Chemical Kinetics
CHEMICAL KINETICS
Solution to Subjective Problems
LEVEL – I
1. The one existing for long time will cause serious effect.
- Use concepts of complex reactions.
- Arrhenius equation.
- Calculate‘t’.
- Use first order kinetics concepts.
- Calculate‘t’.
- Use pressure concepts in first order kinetics.
LEVEL – II
1. Calculate no. of He particles and proceed.
- Write the expression and then calculate (a – x).
- Compare both the isotopes’ rate and proceed.
- Activity α vol. of blood.
- Find a relation between the total rotation and individual rotations.
LEVEL – III
4. Calculate initial nuclei of and proceed.
- Calculate initial moles of Cl238 and proceed.
- Find the relation between total rotation and individual rotations from the graph & find log A.
- Use concepts of complex reactions.
- Each Po nucleus gives one mole of He.
- Use concepts of complex reactions.
- Volume α activity.
Solution to Subjective Problems
LEVEL – I
- 2.303 log =
2.303 log =
K2 = 9.38 × 10–3 sec–1
For I131 time required to fall its activity to 1%
t = = 53.17 days
For Sr90, time required to fall its activity to 1% be, = = 59.8 days
Thus 90Sr which takes longer time to fall its activity to 1% is to be serious
- Time of maximum activity tmax =
λd ⇒ disintegration constant of daughter element
λp ⇒ disintegration constant of parent elements
λd = = 0.01145 min–1
λp = = 0.00109 min–1
tmax =
tmax = 227 min
- Since the plot of logarithms of partial pressure versus time is linear, hence the reaction is of first order
The equation for first reaction is
or t = = –
This is a straight line equation (y = mx + c) hence its slope is equal to – 2.303 / k from which the specific rate reaction can be calculated. The intercept of this equation log a: hence the value of ‘a’ can also be calculated by knowing the value of k.
- i) A nB
The initial concentration of A = 0.6 M
The equilibrium concentration of A = 0.3 M
Hence concentration of A lost = 0.3 M
If x is the concentration of A lost, then more is the concentration, B obtained
Hence, B = nx = n × 0.3 M
B is 0.6 M (from the graph)
0.6 M = n × 0.3 M
∴ n = 2
- ii) K = = 1.2
iii) When t = 0, [A] = 0.6 M
t = 1 hr, [A] = 0.5 M
∴ Rate of conversion = 0.5 – 0.6 = – 0.1 M hr–1
- Given t1/2 = 28.1 years N0 = 10-6 g, t = 20 yrs
t = =
N = 6.1 x 10-7g
- t = = = 491.16 year
∴ the painting was done in the year (1999-491) = 1508 which was within the life span of Raphael (1483-1520)
∴ It was not a forgery
- a) Rate =
- b) Rate =
Change in molarity =
- Rate = k [NO]x [Cl2]y where x = order w.r.t. NO
i.e. R = k [NO]x [Cl2]y y = Order w.r.t. Cl2
Case (1) 8R = k [2NO]x [2Cl2]y
Case (2) 2R = k [NO] x [2Cl2]y
Dividing the equation we have
4 = 2x ∴ x = 2
Similarly 2R = k [NO] x [2Cl2]y
R = k [NO] x [Cl2]y
∴ 2 = 2y ⇒ y =1
∴Overall order = x + y = 2 + 1 = 3
- Pb present =
U present =
No = U present + U decayed
= 0.1 + 0.1= 0.2
and N = 0.1
∴ t = =
t = 4.5 × 109 years
- i) k = =
= 6.2 × 10–3 s–1
- ii) Again t = 100 s, Pt = ?
6.2 × 10–3 = log or Pt = 0.033 atm
LEVEL- II
- No. of α-particles (or) He formed = 2.24 × 1013 min–1
∴ No. of He particles formed in 420 days = 2.24 × 1013 × 420 × 1440 = 1.355 × 1019
Also at 27°C and 750 mm; He = 0.5 mL
Using PV = nRT
= n × 0.0821 × 300 ⇒ n = 2.0 × 10–5 moles
2.0 × 10–5 moles of He = 1.355 × 1019 particles of He
⇒ 1 mole of He = = 6.775 × 1023 particles
∴ Avogadro’s number = 6.775 × 1023 particles/mol
- Let the order of the reaction is ‘n’
∴ = ; =
or, = logK + nlogC1; log = log K + n logC2
∴ n =
n = = = 1.128
So order of the reaction is 1.128
3.
Rearranging and taking indefinite integration within the limit,
When t → 0, then y →0 & t →, t, the y → y
Then this equation
⇒
⇒
When t → t1/2, then y → C0/2
∴
⇒ proved
- [N2O5] =
K = 1.68 × 10–2
t = 1 min = 60 sec
[A]t = ?
K =
1.62 × 10–2 =
on solving, n = 0.185M
= 0.9215 mol.
decomposed = (2.5 – 0.9215) mol = 1.5875 mol
decomposed = mol = 0.7938 mol
- Now,
At the time of earth formation,
For X193, λ1 =
For X192, λ2 =
If ‘t’ is the age of earth
∴ λ1 – λ2 = = log
or, =
∴ t = 32.56 × 107 years
- Let volume of the blood in the animal = V mL
total activity before mixing = 1.8 × 106 cps
total activity after mixing in V mL = × V mL
= 0.6 × 104 V mL
Thus 0.6 × 104 V = 1.8 × 106 (assuming activity remains constant)
V = = 300 mL
- We know,
log
Ea = energy of activation
Given: Ea = 0
∴ = 0
or log = 0
or = 1
Hence K300 = K280
Or K300 = 1.6 × 106 s–1
- K =
Total moles of sucrose =
Sucrose ⎯→ Glucose + Fructose
t = 0 0.38r10 0 0
t = 10 min 0.38 r10 –αr10 αr20 –αr30
t = ∞ 0.38 r20 –0.38 r30
Total rotation at t∞ = –4.7120
K = =
K = 0.0113 min–1 , t1/2 = = 61.16 min
- 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
- According to kinetic equation of first order
K =
a = 5.0 mg/ml (a-x) = 4.2 mg/ml, t = 20 months
K = log =
K = 8.728 × 10–3 months–1
Suppose t months are sufficient for its 30% decomposition, therefore
8.728 × 10–3 =
t = log = 263.86 × 0.1549 = 40.87 months
≈ 41 months
Half life period = = 79.39 months
≈ 80 months
LEVEL – III
- At t = 20 min, [R] = [P] i.e. half of the concentration of R has got consumed during this time interval. So, t1/2 = 20 min
K = = 0.03465 min–1
i)
= 2.303 log (3.8 × 10–18)
Ea = 2.303 × 8.314 × 300 × 17.420 = 100.0 kJ mol–1
- ii) k =
when T →∞, k → A. Thus A is the maximum rate constant of the reaction i.e. the rate constant at temperature approaching infinity.
Putting the value of k and we have
0.03465 = A × 3.8 × 10–18
A = = 9.11 × 1015 min–1
iii)
= 0.561
= antilog (0.561) = 3.65
Thus, the rate will increase 3.65 fold
Alternatively, rate constant at 37°C may also be calculated using the equation
K =
And rate constant at 27°C being known, the ratio may thus be found out.
iv)
log 104=
Ea – EaC= 22.96
EaC = 100 – 22.96 = 77.04 kJ mol–1
(Kcat = Rate constant of catalysed reaction, k = rate constant of uncatalysed reaction, EaC = energy of activation of the catalysed reaction and Ea = energy of activation of the uncatalysed reaction).
- 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
- Let T1 = 300 K and T2 = 310 k
we know that,
log
For (i) reaction
log
or log 2 = —— (i)
For (ii) reaction
log
Since =
Here k2 = 2
k2 = 0.0462 min–1 and Ea = EA
Therefore, log
From equation (i)
log = log2 = log or =
or k1 = = 0.327 min–1
- 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
- 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 years
- A + H2O ⎯⎯→ 2B + C
a excess 0 0
(a-x) const. 2x 2x
0 const. 2a a
After completion of reaction 50 × 29 + (–80) × a = 10, ∴ a = 0.5
After 40 minute. 60(a-x) + 50 × 2x – 80 × x = 26 ⇒ x = 0.1
K = log 5/4
∴ K = 5.57 × 10–3 min–1
∴t1/2 = = = 124.4 min
From Arrhenius equation, K = Ae–Ea/RT
Log K = + log A; Intercept = log A
When T = 27 + 273 = 300K , log K = log (5.57 × 10–3) = –2.254
Putting the value in above equation
– 2.254 = × + 15.046
∴ Ea = 99.37 kJ /mol
In order that t1/2 of reaction may decrease from 124.4 min, rate constant will have to increase 4 time.
Log (4 × 5.57 × 10–3) = × + 15.046
∴ – 1.652 – 15.046 = ×
∴ T = 310.80 K
- Let [A]o = initial concentration of A and
[A], [B], [C] = concentration of A B and C at any time t
k1 + k2 = k =
Now [B] = and [A] = 1
∴calculate & get the answer
- t1/2 = 138.4 days t = 69.2 days
No. of halves
Amount of Po left after ½ halves =
Amount of Po used = 1-0.707 = 0.293g
84Po210 → 82Po206 + 2He4
210g Po on decay will produce 4g the
0.293 Po will produce = = 5.581 × 10-3g He
Volume of He the at STP = = 31.25 ml = 31.25 cm3
- Let initial amount = 20
∴amount left = 20 x = 15
Calculate k at 25°C from the equation k = with t = 20 mins.
Now from Arrhenius equation
Calculate & from this k40 can be calculated, as k25 is already known
Let m = amount left in 30% solution
∴k40 =
From this calculate m
& % decomposition =
10.
Initial amount mo = 1g atom
Let us assure [A]o to be the initial concentration of [A] and [B] to be the concentration of Pb214.
This is an eg. of a consecutive reaction
For the number of nuclei to be maximum, should be equated to zero.
or
or
Now k1 =
k2 =
∴ t = 4.12 mins
- Infected sample activity = 1 × 10–7 Ci × = 3.7 × 103 d/sec.
In sample withdrawn : = 0.33 dis/sec
The ratio of total activity to activity of sample withdrawn is equal to the ratio of volumes, where V is original volume of blood.
Thus, =
∴ V = 1.1 × 103 ml = 1.1L
- N0 = 1g atom t1/2 Ra = 1600 years t = 800 y
t =
800=
N = 0.707g atom
∴ Amount of Ra decayed = 1-0.707
Moles of Ra decayed = 0.293g atom
& Moles of He formed = 0.293
as 80Ra226 → 86Ra222 + 2He4
∴partial pressure of Helium
P = =
P = 1.44 atm
- Total activity of a sample is the sum of the individual activities of all its components.
Let the total mass of the sample be 1 gm and the mass of 239Pu be x gm
∴ × 6.023 ×1023 ×
= 6 × 109 × 365 × 24 ×60 ×60
On calculating, x = 0.3896
∴ 239Pu = 38.96% and 240Pu = 61.04%
- Rate = k [CH3COOC2H5]a[H+]b.
[H+] is constant through out the reaction
k1 = k [H+]b
Hence ,
b = 1
k1 = k [H+]
1.1 × 10–4 = k (10–3)
k = 1.1 × 10–1 dm3mol–1 sec–1
- Let the nuclei of A236 initially be N0
∴ the nuclei of A234 initially will be
Initial activity of A236 = 106 =
∴ Initial activity of A234 =
Now for A236, At = A0 e–λt
∴ At = 106
For A234, At =
Since the activities are to be equal
106
∴ t = 180 min or 3 hrs.
For the number of nuclei to be same,
For A236, Nt = N0
For A234, Nt =
∴ t = 120 min or 2 hrs.
Solution to Objective Problems
LEVEL – II
- Order = + 3/2 – 1 = 1/2
- A(g) + 2B(g) ⎯→ C(g) + D(g)
t = 0 0.60 0.80 0 0
t = t (0.6–0.2) (0.8 – 2×0.2) 0.2 0.2
(Rate)i = k [A] [B]2 = k[0.6] [0.8]2
(Rate)t = k[0.4] [0.4]2
= = 1/6
- For a 3rd order reaction
Rate = k [A]3
=
is k = lit2 mol–2 min–1
- t½ = 140 days
Given 1120 days ∴ x = = 8
∴ After 1120 day, = of it will remain
- Equation of a first order reaction
k =
A plot of log a Vs t for a 1st order reaction is
log [A]t = log [A]o –
y = c + m x
c = log [A]o
m =
Which means a negative slope and non zero intercept
- k = k[2]–1/2 [2]3/2 = k[2]3/2–1/2 = k = [2]1 = 2k
- Average energy of reactants + activation energy = threshold energy
- For a first order reaction
As we know that
k = log
Where No = original amount
N = amount left after time t
When t becomes t3/4 or t0.75
3/4th of the original amount converts into the product.
i.e. left amount N = No – ¾ No = ¼ No
or, k = log
or, k – = 10-2 sec–1
- No. of half lives needed for change in the concentration of reactant from 0.08 to 0.02m is 3
∴time needed = 10 × 3 = 30 min.
- Let original amount, NO = 100
Since 75% completed, so, final amount N = 100-75 = 25
As we know
, where n = no. of half lives
or,
or, 4 = 2n or, 22 or, 22 = 2n
∴n= 2
Since total time = n x t1/2
32 minutes = 2 x t1/2
∴t1/2 = 16 minutes
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