1. IIT-JEE syllabus
Law of mass action; equilibrium constant; exothermic and endothermic reactions;
Le-Chatelier’s Principle and its applications
It is an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion. Under these conditions, a process starts by itself or by initiation, continues for some time at diminishing rate and ultimately appears to stop. The reactants may still be present but they do not appear to change into products any more.
A reaction is said to be reversible if the composition of reaction mixture on the approach of equilibrium at a given temperature is the same irrespective of the initial state of the system, i.e. irrespective of the fact whether we start with reactants or the products. Some examples are:
A + B C + D
H2(g) + I2(g) 2HI(g)
Ag+(aq) + Fe2+(aq) Fe3+(aq) + Ag(s)
2NH3(g) N2(g) + 3H2(g)
2.1 Characteristics of Chemical Equilibrium
∙ The equilibrium is dynamic i.e. the reaction continues in both forward and reverse directions.
∙ The rate of forward reaction equals to the rate of reverse reaction.
∙ The observable properties of the system such as pressure, concentration, density remains invariant with time.
∙ The chemical equilibrium can be approached from either side.
A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.
2.2 Types of Equilibria
There are mainly two types of equilibria
- a) Homogeneous: Equilibrium is said to be homogeneous if reactants and products are in same phase.
H2(g) + I2(g) 2HI(g)
N2(g) + 3H2(g) 2NH3(g)
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
- b) Heterogeneous: Equilibrium is said to be heterogeneous if reactants and products are in different phases
CaCO3(s) CaO(s) + CO2(g)
NH4HS(s) NH3(g) + H2S(g)
NH2CO2NH4(s) 2NH3(g) + CO2(g)
2.3 Law of mass action
Guldberg and Waage proposed that “The rate at which a substance reacts is directly proportional to its active mass and rate of a chemical reaction is directly proportional to product of active masses of reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in the balanced chemical equation”.
For dilute solutions active mass is equal to concentration. Taking the example of the reaction.
N2(g) + 3H2(g) 2NH3(g)
We can write,
Rate of forward reaction rf α [N2] [H2]3 rate of reverse reaction rr α [NH3]2
- Equilibrium Constants, KC & KP
Let us consider a reaction of the type,
A(g) + B(g) C(g) + D(g)
The double headed arrow signifies that the reaction occurs in both the directions in measurable extent. What do you do at the start? You start with pure A & B which are the reactants, in a closed system. Initially A and B reacts very fast to give C and D. Now as soon as C and D are formed they also started giving back A and B. But A and B being present in greater quantity forces the reaction to occur in forward direction much faster than the backward one. So what is the net result? It’s nothing but the net formation of C & D. In this way as the time passes, A & B decreases and hence their strength, on the other hand C & D increases and so is their strength. So ultimately a time comes when the forward reaction is balanced by backward reaction. This state is the equilibrium.
Now, before deriving the different related expression let us know what is Law of Mass Action. It states that,
“The rate of a reaction is directly proportional to the product of concentrations of reactants with the stoichiometric co-efficient being raised to the power”
e.g. for the reaction.
aA(g) + bB(g) cC(g) + dD(g)
for the forward reaction,
Rf ∝ [A]a [B]b
Where Rf denotes rate of forward reaction.
[A] denotes concentration of A
‘a’ , ‘b’ are the powers which are the coefficients of A & B respectively.
If kf be the rate constant for forward reaction the expression can be rewritten as
Rf = kf[A]a [B]b …(1)
Similarly, for the backward reaction:
Rb = kb [C]c [D]d …(2)
Rate of forward reaction = Rate of backward reaction
From equation (1) and (2)
Rf = Rb
or kf [A]a [B]b = kb [C]c [D]d
Rearranging we get,
Now, the expression on the L.H.S. is a constant. So it means the expression on the R.H.S. has to be a constant. In fact it is so for a particular reaction. But obviously at a given temperature. Effect of temperature we will study later. This constant is named as equilibrium constant and is denoted by K.
This implies that no matter what we start with (i.e., A & B or C&D or A+B+C or A+B+D or A+C+D or B+C+D or A+B+C+D) and how much of these we start with, the ratio of is a fixed quantity at a given temperature when the reaction reaches equilibrium. That is , if we assume that this reaction has K = 4 then no matter what we take initially and irrespective of how much we take, once equilibrium is reached the ratio of will always be equal to 4.
Now let us consider the following reaction
A(s) + B(g) C(s) + D(g)
Its equilibrium constant , K would be ; K =
Concentration of C is the number of moles of C per unit volume of solution. Concentration of D is the number of moles of D per unit volume of the container (we can assume that the gas occupies the entire container). The concentration of A is the number of moles of A per unit volume of A. The concentration of all solids and pure liquids is a constant. This is because if initially we take w gm of A, then the moles of A are w/M. The volume of A is w/d where d is the density of A. Therefore, the initial concentration of A is = . We can see that at equilibrium also the concentration of A remains as d/M ( d and M are constants). In fact even if A were a pure liquid, its concentration would have remained constant.
Therefore, we bring all the constant terms on one side and we get
This ratio which is a constant and which involves only those concentration terms which are variables is called Kc, the equilibrium constant in terms of concentration.
∴KC = = K [A]
Now, let us consider the reaction,
A(g) + 2B(g) 3C (g) + 4D (g)
We know that concentration of a gas can be expressed as P/RT as shown below:
PV = nRT
= number of moles per litre = concentration
∴ [C]=; [D] = , [A]= and [B] =
PC = partial pressure of C
PD = partial pressure of D
PA = partial pressure of A
PB = partial pressure of B
∴ KC =
⇒ = KC (RT)(3+4) –(1+2).
Since KC is a constant and RT is also a constant, so the right hand side of the above expression is also a constant. This is called Kp, the equilibrium constant in terms of the partial pressure.
∴ = KP
Therefore, for this particular equilibrium, the ratio of partial pressures is also a constant.
In general, the relation between KP and KC is KP = KC (RT)Δn
Where Δn = number of moles of gaseous products – number of moles of gaseous reactants.
Now Δn can have three possibilities.
Δn < 0, Δn = 0, Δn > 0
Accordingly we can predict, out of Kp and Kc which one will be higher or lower. At Δn = 0 both Kp & Kc are the same.
Now let us assume that A is a solid or pure liquid. The changes now would be that KC would look like this, and following the above given sequence of derivation, Kp would like this
Next we assume that A was a solute present in a solution then Kc would remain the same i.e, KC = . Now if we try to express the concentrations in terms of partial pressures, we would fail to do that for A. It is not possible to express the concentration of a solution in terms of its pressure or vapour pressure and constants.
Therefore [A] remains as such
= KC (RT)(3+4)–(2)
The R.H.S. of the above expression is a constant which implies that the L.H.S is also a constant. This new expression cannot be called as either KC or KP since it contains both concentration terms and pressure terms. We call it KPC. We can also see that if we take [A] to the R.H.S. the L.H.S. contains only pressure terms, but then it is not a constant since [A] is not a constant.
Therefore, we can conclude that for KP to exist for a reaction it must fulfill two conditions:– (i) it must have at least one gas either in the reactants or in the products and (ii) it must not have any component in solution phase.
Illustration 1: Calculate KC for the reaction A + B 2C when the reaction was started with 2.0 mole/litre of A and B. The equilibrium concentration of C is 0.24 mole/litre.
Solution: A(g) + B(g) 2C(g)
2 2 0 at initial
(2 – 0.12) (2 – 0.12) 0.24 at equilibrium
Since the concentration of C at equilibrium is 0.24 mole/litre. It was 0.12 mole/litre of A and B must have reacted as one mole of A combines with one mole of B to produce 2 moles of C.
∴ = 1.63 × 10–2
Exercise 1: The equilibrium concentration of X, Y and L in a reaction of the type of
2X(g) + 2Y(g) 2L(g) + M are respectively 0.002, 0.005, 2.3 mole/litre calculate the initial concentration of X and Y.
3.1 Important relationships involving Equilibrium Constant
If we reverse an equation, Kc or Kp is inverted i.e.
If A(g) + B(g) C(g) + D(g) KC = 10
Then for C(g) + D(g) A(g) + B(g) KC′ = 10–1
If we multiply each of the coefficient in a balanced equation by a factor n, then equilibrium constant is raised to the same factor.
If N2 + O2 NO Kc = 5
Then for N2 + O2 2NO Kc′ = 52 = 25
If we divide each of the coefficients in a balanced equation by the factor n, then new equilibrium constant is nth root of the previous value.
If 2SO2 + O2 2SO3 Kc = 25
Then for SO2 + O2 SO3 KC′ = = 5
When we combine individual equation, we have to multiply their equilibrium constants for net reaction. If K1, K2 and K3 are stepwise equilibrium constant for A B, B C,
C D. Then for A D, equilibrium constant is K = K1 K2 K3.
Significance of the Magnitude of on Equilibrium Constant:
- A very large value of KC or KP signifies that the forward reaction goes to completion or very nearly so.
- A very small value of KC or KP signifies that the forward reaction does not occur to any significant extent.
- A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of Kc or KP is neither very large nor very small.
Illustration 2: From the given data of equilibrium constants of the following reactions
CoO(s) + H2(g) Co(s) + H2O(g) K = 670
CoO(s) + Co(g) Co(s) + CO2(g) K = 490
Calculate KC for the reaction
CO2(g) + H2(g) CO(g) + H2O(g)
Solution: Co(s) + H2(g) Co(s) + H2O(g) K = 670 …(1)
CoO(s) + CO(g) Co(s) + CO2(g) K = 490 …(2)
Substracting (2) from (1) we get
H2(g) + CO2(g) CO(g) + H2O(g) K′ = …(3)
Multiplying equation (3) with 1/2 we get
H2(g) + CO2(g) CO(g) + H2O(g) K″ =
Exercise 2: The equilibrium constant of the reaction H2(g) + I2(g) 2HI(g)
At 426°C is 60, what will be the value of equilibrium constant
- a) If the reaction is reversed
- b) If the given reaction is represented as 4H2(g) + 4I2(g) 8HI(g)
3.2 The Reaction Quotient ‘Q’
Consider the equilibrium
PCl5 (g) PCl3(g) + Cl2 (g)
At equilibrium = KC. When the reaction is not at equilibrium this ratio is called ‘QC’ i.e., QC is the general term used for the above given ratio at any instant of time. And at equilibrium QC becomes KC.
Similarly, is called QP and at equilibrium it becomes KP.
- If the reaction is at equilibrium, Q = Kc
- A net reaction proceeds from left to right (forward direction) if Q < KC.
- A net reaction proceeds from right to left (the reverse direction) if Q >Kc
Illustration 3: For the reaction A(g) + B(g) 3C(g) at 25°C, a 3 litre vessel contains 1, 2, 4 moles A, B and C respectively. Predict the direction of reaction if
- a) KC for the reaction is 10
- b) KC for the reaction is 15
- c) KC for the reaction is 10.66
Solution: A(g) + B(g) 3C(g)
[A] = 1/3 [B] = 2/3, [C] = 4/3 ⇒ Q = = 10.66
- a) Since KC = 10 ⇒ Q > KC, thus reaction will occur in backward direction.
- b) Since KC= 10 ⇒ KC > Q, the reaction will occur in forward direction.
- c) Q = KC reaction is at equilibrium
Exercise 3: Streams of SO3, SO2 and O2 are fed into a previously evacuated vessel at 100K so that atm
In which direction can the process move if the equilibrium constant is 3.4 for the process below.
2SO2(g) + O2(g) 2SO3(g)
4. The Le–Chatelier’s Principle
This principle, which is based on the fundamentals of a stable equilibrium, states that
“When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium shifts in that direction in which the effect of the stress is reduced”.
Confused with “stress”. Well by stress here what I mean is any change of reaction conditions e.g. in temperature, pressure, concentration etc.
This statement will be explained by the following example.
Let us consider the reaction: 2NH3 (g) N2 (g) + 3H2 (g)
Let the moles of N2, H2 and NH3 at equilibrium be a, b and c moles respectively. Since the reaction is at equilibrium,
= KP =
X terms denote respective mole fractions and PT is the total pressure of the system.
⇒ = KP
Here, = mole fraction of N2
= mole fraction of H2
= mole fraction NH3 ⇒ = KP
Since PT = ( assuming all gases to be ideal)
∴ = KP … (1)
Now, let us examine the effect of change in certain parameters such as number of moles, pressure, temperature etc.
If we increase a or b, the left hand side expression becomes QP ( as it is disturbed from equilibrium) and we can see that QP > KP. The reaction therefore moves backward to make QP = KP.
If we increase c, QP < KP and the reaction has to move forward to revert
back to equilibrium.
If we increase the volume of the container (which amounts to decreasing the pressure), QP < KP and the reaction moves forward to attain equilibrium.
If we increase the pressure of the reaction then equilibrium shifts towards backward direction since in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is reduced in backward direction.
If temperature is increased the equilibrium will shift in forward direction since the forward reaction is endothermic and temperature is reduced in this direction.
However from the expression if we increase the temperature of the reaction, the left hand side increases (QP) and therefore does it mean that the reaction goes backward (since QP > KP)?. Does this also mean that if the number of moles of reactant and product gases are equal, no change in the reaction is observed on the changing temperature (as T would not exist on the left hand side). The answer to these questions is No. This is because KP also changes with temperature. Therefore, we need to know the effect of temperature on both QP and KP to decide the course of the reaction.
Illustration 4: What will be the effect on the equilibrium constant for the reaction
A2(g) + 3B2(g) 2AB3(g) ΔH = – xkcal. When (a) pressure is increased (b) concentration of A2 is increased (c) temperature is raised at equilibrium.
Solution: a) No effect
- b) No effect as K doesn’t ’t depend on pressure and concentration.
- c) Since reaction is exothermic the equilibrium constant will decrease as the temperature is increased.
Exercise 4: Carbon monoxide and chlorine are introduced into a vessel at 395 K forming phosgene according to
CO(g) + Cl2(g) COCl2(g)
At equilibrium, the pressure of the three gases measured as 11.8kPa, 13.0 kPa and 33.8 kPa for carbon monoxide chlorine and phosgene respectively. Evaluate K and the total pressure P.
4.1 Effect of Addition of Inert Gases to a Reaction at Equilibrium
- Addition at constant pressure
Let us take a general reaction
aA(g) + bB(g) cC(g) + dD(g)
nC nD, nA, nB denotes the no. of moles of respective components and PT is the total pressure and ∑n = total no. of moles of reactants and products.
Where Δn = (c + d) – (a + b)
Now, Δn can be = 0, < 0 or > 0
Lets take each case separately.
- a) Δn = 0 : No effect
- b) Δn = ‘+ve’ :
Addition of inert gas increases the ∑n i.e. is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.
- c) Δn = ‘–ve’ :
In this case decreases but increases. So products have to decrease and reactants have to increase to maintain constancy of Kp. So the equilibrium moves backward.
- Addition at Constant Volume
Since at constant volume, the pressure increases with addition of inert gas and at the same time ∑n also increases, they almost counter balance each other. So can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.
4.2 Dependence of KP or Kc on Temperature
Now we will derive the dependence of KP on temperature.
Starting with Arrhenius equation of rate constant
Where, kf = rate constant for forward reaction, Af = Arrhenius constant of forward reaction,
= Energy of activation of forward reaction
Dividing (2) by (3) we get,
We know that (equilibrium constant )
∴ K =
At temperature T1
At temperature T2
Dividing (iv) by (iii) we get
|The enthalpy of a reaction is defined in terms of activation energies as = ΔH
∴log … (v)
For an exothermic reaction , ΔH would be negative. If we increase the temperature of the system ( T2>T1), the right hand side of the equations (V) becomes negative.
∴ , that is, the equilibrium constant at the higher temperature would be less than that at the lower temperature. Now let us analyse our question. Will the reaction go forward or backward?
Before answering this, we must first encounter another problem. If temperature is increased, the new KP would either increase or decrease or may remain same. Let us assume it increases.
Now, QP can also increase, decrease or remain unchanged. If KP increases and QP decreases, than , therefore the reaction moves forward. If KP increase and QP remains same, then . Again, the reaction moves forward. What, if KP increase and QP also increases?
Will or or ? This can be answered by simply looking at the dependence of QP and KP on temperature. You can see from the equation (6) that KP depends on temperature exponentially. While Q’s dependence on T would be either to the power g,l,t…….. Therefore the variation in KP due to T would be more than in QP due to T.
∴ KP would still be greater than QP and the reaction moves forward again.
Therefore, to see the temperature effect, we need to look at KP only. If it increases the reaction moves forward, if it decreases, reaction moves backward and if it remains fixed, then, no change at all.
Illustration 5: Would you expect the equilibrium constant for the reaction I2(g) 2I(g) to increase or decrease as temperature increases why?
Solution The forward reaction is endothermic as energy is required to break I2
into I. So the equilibrium constant will increase.
Illustration 6: Kp for the reaction, N2 + 3H2 2NH3 is 1.6 × 10–4 atm–2 at 400°C. What will be Kp at 500°C. Heat of reaction in this temperature range is – 25.14 kcal.
Solution: We know that
⇒ = 1.46 × 10–5 atm–2
Exercise 5: Equilibrium constant Kp for the reaction H2(g) + N2(g) NH3(g) are 0.0266 and 0.0129 at 310°C and 400°C respectively. Calculate the heat of formation of gaseous ammonia.
- Degree of Dissociation (α)
Let us consider the reaction; 2NH3 (g) N2 (g) + 3H2 (g)
Let the initial moles of NH3(g) be ‘a’. Let x moles of NH3 dissociate at equilibrium.
2NH3 (g) N2 (g) + 3H2 (g)
Initial moles a 0 0
At equilibrium a–x
Degree of dissociation (α) of NH3 is defined as the number of moles of NH3 dissociated per mole of NH3.
∴ if x moles dissociate from ‘a’ moles of NH3, then, the degree of dissociation of NH3 would be .
We can also look at the reaction in the following manner.
2NH3 (g) N2 (g) + 3H2 (g)
Initial moles a 0 0
At equilibrium a(1–α)
or a–2x′ x ′ 3 x ′
Here total no. of moles at equilibrium is
a – 2x′ + x′ + 3x′ = a + 2x′
Mole fraction of NH3 =
Mole fraction of N2 =
Mole fraction of H2=
The expression of Kp is
Kp = =
In this way you should calculate the basic equation. So my advice to you is that, while solving problem follow the method given below:
- Write the balanced chemical reaction (mostly it will be given)
- Under each component write the initial no. of moles.
- Do the same for equilibrium condition.
- Then derive the expression.
Do it and you are the winner.
5.1 Dependence of Degree of Dissociation on Density Measurements
The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose KP exist, i.e., reactions having at least one gas and having no solution.
Since PV = nRT
∴ VD =
Since P =
∴ VD =
For a reaction at eqb., V is a constant and ρ is a constant. ∴ vapour Density
( molecular weight = 2 × V.D)
Here M = molecular weight initial
m = molecular weight at equilibrium
Let us take a reaction
PCl5 PCl3 + Cl2
Initial moles C 0 0
At eqb. C(1−α) Cα Cα
Knowing D and d, α can be calculated and so for M and m.
Illustration 7: The degree of dissociation of PCl5 is 70% at 250°C. Find out its vapour density.
Solution: PCl(g) PCl3(g) + Cl2(g)
1 0 0 at initial
1 – x x x at equilibrium
Moles at equilibrium = 1 – x + x + x = 1 + x = 1 + 0.7 = 1.7
x = 0.7
We know that
⇒ (V.D.)final = 61.32
Exercise 6: Vapour density of N2O4 which dissociates according to the equilibrium N2O4(g) 2NO2(g) is 25.67 at 100°C and a pressure of 1 atm. Calculate the degree of dissociation and Kp for the reaction.
- Relationship Between ΔG° and K
Free energy, G, denotes the self intrinsic electrostatic potential energy of a system. This means that in any molecule if we calculate the total electrostatic potential energy of all the charges due to all the other charges, we get what is called the free energy of the molecule. It tells about the stability of a molecule with respect to another molecule. Lesser the free energy of a molecule more stable it is.
Every reaction proceeds with a decrease in free energy. The free energy change in a process is expressed by ΔG. If it is negative, it means that products have lesser G than reactants, so the reaction goes forward. If it is positive the reaction goes reverse and if it is zero the reaction is at equilibrium.
ΔG is the free energy change at any given concentration of reactants and products. If all the reactants and products are taken at a concentration of 1 mole per liter, the free energy change of the reaction is called ΔGo (standard free energy change).
One must understand that ΔGo is not the free energy change at equilibrium. It is the free energy change when all the reactants and products are at a concentration of 1 mole/L. ΔGo is related to K (equilibrium constant) by the relation, ΔGo = −RTln K. K may either be KC or KP. Accordingly we get .
The units of ΔGo depends only on RT. T is always in Kelvin, and if R is in Joules, ΔGo will be in joules, and if R is calories then ΔGo will be in calories.
Illustration 8: ΔG0for N2(g) + H2(g) NH3(g) is – 16.5 kJ/mol. Find out Kp for the reaction. Also report Kp and ΔG0 for N2 + 3H2 2NH3 at 25°C.
Solution: – ΔG0 = 2.303 RT log Kp
– (16.5 × 103) = 2.303 × 8.314 × 298 log Kp
⇒ Kp = 779.4 atm–1
Also for N2 + 3H2 2NH3
kp1 = (kp)2 = (779.4)2
= 6.06 × 105 atm–2
Also ΔG0 = 2.303 RT log
= 2.303 × 8.314 × 298 log (6.07 × 105) ⇒ – 32.989 kJ/mole
Exercise 7: The value of Kp at 298 K for the reaction N2 + H2 2NH3 is found to be 826.0, partial pressure being measured in atm units calculate ΔG0 at 298K.
- Solution to Exercises
Exercise 1: 2X(g) + Y(g) 2L(g) + M(g)
a b 0 0 at initial
(a – 2x) (b – x) 2x x at equilibrium
a – 2x = 0.002
b – x = 0.005
2x = 2.3 ⇒ x = 1.15
⇒ a = 2.302 moles/litre
b = 1.155 moles/litre
Exercise 2: a) The reverse reaction of the given reaction is
2HI(g) H2(g) + I2(g0
∴ equilibrium constant = = 0.966
- b) For the reaction
4H2(g) + 4I2(g) 8HI
K = (60)4 = 1.296 × 107
Exercise 3: = 0.495
Q < K. Hence equilibrium will move towards right.
Exercise 4: = 0.117 atm
= 0.129 atm
= 0.335 atm
Kp = = 22.2
Total pressure = 0.117 + 0.129 + 0.335 = 0.581 atm
ΔH = – 12140 cals
Exercise 6: N2O4(g) 2NO2(g)
1 0 at initial
1 – x 2x at equilibrium
Total no. of moles at equilibrium = 1 + x
As we know that
⇒ x = 0.792
Kp = = 6.7 atm
Exercise 7: ΔG0 = 2.330 RT logKp
= – 2.303 × 1.98 × 298 × log826
⇒ – 3980 calories
8. Solved Problems
Problem 1: KC for PCl5(g) PCl3(g) + Cl2(g) is 0.04 at 250°C. How many mole of PCl5 must be added to a 3 litre flask to obtain a Cl2 concentration of 0.15 M?
Solution: At equilibrium number of mole of Cl2 in 3 litres is 0.15 × 3 i.e. 0.45.
PCl5 PCl3 + Cl2
Thus x(say) 0 0 Initial mole
(x – 0.45) 0.45 0.45 Mole at equilibrium
0.15 0.15 Equilibrium concentration
KC = = 0.04 (given)
X = 2.1 mole
Problem 2: A mixture of SO2, SO3 and O2 gas is maintained in a 10 litre flask at a temperature at which KC = 100 for the reaction
2SO2 + O2 2SO3
- a) If the number of moles of SO2 and SO3 in the flask are equal, how much O2 is present?
- b) If the number of moles of SO3 in the flask is twice the number of moles of SO2. How much O2 is present?
Solution: a) We know, at equilibrium
KC = = 100
As mole of SO2 is equal to that of SO3
[SO2] = [SO3]
(volume = 10 litre)
∴ mole of O2 = 0.1
- b) Further, if mole of SO3 is twice that of SO2
= 2 …(1)
We again have,
or = 100
or = 100 …(2)
from (1) and (2), we get,
= = 100
Mole of O2 = 0.4
Problem 3: KC for N2O4(g) 2NO2(g) is 0.00466 at 298 K. If a one litre container initially contained 0.8 mole of N2O4, what would be the concentrations of N2O4 and NO2 at equilibrium? Also calculate equilibrium concentration of N2O4 and NO2 if the volume is halved at the same temperature.
Solution: Suppose x mole of N2O4 changes to NO2 at equilibrium
0.8 – x 2x mole/1 at equilibrium
∴ KC = = 0.00466, x = 0.03 M
∴ at equilibrium [N2O4] = 0.8 – x = 0.8 – 0.03 = 0.77 M
[NO2] = 2x = 2 × 0.03 = 0.06 M
Now when the volume of halved, pressure will increase. From the Le-Chatlier’s Principle we know that the equilibrium will shift to left hand side with the increase in pressure (or decrease in volume). Further when volume is halved, concentration will be doubled.
Concentration of N2O4 = 0.77 × 2 = 1.54 M
And concentration of NO2 = 0.06 × 2 = 0.12 M
The equilibrium concentration now will be
(1.54 – y) (0.12 – 2y)
KC = = 0.00466, y = 0.104 and 0.017
y = 0.104 is unacceptable as in this case 2y > 0.12, y = 0.017
∴ At new equilibrium
[N2O4] = 1.54 + 0.017 = 1.557 M
[NO2] = 0.12 – 2 × 0.017 = 0.086 M
Problem 4: At 800K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 3 mole of SO3 at equilibrium. KC for the equilibrium 2SO2 + O2 2SO3 is 833 lit/mol. If the volume of the container is 1 litre. Calculate how much O2 is to be added at this equilibrium in order to get 5.2 mole of SO3 at the same temperature.
Solution: Suppose x mole of O2 is added by which equilibrium shifts to right hand side and y mole of O2 changes to SO3. The new equilibrium concentration may be represented as
2SO2 + O2 2SO3
First equilibrium 0.5 mole 0.12 mole 5 mole
Second equilibrium (0.5 – 2y) (0.12 + x – y) (5 + 2y)
But 5 + 2y = 5.2 given ∴ y = 0.1
Now, KC = = 833 (volume = 1 lit)
Substituting y = 0.1, we get
x = 0.34 mole
Problem 5: Would 1% CO2 in the air be sufficient to prevent any lose in weight when Ag2CO3 is dried at 120°C? Ag2CO3(s) Ag2O(s) + CO2(g), Kp = 0.0095 atm at 120°C. How low would the partial pressure of CO2 have to be promote this reaction at 120°C?
∴ = 0.0095 atm = constant at 120°C
Thus if Ag2CO3 is taken in a closed container, a small amount of it would decompose to give CO2 gas until the partial pressure of CO2 reaches 0.0095 atm. As this is equilibrium pressure of CO2, the decomposition would then stop.
Now since partial pressure of CO2 in air is 0.01 atm (since CO2 is 1% in air) which is much greater than 0.0095 atm, the equilibrium would practically shift to left hand side completely on in other wards, there would be no loss in weight of Ag2CO3 (by decomposition) if placed in air containing 1% CO2.
Further, if the partial pressure of CO2 in air is less than the equilibrium pressure of 0.0095 atm the decomposition of Ag2CO3 would continue till the CO2 pressure around Ag2CO3 becomes 0.0095 atm.
Problem 6: The value of KC for the reaction: A2(g) + B2(g) 2AB(g) at 100°C is 50. If 1.0 L flask containing one mole of A2 is connected with a 2.0 L flask containing two moles of B2, how many moles of AB will be formed
Solution: A2(g) + B2(g) 2AB(g)
As the two vessels are connected, the final volume for the contents is now 3.0 L. Let x mole each of A2 and B2 react to form 2x moles of AB2 (from stoichiometry of reaction)
KC = ; concentration of species at equilibrium are:
[A2] = (1 – x)/3, [B2] = (2-x)/3, [AB] = 2x/3
KC = ⇒ 46x2 + 150x + 100 = 0
⇒ x = 0.93 or 2.33 (neglecting this value)
⇒ moles of AB(g) formed at equilibrium = 2x = 1.86
Problem 7: Variation of equilibrium constant K with temperature T is given by Van’t Hoff equation
log K = log A –
A graph between log K and T–1 was a straight line as shown and having OP = 10 and tan θ = 0.5. Calculate
|i) equilibrium constant at 298 K, and
ii) and equilibrium constant at 798 K, assuming ΔH° to be independent of temperature.
Solution: To calculate equilibrium constant, we need to know A and ΔH°, which are calculated as –
The given equation represents a straight line of slope
= = –tan θ = –0.5
∴ ΔH° = 2.303 ×8.314 ×0.5 = 9.574 J/mol
Intercept = log A = OP = 10
∴ log K = log A–
∴ K = 9.96 ×109
Now, to calculate equilibrium constant at some other temperature, we will use the expression
∴ K2 (equilibrium constant at 798 = 9.98 × 109
Problem 8: The equilibrium constant for the reaction
C(s) + CO2(g) 2CO(g) as a function of temperature can be expressed as
KT ln kp = – 40900 + 4.9T ln T – 4.95 × 10–2 T2 + 5.1 × 10–7 T3 + 12.66T
In the range of 1100 to 1500 K. If the at 1200 K is 1.2 atm
Solution: For the given reaction
kp = or ⇒ =
kp at 1200 K can be calculated from the reaction
lnkp = [–40900 + 4.9 (1200) ln1200 – (4.95 × 10–2) (1200)2
+ (5.1 × 10–7) (1200)3 + (12.66) (1200)]
lnkp = – 5.454
⇒ kp= 4.28 × 10–3
putting the value of kp
= (4.28 × 10–3 × 1.2)1/2 = 0.072 atm
Problem 9: Ammonium hydrogen sulphide dissociated according to the equation
NH4HS(s) NH3(g) + H2S(g). If the observed pressure of the mixture is 1.12 atm at 106oC, what is the equilibrium constant Kp of the reaction ?
Solution: The reaction is
NH4HS(s) NH3(g) + H2S(g)
If α is the degree of dissociation of equilibrium,
Total moles of NH3 + H2S = 2α.
Partial pressure =
∴ pNH3 = × P = 0.5P; pH2S = × P = 0.5P
Kp = pNH3 × pH2S = 0.5P × 0.5P = 0.25P2
Substituting the value of P = 1.12 atm.,
Kp = 0.25 × 1.12 × 1.12 = 0.3136 atm2.
At equilibrium = 1.12 atm
∴ = 0.56 atm
∴KP = 0.56 × 0.56 = 0.3136 atm2.
Problem 10: Consider the equilibrium: P(g) + 2Q(g) R(g). When the reaction between P and Q is carried out at a certain temperature, the equilibrium concentration of P and Q are 3 M and 4 M respectively. When the volume of the vessel is doubled and the equilibrium is allowed to re-established, the concentration of Q is found to be 3 M. Find the
- a) value of KC
- b) concentration of R at new equilibrium stages
Solution: P(g) + 2Q(g) R(g)
At equilibrium, [P] = 3 M, [Q] = 4 M & let [R] = x M,
KC = (1)
Now the volume is doubled, hence the concentrations are halved and a new equilibrium will be re-established with same value of KC., Calculate Q and determine the direction of equilibrium.
⇒ Q> Kc Hence the system will predominantely move in backward direction so as to achieve new equilibrium state. Let y M be the decrease in concentration of R.
|At new equilibrium||1.5 + y||2 + 2y||x/2 – y|
Given: [Q] = 3 M at new equilibrium ⇒ 2 + 2y = 3 ⇒ y = 0.5 M
⇒ at new equilibrium, [P] = 1.5 + 0.5 = 2 M;
[Q] = 3M (given); [R] = x/2 – 0.5 M
⇒ Q =
equating this value of KC with (I)
⇒ ⇒ x = 4 M
Hence [R] = 4M and at new equilibrium [R] = x/2 – 0.5 = 1.5 M
and KC =
Problem 11: In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC, the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (KP) of the mixture, for the reaction,
N2(g) + 3H2(g) 2NH3(g)
Solution: Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as KP will not depend on the exact no. of moles of N2 and H2. One can even start with x and 3x).
N2(g) + 3H2(g) 2NH3
Initially 1 3 0
At equib 1−x 3−3x 2x
Since % by volume of a gas is same as % by mole,
∴ Mole fraction of H2 at equilibrium =
Mole fraction of N2 at equilibrium
= 1 − 0.6165 − 0.178 = 0.2055
KP = 7.31 × 10-4 atm-2.
N2(g) + 3H2(g) 2NH3(g)
Initially x 3x 0
At equib. x−a 3x−3a 2a
∴ = 0.178
∴ = 0.302
Similarly we can calculate the mole fraction of N2(g) and H2(g) at equilibrium and finally KP comes out to be 7.31 ×10–4 atm–2
Problem 12: Following equilibrium are established by mixing NO and NO2
- i) 2NO2(g) N2O4(g) ; Kp = 6.67 atm–1
- ii) NO(g) + NO2(g) N2O3(g)
If equimolar amounts of NO and NO2 are mixed, calculate
(A) the equilibrium partial pressure of nitric oxide (NO)
(B) the equilibrium constant for the formation of N2O3.
Given final total pressure is 3.9 atm. and the equilibrium partial pressure of NO2 is 0.3 atm
Solution: For the first reaction,
2NO2(g) N2O4(g) Kp = 6.67 atm–1
t -teq .3 atm x atm (let)
Kp = = 6.67 atm–1
∴ x = 0.6 atm
The second equilibrium is
NO(g) + NO2(g) N2O3 (g)
Applying Dalton’s law of partial pressures, PT =
Or 3.9 atm = .3 atm + .6 atm + PNO +
∴ PNO + = 3 atm
[initially NO2 and NO are equimolar]
∴ = 3 atm
reacted in first equilibria = 2 ×
(initial) – = (at equil)
∴ 3 – 1.2 – Y = 3
∴Y = 1.5 atm
Kp (II) = = = 3.33 atm–1
Problem 13: SO3 decomposes at a temperature of 900 K and at a total pressure of
1.5 atm. At equilibrium, the density of mixture is found to be 1.1 g/l in a vessel of 90 litres. Find the degree of dissociation of SO3 and Kp for
SO3 SO2 + I/2O2
Solution: SO3 decomposes as follows
SO3(g) SO2(g) +
Let x be the degree of dissociation
⇒ total moles = 1 – x + x + x/2
First calculate d0 by using PM = d0 RT
⇒ do = 1.624 g/lt
⇒ ⇒ x = 0.95 ≡ 95% dissociation
For calculation of partial pressures, first calculate P at equilibrium and total moles.
Total moles = 1 + = 1.475
Peq = 1.21 atm
= 0.389 atm
= 0.779 atm
= 0.041 atm
⇒ Kp = = 11.85 atm
Problem 14: For the reaction NH3 (g) N2 (g) + H2 (g)
Show that the degree of dissociation of NH3 is given as
where P is the equilibrium pressure. If KP of the above reaction is 78.1 atm at 400°C, determine the value of KC.
Solution: Let n and α be the initial number of moles and degree of dissociation of NH3 (g) respectively.
NH3 (g) N2 (g) + H2 (g)
Initially n 0 0
At equilibrium n(1–α)
Total number of moles = n(1+α)
Partial pressure of NH3, =
Partial pressure of N2, =
Partial pressure of H2, = × P
∴ KP =
= or, α =
Δn, change in number of the moles of the given reaction = +1
KP = KC(RT)Δn
or, KC= KP (RT)–Δn
KC = 78.1 [0.0821×673]–1 = 1.413 mol/l
Problem 15: In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were introduced and equilibriated at a temperature at 250oC. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, α of PCl5 and KP for the reaction
PCl5 PCl3 + Cl2 at this temperature.
Solution: PCl5(g) PCl3(g) + Cl2(g)
Initial moles 5 0 0
At equilibrium 5−x x x
Total moles = 5 + x + 4 (including moles of Argon)
= 9 + x
Since total moles = = 11.99 12
∴ x = 3
∴ α = = 0.6
KP = = 1.75 atm.
The degree of dissociation of PCl5 would have been 0.6 even in the absence of Argon. As one can see, the total pressure of gases constituting equilibrium is equal to 3.11 atm. The observed equilibrium pressure is 4.678 atm which means by the addition of 4 moles of Argon, the total pressure increases. This implies that addition of Argon has been done at constant volume which doesnot result in any change in degree of dissociation.
Problem 1: For which of the following Kp may be equal to 0.5 atm
(A) 2HI H2 + I2 (B) PCl5 PCl3 + Cl2
(C) N2 + 3H2 2NH3 (D) 2NO2 N2O4
Solution: For Kp = 0.5 atm
Δn = 1 (since the unit is atm)
and PCl5 PCl3 + Cl2
Δn = 1
Problem 2: The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final temperature is
(A) 80 (B) 60
(C) 40 (D) 70
Solution: N2O4 2NO2
1 0 at initial
1 – α 2α at equilibrium
1.8 = 1 + α ⇒ α = 0.8
Problem 3: For the reaction PCl5(g) PCl3(g) + Cl2(g), the forward reaction at constant temperatured is favoured by
(A) Introducing an inert gas at constant volume
(B) Introducing Cl2 gas at constant pressure
(C) Introducing an inert gas at constant pressure
(D) Increasing the volume of the container
Solution: Introducing the inert gas at constant pressure will increase the volume of container and when the volume of container is increased equilibrium shift in the direction where no. of moles are increasing i.e. toward forward direction.
∴(C) and (D)
Problem 4: If pressure is applied to the following equilibrium, liquid vapours the boiling point of liquid
(A) will increase (B) will decrease
(C) may increase or decrease (D) will not change
Solution: Boiling point of a liquid is the temperature at which vapour pressure became equal to atm pressure. If the pressure is applied to the above equilibrium the reaction will go to the backward direction, i.e. vapour pressure decrease hence the boiling point increase.
Problem 5: For the reaction
A(g) + B(g) 3C(g) at 250°C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C respectively. If KC for the reaction is 10, the reaction will proceed in
(A) Forward direction (B) Backward direction
(C) In either direction (D) In equilibrium
Solution: = 10.66
KC = 10, and Q > KC
∴ reaction will proceed in backward direction
Problem 6: The equilibrium constant for the reaction N2(s) + O2(g) 2NO(g) is 4 × 10–4 at 200 K. In presence of a catalyst the equilibrium is attained 10 times faster. Therefore, the equilibrium constant in presence of catalyst at
(A) 40 × 10–4 (B) 4 × 10–4
(C) 4 × 10–3 (D) Can’t be calculated
Solution: Kp and KC value doesn’t depend on catalyst
Problem 7: In a system A(s) 2B(g) + 3C(g). If the concentration of C at equilibrium is increased by a factor 2, it will cause the equilibrium concentrations of B to change to
(A) Two times of its original value
(B) One half of its original value
(C) 2√2 time of its original value
(D) times of its original value
Solution: A(s) 2B(g) + 3C(g)
KC = [C]3 [B]2
If (C) becomes twice, then let the concentration of B is B′ then
KC = [2C]3 [B′]2 = [C]3 [B]2
or [B′]2 =
or [B′] =
Problem 8: In a reaction at equilibrium ‘X’ moles of the reactant A decomposes to give 1 mole each of C and D. If the fraction of A decomposed at equilibrium is independent of initial concentration of A then the value ‘X’ is
(A) 1 (B) 3
(C) 2 (D) 4
Solution: xA C + D
a 0 0 at initial
a(1 – α) at equilibrium
K = =
α is independent of A, ⇒ 2 – x = 0 or x = 2
Problem 9: If CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l) Kp = 1.086 × 10–4 atm2 at 25°C. The efflorescent nature of CuSO4.5H2O can be noticed when vapour pressure of H2O in atmosphere is
(A) > 7.29 mm (B) < 7.92 mm
(C) ≥ 7.92 mm (D) None
Solution: An efflorescent salt is one that loses H2O to atmosphere.
For the reaction
CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l)
Kp = ()2 = 1.086 × 10–4
= 1.042 × 10–2 atm = 7.92 mm
If at 25°C < 7.92 mm only then, reaction will proceed in forward direction.
Problem 10: In the system, LaCl3(s) + H2O (g)+heat⎯→LaClO(s)+2HCl (g), equilibrium is established. More water vapour is added to reestablish the equilibrium. The pressure of water vapour is doubled. The factor by which pressure of HCl is changed is
(A) 2 (B)
Solution LaCl3(s) + H2O(g) + heat LaClO(s) + 2HCl(g)
At eq. P1 P2
When = 2P1 , Let pHCl =
Then KP =
therefore pressure of HCl is changed by factor.
Problem 11: The equilibrium constant for the reaction
N2(g) + O2(g) 2NO(g) is 4 ×10–4 at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibirum constant in presence of the catalyst at 200 K is
(A) 4 ×10–3 (B) 4 ×10–-4
(C) 4 ×10–5 (D) None
Solution: Equilibrium constant of a reaction remains unchanged when the same reaction remains unchanged when the same reaction is carried out in presence of a catalyst because the catalyst will catalyse both the forward and backward reactions equally.
Problem 12: For the decomposition reaction
NH2COONH4 (s) 2NH3(g) + CO2(g)
The KP = 2.9 ×10–5 atm3. The total pressure of gases at equilibrium when 1 mole of NH2COONH4 (s) was taken to start with would be
(A) 0.0194 atm (B) 0.0388 atm
(C) 0.0582 atm (D) 0.0766 atm
Solution: NH2COONH4(s) 2NH3(g) + CO2(g)
1 2 1
KP = 2.9 × 10–5 atm3
If the P is the total pressure at equilibrium
KP = ∴P3 = = 1.9575
P = = 0.058
Problem 13: For a reaction A(g) + B(g) C(g) + D(g), the initial concentration of A and B are equals but the equilibrium concentration of C is twice that of equilibrium concentration of A. Then K is
(A) 4 (B) 9
(C) 1/4 (D) 1/9
Solution: A(g) + B(g) C(g) + D(g)
Initial conc. a a 0 0
Eq. conc. a – x a – x x x
Given x =2(a-x) or x =
Kc = = 4
Problem 14: The partial pressure of CH3OH (g), CO(g) and H2(g) in equilibrium mixture for the reaction CO(g) + 2H2(g) CH3OH(g) are 2.0, 1.0, and 0.1 atm respectively at 427°C. The value of Kp for decomposition of CH3OH to CO and H2 is
(A) 102 atm (B) 2 ×102 atm–1
(C) 50 atm2 (D) 5 × 10–3 atm2
Solution: Kp = = 200
Kp for reverse reaction will be
= 5 × 10–3 atm2
A sample of air consisting of N2 and O2 was heated to 2500 K until the equilibrium N2(g) + O2(g) 2NO(g) was established with an equilibrium constant Kc = 2.1 × 10–3. At equilibrium, the mole of NO was 1.8. Estimate the initial composition of air in mole fraction of N2 and O2.
Problem 15: For reaction A(g) + B(g) AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB
(A) 20% (B) 40%
(C) 60% (D) 4%
Solution: A(g) + B(g) AB(g)
Initial moles 2 2 0
At equilibrium 2 – x 2 – x x = 0.8
∴ % of A changed to AB = × 100 = 40%
- Assignments (Subjective Problems)
LEVEL – I
- In a gaseous reaction of the type A(g) + 2B(g) 2C(g) + D(g), the initial concentration of B was 1.5 times that of A. At equilibrium the equilibrium concentration of A and D were equal. Calculate the equilibrium constant?
- At a certain temperature, equilibrium constant (KC) is 16 for the reaction:
SO2(g) + NO2 (g) SO3(g) + NO(g)
If we take one mole of each of the four gases in one litre container, what would be the equilibrium concentration of NO and NO2?
- The equilibrium constant is 0.403 at 1000K for the process
FeO(s) + CO(g) Fe(s)+ CO2)g)
A steam of pure CO is passed over powdered FeO at 1000K until equilibrium is reached. What is the mole fraction of CO in the gas stream leaving the reaction zone?
- A reaction vessel is charged with steam and ethene in a molar ratio of 10:1. Pressure is maintained at 1.0 MPaand temperature at 400K. Assuming that the only process is C2H4(g) + H2O(g) C2H5OH(g)
What fraction of ethene is converted to ethanol at equilibrium? K is 0.306.
- To 500 ml of 0.150 M AgNO3 solution we add 500 ml of 1.09 M Fe2+ solution and the reaction is allowed to reach equilibrium at 25°C.
Ag+ (aq) + Fe2+ (aq) Fe3+ (aq) + Ag(s)
For 25 ml. of the solution, 30 ml. of 0.0832 M KMnO4 were required for oxidation under acidic conditions. Calculate KC for the reaction.
- When one mole of benzoic acid (C6H5COOH) and three moles of ethanol (C2H5OH) are mixed and kept at 200°C until equilibrium is reached, it is found that 87% of the acid is consumed by the reaction.
C6H5COOH(l) + C2H5OH (l) C6H5COOC2H5(l) + H2O(l)
Find out the percentage of the acid consumed when one mole of the benzoic acid is mixed with four moles of ethanol and treated in the same way.
- N2O4 is 25% dissociated at 37°C and one atmospheric pressure. Calculate (i) KP and (ii) percent dissociation at 0.1 atmospheric pressure and 37°C.
- Kp for the reaction PCl5 PCl3 + Cl2 at 250°C is 0.82. Calculate the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.
- a) Calculate the equilibrium constant for the process
At a temperature 350 K, given that K = 0.14 at 298 K and ΔH = 57.2 kJ
- b) The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10-4 atm at 400o What will be the equilibrium constant at 500oC if heat of the reaction in this temperature range is −25.14 k cal?
- a) H2 and I2 are mixed at 400°C in a 1.0 L container and when equilibrium established, the following concentrations are present: [HI] = 0.49 M, [H2] = 0.08 M and [I2] = 0.06 M. If now an additional 0.3 mol of HI are added, what are the new equilibrium concentrations, when the new equilibrium H2 + I2 2HI is re-established?
- b) The equilibrium constant is 0.4 at 300K for the process N2O4(g) 2NO2(g) and it was found that at equilibrium partial pressure of N2O4 and NO2 were 18.9 and 16.5 atm respectively and no. of moles of N2O4 = 7.6 × 10–3mol and NO2 = 6.6 × 10–3. In what direction would the process move, if any, on adding 1.42 × 10–2 moles of argon
- i) If the volume were simultaneously doubled to two litres
- ii) If the volume were increased so that total pressure was held constant at
iii) If the volume were held constant? (initial amount of N2O4 = 1 gm).
LEVEL – II
- An air sample containing 21:79 of O2 and N2 (molar ratio) is heated to 2400°C. If the mole percent of NO at equilibrium is 1.8%. Calculate the Kp of the reaction
N2 + O2 2NO
- For the reaction F2 2F. Calculate the degree of dissociation of fluorine at 4 atm and 1000K, when Kp = 1.4 × 10–2 atm.
- The equilibrium constant of ester formation of propionic acid with ethyl alcohol is 7.36 at 50°C. Calculate the weight of ethyl propionate in gram existing in an equilibrium mixture when 0.5 mole of propionic acid is heated with 0.5 mole of ethyl alcohol at 50°C.
- 0.1 mole of ethanol and 0.1 mole of butanoic acid are allowed to react. At equilibrium, the mixture is titrated with 0.85 M NaOH solution and the titre value was 100 ml. Assuming that no ester was hydrolysed by the base, calculate K for the reaction. C2H 5OH + C3H7COOH C3H7COOC2H5 + H2O
- At temperature T, a compound AB2(g) dissociates according to the reaction,
2AB2(g) 2AB(g) + B2(g) with degree of dissociation, α, which is small compared to unity. Deduce the expression for α in terms of the equilibrium constant KP and the total pressure P.
- Show that KP for the reaction
2H2S(g) 2H2 (g) + S2 (g) is given by the expression
Where α is the degree of dissociation and P is the total equilibrium pressure. Calculate KC of the reaction if α at 298 K and 1 atm pressure is 0.055.
- The equilibrium constant at 375 K is 2.4 for the process
SO2Cl2(g) SO2(g) + Cl2(g)
Suppose that 6.745g of SO2Cl2 is placed in a previously evacuated one litre bulb and the temperature is raised to 375K. What would be the pressure of SO2Cl2 if none dissociated? What the partial pressure taking dissociation into account?
- When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate KC and KP for the reaction at 27°C. (ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask?
- a) NO and Br2 at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of the equilibrium constant and the standard free energy change at 300K for the reaction: 2NO(g) + Br2(g) 2NOBr(g).
- b) When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2-pentadiene (C). The equilibrium was maintained at 175° Calculate ΔG0 for the following equilibria.
B =A ΔG10 = ?, B =C ΔG20 = ?
From the calculated value of ΔG10 and ΔG20 indicate the order of stability of A, B and C. Write a reaction mechanism showing all intermediates leading to
A, B and C.
- a) When PCl5 is heated it dissociates into PCl3 and Cl2. The density of the gas mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC and 250oC.
- b) The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 g dm-3. Calculate the equilibrium constant of the reaction,
LEVEL – III
- Equilibrium constants are given in (atm) for the following reactions at 0°C.
- i) SnCl2⋅6H2O(s) SnCl2⋅2H2O(s) + 4H2O(g) KP = 6.9 × 10–12
- ii) Na2HPO4⋅12H2O(s) Na2HPO4⋅7H2O(s)+ 5H2O(g) ; KP= 5.25 × 10–13
iii) Na2SO4.10H2O(s) Na2SO4(s) + 10H2O(g) KP = 4.08 × 10–25
The vapour pressure of water at 0°C is 4.58 torr
(A) Calculate the pressure of water vapour in equilibrium at 0°C with each
(i), (ii) and (iii)
(B) Which is the most effective drying agent at 0°C
(C) At what relative humidities will Na2SO4.H2O be efflorescent when exposed to
air at 0°C
(D) At what relative humidities will Na2SO4 be deliquescent (i.e. absorb moisture) when exposed to air at 0°C.
- Solid NH4I on rapid heating in a closed vessel at 357°C develops a constant pressure of 275 mm of Hg owing to the partial decomposition of NH4I into NH3 and HI but the pressure gradually increases further (when excess solid residually remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed under equilibrium.
NH4I(s) NH3(g) + HI(g)
2HI(g) H2(g) + I2(g) Kc = 0.015 at 357°C
- In the manufacture of SO3 from SO2 by the reaction
2SO2(g) + O2(g) 2SO3(g)
It has been found that variation of equilibrium constant with temperature in the range 800 to 1170°C can be expressed as
logkp = – 9.346 +
Calculate ΔH of this reaction
- 20g solid NH4HS is placed in a 5 litre flask containing 0.5 mole of NH3 and 1.2 moles of H2S. Kp for the reaction is 0.11. What are the pressure of NH3 and H2S at 270K when equilibrium is reached.
- 0.0755 g of selenium vapour occupying a volume of 114.2 ml at 700°C exerts a pressure of 185 mm. The selenium is in a state of equilibrium according to the reaction Se6(g) 3Se2(g). Calculate (i) degree of dissociation of selenium; (ii) Kp and (iii) KC (Se = 79).
- Two solids X and Y dissociates into gaseous products at a certain temperature as follows
X(s) A(g) + C(g), and Y(s) B(g) + C(g). At a given temperature, pressure over excess solid X is 40 mm and total pressure over solid Y is 60 mm. Calculate.
- i) the values of Kp for two reactions
- ii) the ratio of mole sof A and B in the vapour state over a mixture of X and Y
iii) the total pressure of gases over a mixture of X and Y
- A saturated solution of I2 in water contains 0.33 g of I2 L–1. More than this can be dissolved in a KI solution because of the following equilibrium : I2 + I–
A 0.1 M KI solution (0.1M I–) actually dissolves 12.5 g I2/ litre, most of which is converted to , assuming that the concentration of I2 in all saturated solution is the same, calculate the equilibrium constant (KC) for the above reaction. What is the effect of adding water to a clear saturated solution of I2 in the KI solution?
- 0.15 mole of CO taken in a 2.5L flask is maintained at 750 K along with a catalyst so that the following reaction can take place.
Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (a) KP and KC and (b) the final pressure if the same amount of CO and H2 as before are used, but with no catalyst so that reaction does not take place.
- For the equilibrium
LiCl.3NH3 (s) LiCl. NH3 (s) + 2NH3,KP = 9 atm2 at 40°C. A five litre vessel contains 0.1 mole of LiCl.NH3. How many minimum moles of NH3 should be added to the flask at this temperature to derive the backward reaction for completion?
- 16 moles of H2 and 4 moles of N2 are sealed in a one litre vessel. The vessel is heated at a constant temperature until the equilibrium is established, when it is found that the pressure in the vessel has fallen 9/10 of its original value. Calculate Kc for the reaction.
N2 + 3H2 2NH3
- For the dissociation of phosgene gas the value of Kp at 100°C is 6.7 × 10–9 atm. Find the fraction of phosgene dissociated at this temperature when 1 mole of phosgene is placed in 100 litre vessel containing N2 gas at partial pressure of 1 atm
- The degree of dissociation of HI at a particular temperature is 0.8. Find the volume of 1.5 M sodium thiosulphate solution required to react completely with the iodine present at equilibrium in acidic condition, when 0.135 mol each of H2 and I2 are heated at 440 K in a closed vessel of capacity 2.0 L.
- For the gaseous reaction: C2H2 + D2O C2D2 + H2O
ΔH is 530 cal. At 25°C, KP = 0.82. Calculate how much C2D2 will be formed if
1 mole of C2H2 and 2 moles of D2O are put together at a total pressure of 1 atm
- Solid Ammonium carbamate dissociates as:
NH2COONH4(s) 2NH3(g) +CO2(g). In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium ammonia is added such that the partial pressure of NH3 at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.
- The equilibrium constant KP of the reaction
2SO2(g) + O2(g) 2SO3(g)
is 900 atm–1 at 800°C. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800°C.
- Assignments (Objective Problems)
LEVEL – I
- When KOH is dissolved in water, heat is evolved. If the temperature is raised, the solubility of KOH.
(A) Increases (B) Decreases
(C) Remains the same (D) Cannot be predicted
- For the liquefaction of gas, the favourable conditions are
(A) Low T and high P (B) Low T and low P
(C) Low T and high P and a catalyst (D) Low T and catalyst
- Consider the water gas equilibrium reaction
C(s) + H2O(g) CO(g) + H2(g)
Which of the following statement is true at equilibrium
(A) If the amount of C(s) is increased, less water would be formed
(B) If the amount of C(s) is increased, more CO and H2 would be formed
(C) If the pressure on the system is increased by halving the volume, more water would be formed.
(D) If the pressure on the system is increased by halving the volume, more CO and H2 would be formed.
- A reaction takes place in two steps with equilibrium constants 10–2 for slow step and 102 for last step. The equilibrium constant of the overall reaction will be
(A) 104 (B) 10—4
(C) 1 (D) 10–2
- For the reaction PCl3(g) + Cl2(g) PCl5(g), the value of KC at 250°C is 26 mol–1/litre. The value of Kp at this temperature will be
(A) 0.61 atm–1 (B) 0.57 atm–1
(C) 0.85 atm–1 (D) 0.46 atm–1
- According to Le Chatlier’s principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the
(A) Amount of solid to decrease (B) Amount of liquid to decrease
(C) Temperature to rise (D) Temperature to fall
- An aqueous solution of hydrogen sulphide shows the equilibrium
H2S H+ + HS–
If dilute hydrochloric acid is added to an aqueous solution of H2S, without any change in temperature, then
(A) The equilibrium constant will change
(B) The concentration of HS– will increase
(C) The concentration of undissociated hydrogen sulphide will decrease
(D) The concentration of HS– will decrease
- What would happen to a reversible reaction at equilibrium when temperature is raised, given that its ΔH is positive
(A) More of the products are formed (B) Less of the products are formed
(C) More of the reactants are formed (D) It remains in equilibrium
- Applying the law of mass action to the dissociation of hydrogen Iodide
2HI(g) H2 + I2
We get the following expression
Where a is original concentration of H2, b is original concentration of I2, x is the number of molecules of H2 and I2 reacted with each other. If the pressure is increased in such a reaction then
(A) K = (B) K >
(C) K < (D) none of these
- One mole of ethanol is treated with one mole of ethanoic acid at 25°C.
One–fourth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be
(A) 1/9 (B) 4/9
(C) 9 (D) 9/4
- The equilibrium, PCl5(g) PCl3(g) + Cl2(g) is attained at 25°C in a closed container and an inert gas He is introduced. Which of the following statements are correct.
(A) concentration of PCl5, PCl3 and Cl2 are changed
(B) more Cl2 is formed
(C) concentration of PCl3 is reduced
(D) Nothing happens
- In which of the following equilibrium, the value of KP is less than KC?
(A) N2O4 2NO2 (B) N2 + O2 2NO
(C) N2 + 3H2 2NH3 (D) 2SO2 + O2 2SO3
- In the reaction A2(g) + 4B2 (g) 2AB4(g) , ΔH > 0. The decomposition of AB4 (g) will be favoured at
(A) low temperature and high pressure
(B) high temperature and low pressure
(C) low temperature and low pressure
(D) high temperature and high pressure
- For the reaction, 2NO2 (g) 2NO(g) + O2 (g), KC = 1.8 ×10–6 at 185°C. At 185°C, the value of KC for the reaction:
NO(g) + O2 (g) NO2 (g) is
(A) 0.9 × 106 (B) 7.5 ×102
(C) 1.95 ×10–3 (D) 1.95 ×103
- For the gaseous phase reaction, 2A B + C, ΔH° = –40 Kcal mol–1 which statement is correct for
(A) K is independent of temperature
(B) K increase as temperature decrease
(C) K increase as temperature increases
(D) K varies with addition of A
LEVEL – II
- On applying pressure to the equilibrium
Which phenomenon will happen
(A) More ice will be formed (B) More water will be formed
(C) Equilibrium will not be disturbed (D) Water will evaporate
- Densities of diamond and graphite are 3.5 and 2.3 grams respectively. Increase of pressure on the equilibrium Cdiamond Cgraphite
(A) Favours backward reaction (B) Favours forward reaction
(C) Have no effect (D) Increase the reaction rate
- For the reaction N2 + 3H2 2NH3 in a vessel after the addition of equal number of mole of N2 and H2 equilibrium state is formed which of the following is correct?
(A) [H2] = [N2] (B) [H2] < [H2]
(C) [H2] > [N2] (D) [H2] > [NH3]
- One mole of N2O4(g) at 300K is kept in a closed container under one atm. It is heated to 600K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is
(A) 1.2 atm (B) 2.4 atm
(C) 2.0 atm (D) 1.0 atm
- At 30°C, Kp for the dissociation reaction
SO2(g) SO2(g) + Cl2(g)
is 2.9 × 10–2 atm. If the total pressure is 1 atm, the degree of dissociation of SO2Cl2 is
(A) 87% (B) 13%
(C) 17% (D) 29%
- A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm is
(A) 1.8 atm (B) 3 atm
(C) 0.3 atm (D) 0.18 atm
- Vapour density of PCl5 is 104.16 but when heated at 230°C its vapour density is reduced to 62. The degree of dissociation of PCl5 at this temperature will be
(A) 6.8% (B) 68%
(C) 46% (D) 64%
- For the reaction SO2(g) + O2(g) SO3(g) Kp = 1.7 × 1012 at 20°C and 1 atm pressure. Calculate Kc.
(A) 1.7 × 1012 (B) 0.7 × 1012
(C) 8.33 × 1012 (D) 1.2 × 1012
- For a gaseous equilibrium
2A(g) 2B(g) + C(g) , Kp has a value 1.8 at 700°K. What is the value of Kc for the equilibrium 2B(g) + C(g) 2A at that temperature
(A) ≈ 0.031 (B) ≈ 32
(C) ≈ 44.4 (D) ≈ 1.3 × 10–3
- For the reaction N2O4(g) 2NO2(g) the reaction connecting the degree of dissociation (α) of N2O4(g) with its equilibrium constant KP is
(A) α = (B) α =
(C) α = (D) α =
- In a closed container at 1 atm pressure 2 moles of SO2(g) and 1 mole of O2(g) were allowed to react to form SO3(g) under the influence of acetalyst. Reaction 2SO2(g) + O2(g) 2SO3(g) occurred
At equilibrium it was found that 50% of SO2(g) was converted to SO3(g). The formal pressure of O2 (g) at equilibrium will be
(A) 0.66 atm (B) 0.493 atm
(C) 0.33 atm (D) 0.20 atm
- KP for a reaction at 25°C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ / mol respectively. The KC for the reaction at 40°C will be
(A) 4.33 ×10–1 M (B ) 3.33 ×10–2 M
(C) 3.33 ×10–1 M (D) 4.33 ×10–2 M
- K for the synthesis of HI (g) is 50. The degree of dissociation of HI is
(A) 0.10 (B) 0.14
(C) 0.18 (D) 0.22
- One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N2O4 (g) decomposes to NO2(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N2O4 (g) is
(A) 10% (B) 20%
(C) 30% (D) 40%
- 40% of a mixture of 0.2 mol of N2 and 0.6 mol of H2 react to give NH3 according to the equation: N2(g) + 3H2(g) 2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are
(A) 4:5 (B) 5:4
(C) 7:10 (D) 8:5
- Answers to Subjective Assignments
LEVEL – I
- 4 2. 1.6M, 0.4M
- 0.7113 4. 73.2%
- 3.178 6. 90%
- i) 0.267 atm; ii) 63.25% 8. 27.5%
- a) 4.3 b) 1.462 × 10–5 atm 10. a) [HI] = 0.724 M
- b) [I2] = 0.093
LEVEL – II
- 2.1 × 10–3 2. 0.65
- 36.12g 4. 9
- 7. i) 1.53 atm ii) 0.03 atm
- 0.19 atm no effect 9. a) 134, 12.2 kJ b) B > C > A
- a) 0.8 b) 5.3 atm
LEVEL – III
- A) 1.62 × 10–3 atm, 3.5 × 10–3 atm
3.64 × 10–3
- B) SnCl2.6H2O C) 5%
- D) Above 60.5%
- 314.64 mm Hg 3. 189.24 kJ
- 0.0257 atm, 4.278 atm 5. i) 0.58 ii) 0.1687
iii) 2.645 × 10–5
- i) 400, 900; ii) 4/9; iii) 72.15 7. 70.72
- KP = 0.047, KC= 178.57, P=12.546 atm 9. 0.7837
- 6.07 × 10–4 11. 1.72 × 10–4
- 144 ml 13. 0.75
- 31:27 15. = 0.0236 atm
= 0.9764 atm
- Answers to Objective Assignments
LEVEL – I
- B 2. A
- C 4. C
- A 6. A
- D 8. A
- A 10. A
- D 12. C, D
- C 14. B
LEVEL – II
- B 2. A
- B 4. B
- C 6. A
- B 8. C
- C 10. C
- D 12. C
- D 14. B