (activation energy) for conversion of lactose to lactic acid.
- A drug is known to be ineffective after it has decomposed to the extent of 30 %. The original concentration of the sample was 500 units per ml. When analysed 20 months later the concentration was found to be 420 units/ml. Assuming that the decomposition is of first order, what will be the expiration time of the drug sample. What is half life of drug.
LEVEL – III
- In 80% ethanol isopropyl bromide reacts with hydroxide ion and reaction follows two parallel paths. The rate of reaction of 55°C (in moles/lit sec) can be given by the following kinetic equation.
Rate = 4.7 × 10–5 [RX][OH–] + 0.24 × 10–5 [RX}
What percentage of isopropyl bromide will react by second order mechanism when [–OH–] is
- a) 001 molar b) 0.1 molar
- c) 1 molar d) 1.0 molar
- e) 0 molar
- A certain organic compound A decomposes by two parallel
first order mechanisms.
If k1 : k2 = 1:9 and k1 = 1.3 ×10–5 is–1. Calculate the concentration ratio of C to A if an experiment is started with only A and allowed to run for one hour.
- 84Po210 decays with α particle to 82Pb206 with a half life of 138.4 days. If 1.0 g of 84Po210 is placed in a sealed tube, how much helium will accumulate in
69.2 days. Express the answer in cm3 at STP. - A first order reaction A ⎯→ B requires activation energy of 73.04 kJ mol–1. When a 20% solution of A was kept at 25°C for 20 minutes, 25% decomposition took place. What will be the percent decomposition in the same time in a 40% solution maintained at 40°C? Assume that activation energy remains constant in this range of temperature.
- (a) The rate constant of a reaction is 1.5 ×107 s–1 at 50°C and 4.5 ×107 s–1 at 100°C. Evaluate the Arrhenius parameters A & Ea.
(b) For the reaction N2O5(g) ⎯→ 2NO2(g) + 1/2 O2 (g), calculate the mole fraction of N2O5 (g) decomposed at a constant volume and temperature, if the initial pressure in 600 mm Hg and the pressure at any time is 960 mm Hg. Assume ideal gas behaviour. [IIT–JEE ‘98]
- A sample of , as iodide ion was administered to a patient in a carrier consisting of 0.10 mg of stable iodide ion. After 4 days, 67.7% of the initial radioactivity was detected in the thyroid gland of the patient. What mass of the stable iodide ion had migrated to the thyroid gland (t1/2 of = 8 day).
- (t1/2 = 3.05 mins) decays to (t1/2 = 2.68 min) by α– emission, while Pb214 is a β–emitter. In an experiment starting with 1 g atom of pure Po218 how much time would be required for the number of nuclei of to reach maximum.
- A 32P radio nuclide with half life T = 14.3 days is produced in a reactor at a constant rate q = 2.7 × 109 nuclei per sec. How soon after the beginning of production of that radionuclide will its activity be equal to A = 1.0 ×109 dis/s.
- A weak complex AB is formed in the reaction
AB C
Assuming that K-1 >> K2 derive the rate equation for formation of C
- 1 g atom of 226Ra is placed in an evacuated tube of volume 5 litres. Assuming that each nucleus is an α– emitter and all the contents are present in tube, calculate the partial pressure of He collected in the tube at 27°C after the end of 800 years. t1/2 of Ra = 1600 years. Neglect the volume occupied
by undecayed Ra. - A mixture of 239Pu and 240Pu has a specific activity of 6.0 × 109 dis/s. The half lives of isotopes are 2.44 ×104 and 6.58 ×103 years, respectively. Calculate the isotopic composition of this sample.
- To 50.00 ml of a solution containing an unknown concentration of Zinc ion was added 0.100 μCi of 62Zn2+ in 10 ml solution and the total volume was diluted to 100 ml with water. Precipitation of Zinc salt yielded 0.2 g Zinc in the solid phase with an activity of 0.0823 μCi. What was the original concentration of the Zinc ion.
- The hydrolysis of ethyl acetate
CH3COOC2H5 + H2O CH3COOH + C2H5OH
in aqueous solution is first order with respect to ethyl acetate. Upon varying the pH of the solution the first order rate constant varies as follows.
pH 3 2 1
k1/10–4s–1 1.1 11 110
what is the order of the reaction with respect to H+ and the value of the rate constant?
- The nuclei of two radioactive isotopes of same substance A236 and A234 are present in the ratio of 4:1 in an ore obtained from some other planet. Their half lives are 30 min. and 60 min. respectively. Both isotopes are alpha emitters and activity of isotope with half – life 30 min. is one Rutherford (106 d.p.s.). Calculate after how much time their activities will become identical. Also calculate the time required to bring the ratio of their atoms to 1:1.
- In nature, a decay chain series starts with 90Th232 finally terminates at 82Pb208. A Thorium ore sample was found to contain 8 × 10–5 ml of He at STP and 5 × 10–7 g of Th232. Find the age of ore sample assuming that source of He to be only due to decay to Th232. Also assume complete retention of He within the ore [t½ for Th232 = 1.39 × 1010 years].
- Assignments (Objective Problems)
LEVEL – I
Instruction: In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required to do this assignment in 20 minutes.
- In the inversion of cane-sugar
C12H22O11 + H2O ⎯→ C6H12O6 + C6H12O6, the molecularity of reaction will be
(A) 1 (B) 2
(C) 3 (D) 0
- In a first order reaction if the concentration units are changed to some other units, which are (say) m times the first. Then the value of velocity co-efficient will
(A) Increase m – times (B) Decrease m – times
(C) Not alter (D) None of these
- In decomposition of carbonyl sulphide in water COS + H2O ⎯→ CO2 + H2S. The rate is found to be dependent on carbonyl sulphide and independent of water. Such reaction will be of
(A) Zero order (B) Second order
(C) Pseudo-unimolecular (D) Unimolecular
- Rate constant depends upon
(A) Temperature (B) Concentration of reactants
(C) Activation energy of reaction (D) None of the above
- Decomposition of N2O5 occurs in the following manner. 2N2O5 ⎯→ 4NO2 + O2its rate is expressed in three ways.
- i) ii) iii) (N2O5)
What is the relation between k & k′
(A) k = k′ (B) k = 2k′
(C) k′ = 2k (D) k′ = 1/3 k
- All chemical reactions take place at a definite rate depending on the conditions, of which the most important are
(A) concentration of reactants (B) temperature
(C) radiation (D) presence of a catalyst
- The rate constant of a reaction depends on
(A) temperature (B) mass
(C) density (D) time
- The rate at which a substance reacts depends on its
(A) atomic mass (B) equivalent mass
(C) molecular mass (D) active mass
- The rate expression for a reaction is
rate = k[A]3/2 [B]–1, the order of reaction is
(A) 0 (B) 1/2
(C) 3/2 (D) 5/2
- Which of the following rate law has an overall order of 0.5 for the reactions involving substances x,y,z?
(A) Rate = k(Cx) (Cy) (Cz) (B) Rate = k (Cx)0.5 (Cy)0.5 (Cz)0.5
(C) Rate = k(Cx)1.5 (Cy)–1(Cz)0 (D) Rate =
- The reaction A(g) + 2B(g) ⎯→ C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A & B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is
(A) 1/48 (B) 1/24
(C) 9/16 (D) 1/6
- If concentration are measured in mole/lit and time in minutes, the unit for the rate constant of a 3rd order reaction are
(A) mol lit–1 min–1 (B) lit2 mol–2min–1
(C) lit.mol–1 min–1 (D) min–1
- A radioactive element has a half life period of 4 days. How much of it will remain after 16 days
(A) (B)
(C) (D)
- For a first order reaction the plot of log [A]t Vs t is linear with a
(A) positive slope and zero intercept
(B) positive slope and non zero intercept
(C) negative slope and zero intercept
(D) negative slope and non zero intercept
- For a hypothetical reaction A + B ⎯→ C+D, the rate = k[A]–1/2 [B]3/2. On doubling the concentration of A and B the rate will be
(A) 4 times (B) 2 times
(C) 3 times (D) none of these
- In a reaction the threshold energy is to equal to
(A) average energy of the reactants
(B) activation energy
(C) activation energy + average energy of the reactants
(D) activation energy – average energy of the reactants
- The temperature coefficient of most of the reactions lies between
(A) 1 and 2 (B) 2 and 3
(C) 3 and 4 (D) 2 and 4
- Select the rate law that corresponds to the data shown for the following reaction
A + B ⎯→ C.
Experiment [A] [B] Initial Rate
1 0.012 0.035 0.10
2 0.024 0.070 0.80
3 0.024 0.035 0.10
4 0.012 0.070 0.80
(A) Rate = k [B]4 (B) Rate = k [A] [B]3
(C) Rate = k [A]2 [B]2 (D) Rate = k [B]3
- For a first reaction t0.75 is 138.6 seconds. Its specific rate constant ( in sec–1) is
(A) 10–2 (B) 10–4
(C) 10–5 (D) 10–6
- Half life period for a first–order reaction is 20 minutes. How much time is required to change the concentration of the reactants from 0.08M to 0.01M
(A) 20 min (B) 60 min
(C) 40 min (D) 50 min
LEVEL – II
- 75% of a Ist order reaction was completed in 32 minutes. When was 50% of the reaction completed
(A) 24 min (B) 4 min
(C) 16 min (D) 8 min
(E) none of these
- For a general reaction,
aA + bB ⎯→ cC + dD
the rate of reaction may be give as
(A) r = (B)
(C) r = (D) r =
- The half life of a second order reaction is
(A) t1/2 = 0.693 / k (B) t1/2 = k/[A]o
(C) t1/2 = [A]0/k (D) t1/2 =
- According to Arrhenius the relationship between rate constant, k and temperature can be given by
(A) (B) log
(C) log k = log A– (D) ΔG° = –2.303 RT log K
- The concept of mechanical velocity or speed cannot be used in measuring the rate of reactions, because
(A) rate of reaction depends on active mass of reactants and it decreases with time
(B) rate of reaction depends on molar concentration of reactants and it decreases with time.
(C) rate of reaction depends on molar concentration of products and it decreases with time
(D) rate of reaction depends on active mass of products and its decreases with time.
- What will be the fraction of (t1/2 = 25 min) left after 100 minutes?
(A) 1/2 (B) 1/4
(C) 1/3 (D) 1/16
- What is the activation energy for the decomposition of N2O5 as
N2O5 2NO2 +
If the values of the rate constants are 3.45 × 10–5 and 6.9 × 10–3 at 27°C and 67°C respectively
(A) 102 × 102 kJ (B) 488.5 kJ
(C) 112 kJ (D) 112.5 kJ
- 50% of the amount of a radioactive substance decomposes in 5 years. The time required for the decomposition of 99.9% of the substance will be
(A) 10 years (B) between 10 and 50 years
(C) less than 10 years (D) between 49 and 50 years
- Which one of the following statements is wrong regarding molecularity of the reaction?
(A) It is calculated from the reaction mechanism
(B) It may be either a whole number or fraction
(C) It is the number of molecules of the reactants taking part in a single step chemical reaction
(D) It depends on the rate determining step of the reaction.
- 1.386 hours are required for the disappearance of 75% of a reactant of
first– order reaction. What is the rate constant of the reaction?
(A) 3.6 ×10–3 s–1 (B) 7.2 × 10–3 s–1
(C) 2.7 × 10–4 s–1 (D) 1.8 × 10–3 s–1
- For a first order reaction, the ratio of time for the completion of 99.9% and half of the reaction is
(A) 8 (B) 10
(C) 9 (D) 12
- A certain radioactive element A, has a half – life = t seconds. In (t/2) seconds the fraction of the initial quantity of the element so far decayed is nearly
(A) 25% (B) 29%
(C) 21% (D) 17%
- The terms rate of reaction and rate of appearance (or disappearance) of reactant (or product)
(A) represent one and the same physical quantity
(B) differ by constant factor
(C) are positive parameters and have same value
(D) may or may not have same value depending upon the stoichiometric coefficient of reactants (or products) in the balanced chemical equation.
- For the first order reaction A(g) ⎯→ 2B(g) + C(g), the initial pressure is PA = 90 mm Hg, the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is
(A) 1.15 × 10–3 sec–1 (B) 2.3 × 10–3 sec–1
(C) 3.45 × 10–3 sec–1 (D) 6 × 10–3 sec–1
- The rate law for the reaction
RCl + NaOH (aq) ⎯→ ROH + NaCl
is given by Rate = k[RCl]. The rate of the reaction will be
(A) unaffected by increasing temperature of the reaction
(B) doubled on doubling the concentration of NaOH
(C) halved on reducing the concentration of NaOH to one half
(D) halved on reducing the concentration of RCl to one half
- For a given reaction of first order, it takes 20 minutes for the concentration to drop from 1.0 mol litre–1 to 0.6 mol litre–1. The time required for the concentration to drop from 0.6 mol litre–1 to 0.36 mol litre–1 will be
(A) more than 20 minutes (B) less than 20 minutes
(C) equal to 20 minutes (D) infinity
- The unit of rate constant of a reaction having order 1.5 would be
(A) (mol L–1)–1/2s–1 (B) (mol L–1)–3/2 s–1
(C) (conc)–0.5 time –1 (D) (conc)–0.75 time–1
- A catalyst lowers the activation energy of the forward reaction by 20 kJ mol–1. It also changes the activation energy of the backward reaction by an amount.
(A) equal to that of forward reaction
(B) equal to twice that of the forward reaction
(C) which is determined only by the average energy of products
(D) which is determined by the average energy of products relative to that of reactants.
- The half–life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to
(A) (1/2) g (B) (1/4) g
(C) (1/8) g (D) (1/16) g
- For a second order reaction of the type rate = k [A]2, the plot of [A]t versus t is linear with a
(A) positive slope and zero intercept
(B) positive slope and non zero intercept
(C) negative slope and zero intercept
(D) negative slope and non zero intercept
- Answers to Subjective Assignments
LEVEL – I
- i) 1 and 0
- ii) 2 min-1
iii) 3.465 min
- iv) 06 mol/litre min
- K = 4.011 × 10–2 hr–1, t1/2 = 17.28 hr
- n = 1
- 0.262 min-1
- Reaction is first order
- a) 1.3256 × 10-3 sec-1
- b) 908 × 103 sec
- c) 737 × 103 sec
- Initial rate = 2.0 × 10-4 moles/l/min
% conversion = 18 %
- (i) n = 2
(ii) k = 1.2
(iii) -0.1 M hr–1
- 0.35 mol-1 litre sec-1
- 3
- 6.1 × 10-7 g
- No it was not a forgery
- 0.1122 min-1
- 4.5 × 109 years
LEVEL – II
- K300 = 1.6 × 106 S–1
- a) 1.386 × 10-3 sec-1
- b) 021976 sec-1
- c) 01609 sec-1
- 0.327 min-1
- Ea = 76.65 kJ
K318 = 9.036 × 10–4 S–1
- 5
- 20.33 min
- (i) 239.34 kJ mol–1
(ii) 669 K
- 1.7745 kcal mol-1
- 1014.57
- a) 0.0197 min-1
- b) 18 min
- c) PT = 224.3 torr
- a) 0.0959 min-1
- b) 228 min
- 0.7488 atm
- 3.3 × 108 years
- 43.16 KJ/mol
- 40.9 months, 79.49 months
LEVEL – III
- a) 1.9 %
- b) 4 %
- c) 2 %
- d) 1 %
- e) 99 %
- 0.525
- 31.25 cm3
- 60 %
- (a) 22.013 kJ mol–1, 5.45 × 1010 S–1
(b) 0.407
- 0.0957 mg
- 4.125 min
- 10.7 days
- (a) 595 × 103 ml
(b) 0.2118 dps/ml
- 1.443 atm
- 239Pu = 38.96%
240Pu = 61.04%
- 0.0744 M Zn+2
- 1, 1.1 × 10–1 dm3 mol–1 S–1
- t = 3 hrs., t = 2 hrs
- 4.89 × 109
- Answers to Objective Assignments
LEVEL – I
- B 2. C
- C 4. A, C
- C 6. A,B,C,D
- A 8. D 9. B 10. C 11. D 12. B
- B 14. D
- B 16. C 17. B 18. D
- A 20. B
LEVEL –II
- C 2. A,B,C
- D 4. A,B,C
- A,B 6. D
- D 8. D
- B 10. C
- B 12. B
- D 14. A
- D 16. C
- C 18. A,D
- D 20. B