CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. After five half-life periods for a first order reaction, what fraction of reactant remains?
Answer:
\( \frac{1}{32} \)

Teacher's Note:
1. Fraction left after n half-lives = (1/2)ⁿ; here (1/2)⁵ = 1/32.

Question 2. What is the effect of adding catalyst on the free energy of a reaction?
Answer:
No change in ΔG.

Teacher's Note:
1. A catalyst cannot make a non-spontaneous reaction spontaneous; it only speeds up a reaction that is already feasible.

Question 3. What value of k is predicted for the rate constant by Arrhenius equation if T → ∞? Is this value physically reasonable?
Answer:
From the equation k = Ae^−E_a/RT, if T → ∞, E_a/RT → 0 and k → A, as if E_a = 0. This is not feasible.

Teacher's Note:
1. The answer k = A means every collision would be effective, which cannot happen at any real temperature.

Question 4. Determine the order of reaction:
Step 1. 2NO + H₂ → N₂ + H₂O₂ (Slow)
Step 2. H₂O₂ + H₂ → 2H₂O (Fast)
Answer:
Rate = k[NO]²[H₂]
Order = 2 + 1 = 3

Teacher's Note:
1. The rate law is written from the slow (rate determining) step only.

Question 5. What is the order of reaction whose rate constant has the same units as the rate of reaction?
Answer:
Zero order

Teacher's Note:
1. For zero order, rate = k[R]⁰ = k, so both have the unit mol L⁻¹ s⁻¹.

Question 6. Why are reactions of higher order less in number?
Answer:
A reaction takes place due to collision of molecules. The chances for a large number of molecules or ions to collide simultaneously are less. Hence, the reactions of higher order are less.

Teacher's Note:
1. Reactions of order more than three are very rare for exactly this reason.

Question 7. What will be the effect of temperature on rate constant?
Answer:
Rate constant of a reaction is nearly doubled with rise in temperature by 10°.

Teacher's Note:
1. The exact dependence is given by the Arrhenius equation, k = Ae^−E_a/RT; "doubling per 10°" is only a rough rule.

Question 8. Why can’t molecularity of any reaction be equal to zero?
Answer:
Molecularity of a reaction means the number of molecules of the reactants taking part in an elementary reaction. Since at least one molecule must be present, the molecularity will be at least one.

Teacher's Note:
1. Order can be zero, but molecularity cannot – a common 1-mark distinction.

Question 9. The rate constant of a reaction is 3 × 10² min⁻¹. What is its order of reaction? (On the basis of units of rate constant)
Answer:
First order reaction.

Teacher's Note:
1. A rate constant with unit time⁻¹ only (no concentration term) always means first order.

Question 10. What is meant by an elementary reaction?
Answer:
A reaction which takes place in one step is called an elementary reaction. For example: H₂ + I₂ → 2HI.

Teacher's Note:
1. A complex reaction is a series of elementary steps; its slowest step decides the rate.

Question 11. Give one example of a reaction where order and molecularity are equal.
Answer:
2HI → H₂ + I₂ (Order = Molecularity = 2)

Teacher's Note:
1. Order and molecularity are equal for elementary reactions.

Question 12. For a reaction R → P, the rate becomes 2 times when the concentration of the reactant is increased 4 times. What is the order of reaction?
Answer:
r = k(a)ⁿ ⇒ 2r = k(4a)ⁿ ⇒ 2 = 4ⁿ ⇒ n = 0.5

Teacher's Note:
1. Divide the two rate expressions so that k and a cancel; then solve 2 = 4ⁿ.

Short Answer Questions (2 Marks)

Question 13. After 24 hrs, only 0.125 gm out of the initial quantity of 1 gm of a radioactive isotope remains behind. What is its half life period? [CBSE Sample Paper 2017]
Answer:
Here, [R]₀ = 1 g, [R] = 0.125 g, t = 24 h
k = \( \frac{2.303}{t} \) log \( \frac{[R]_0}{[R]} \)
k = \( \frac{2.303}{24} \) log \( \frac{1}{0.125} \) ⇒ k = \( \frac{2.303}{24} \) log 8
k = \( \frac{2.303}{24} \) × 0.9031 ⇒ k = 0.0866 h⁻¹
t_1/2 = \( \frac{0.693}{k} \) ⇒ t_1/2 = \( \frac{0.693}{0.0866\ h^{-1}} \) or t_1/2 = 8 h

Teacher's Note:
1. Quick check: 1 → 0.5 → 0.25 → 0.125 g is three half-lives, so t_1/2 = 24/3 = 8 h.

Question 14. Show that in a first order reaction, time required for completion of 99.9% is 10 times that of half-life (t_1/2) of the reaction. [CBSE (F) 2016]
Answer:
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)
\( \frac{t_{99.9\%}}{t_{50\%}} = \frac{\frac{2.303}{k}\log\frac{100}{100-99.9}}{\frac{2.303}{k}\log\frac{100}{100-50}} = \frac{\log\frac{100}{0.1}}{\log\frac{100}{50}} = \frac{\log 10^3}{\log 2} = \frac{3\log 10}{0.3010} \)
\( \frac{t_{99.9\%}}{t_{50\%}} = \frac{30}{3.01} \simeq 10 \)
t_99.9% = 10t_50%

Teacher's Note:
1. Taking the ratio cancels k, so the result holds for every first order reaction.

Question 15. Rate constant k for first order reaction has been found to be 2.54 × 10⁻³ s⁻¹. Calculate its three-fourth life. [CBSE Sample Paper 2013]
Answer:
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)   …(i)
k = 2.54 × 10⁻³ s⁻¹; [R] = \( \frac{[R]_0}{4} \)
Substituting these values in equation (i), we get
t_3/4 = \( \frac{2.303}{2.54\times10^{-3}} \) log \( \frac{[R]_0}{[R]_0/4} \) = 0.9066 × 10³ log 4
t_3/4 = 0.9066 × 10³ × 0.6021 s
= 5.46 × 10² s

Teacher's Note:
1. Three-fourth life means 3/4 of the reactant has reacted, so only 1/4 is left; it equals two half-lives for a first order reaction.

Question 16. Answer the following questions on the basis of the given plot of potential energy vs reaction coordinate:
(i) What is the threshold energy for the reaction?
(ii) What is the activation energy for forward reaction?
(iii) What is the activation energy for backward reaction?
(iv) What is enthalpy change for the forward reaction?

Answer:
(i) Threshold energy for the reaction = 300 kJ mol⁻¹
(ii) Activation energy for the forward reaction = 300 − 150 = 150 kJ mol⁻¹
(iii) Activation energy for the backward reaction = 300 − 100 = 200 kJ mol⁻¹
(iv) Enthalpy change for the forward reaction Δ_rH = 100 − 150 = −50 kJ mol⁻¹

Teacher's Note:
1. Check: E_a(forward) − E_a(backward) = 150 − 200 = −50 kJ mol⁻¹ = ΔH, which confirms the answers.

Question 17. A graph between ln k and \( \frac{1}{T} \) for a reaction is given. Here k is rate constant and T is temperature in kelvin. If OA = a and OB = b, answer the following:
(i) What is the activation energy (E_a) of the reaction?
(ii) What is the frequency factor (A) for the reaction? [HOTS]

Answer:
According to Arrhenius equation, ln k = −\( \frac{E_a}{RT} \) + ln A
(i) Slope = −\( \frac{OB}{OA} \) = −\( \frac{b}{a} \) = −\( \frac{E_a}{R} \) or E_a = \( \frac{b}{a} \) R
(ii) Intercept on y-axis = OB = b = ln A or A = e^b

Teacher's Note:
1. If the graph were log k vs 1/T instead, E_a would be 2.303R(b/a) and A would be 10^b.

Question 18. With the help of a diagram, explain the physical significance of energy of activation (E_a) in chemical reactions. [CBSE Comptt. OD 2013]
Answer:
The excess energy which must be supplied to the reactants to undergo chemical reactions is called activation energy E_a. It is equal to the difference between the threshold energy needed for the reaction and the average energy of all the reacting molecules.
Activation energy = Threshold energy − Average kinetic energy of the reacting molecules

Teacher's Note:
1. Label both axes and mark E_a clearly from the reactant level to the peak – marks are lost for an unlabelled sketch.
2. Physical significance: the higher the activation energy, the fewer the molecules that can react, and the slower the reaction.

Question 19. The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature.
[R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.6021] [CBSE OD 2013]
Answer:
Given: if rate at 293 K is R, then at 313 K rate becomes 4R.
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \) \( \left[\frac{T_2 - T_1}{T_1 T_2}\right] \)
⇒ log \( \frac{4R}{R} \) = \( \frac{E_a}{2.303\times8.314} \) \( \left[\frac{313 - 293}{293\times313}\right] \)
⇒ log 4 = \( \frac{E_a}{19.1471} \) \( \left[\frac{20}{91709}\right] \)
⇒ 0.6021 = \( \frac{E_a}{19.1471} \) \( \left[\frac{20}{91709}\right] \)
⇒ E_a = \( \frac{0.6021\times19.1471\times91709}{20} \)
⇒ E_a = 52863.2 J mol⁻¹ = 52.863 kJ mol⁻¹

Teacher's Note:
1. The ratio of rates equals the ratio of rate constants (same concentrations), so k₂/k₁ = 4.
2. Keep T in kelvin and R in J K⁻¹ mol⁻¹; E_a then comes out in J mol⁻¹.

Question 20. The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate activation energy.
Answer:
K₂/K₁ = 4
T₁ = 293 K, T₂ = 313 K
log \( \frac{K_2}{K_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{1}{T_1} - \frac{1}{T_2} \right] \)
Thus, on calculating and substituting values, we get:
E_a = 52.86 kJ mol⁻¹

Teacher's Note:
1. Here log 4 = 0.602 and (1/T₁ − 1/T₂) = 20/(293 × 313); keep R = 8.314 J K⁻¹ mol⁻¹ and convert J to kJ at the end.

Question 21. If the decomposition of nitrogen oxide as
2N₂O₅ → 4NO₂ + O₂
follows a first order kinetics.
(a) Calculate the rate constant for a 0.05 M solution if the instantaneous rate is 1.5 × 10⁻⁶ mol L⁻¹ s⁻¹.
(b) What concentration of N₂O₅ would give a rate of 2.45 × 10⁻⁵ mol L⁻¹ s⁻¹?
Answer:
(a) Rate = K[N₂O₅]
K = \( \frac{Rate}{[N_2O_5]} \) = \( \frac{1.5 \times 10^{-6}}{0.05} \)
K = 3.0 × 10⁻⁵ s⁻¹
(b) Rate = 2.45 × 10⁻⁵ mol L⁻¹ s⁻¹
[N₂O₅] = \( \frac{Rate}{K} \) = \( \frac{2.45 \times 10^{-5}}{3.0 \times 10^{-5}} \) = 0.82 M

Teacher's Note:
1. For a first order reaction, rate is directly proportional to concentration, so the rate constant found in (a) is reused in (b).

Teacher's Note:
1. Molecularity applies only to elementary reactions; for a complex reaction, it is defined for each step separately.

Question 23. Consider the decomposition reaction:
2H₂O₂ \( \xrightarrow{OH^-/I^-} \) 2H₂O + O₂
This reaction takes place in two steps as given below:
Step 1. H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
Step 2. H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast)
(a) Determine rate law expression.
(b) Determine the order of reaction.
Answer:
(a) Rate = K[H₂O₂][I⁻] because the first (slow) step is the rate determining step.
(b) Order = 1 + 1 = 2

Teacher's Note:
1. The slowest step always decides the rate law; species formed and used up in between (like IO⁻) do not appear in it.

Question 24. The decomposition of hydrocarbon follows the equation K = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T}. Calculate E_a.
Answer:
K = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T}
Comparing the equation with Arrhenius equation,
K = Ae^−E_a/RT
−\( \frac{E_a}{R} \) = −28000 K
E_a = 28000 × 8.314
= 232792 J mol⁻¹

Teacher's Note:
1. Match term by term: the number multiplying 1/T in the exponent is E_a/R, so multiply it by R to get E_a.

Question 25. For a first order reaction time taken for half of the reaction to complete is t₁ and ¾ of the reaction to complete is t₂. How are t₁ and t₂ related?
Answer:
t₂ = 2t₁ because for 3/4th of the reaction to complete time required is equal to two half lives.

Teacher's Note:
1. After one half-life 1/2 is left, after two half-lives 1/4 is left – i.e. 3/4 has reacted.

Question 26. For the reaction A → B, the graph of concentration of A against time is given below.
(a) What is the order of the reaction?
(b) What is the slope of the curve?

Answer:
(a) Zero order reaction.
(b) [R] = [R]₀ − kt   ∴ Slope = −k

Teacher's Note:
1. Compare the graph with y = c + mx: here y = [R], x = t, intercept = [R]₀ and slope = −k.

Question 27. Derive an expression to calculate time required for completion of zero order reaction.
Answer:
For a zero order reaction,
[R] = [R]₀ − kt
For completion of the reaction [R] = 0
∴ kt = [R]₀
Or t = \( \frac{[R]_0}{k} \)

Teacher's Note:
1. Only a zero order reaction can actually reach completion in finite time; a first order reaction never reaches [R] = 0 exactly.

Short Answer Questions (3 Marks)

Teacher's Note:
1. The volume of KMnO₄ used is proportional to the H₂O₂ left, so volumes can be used directly in place of concentrations.

Question 29. Define half-life of a reaction. Write the expression of half-life for
(i) zero order reaction and (ii) first order reaction. [CBSE (F) 2014]
Answer:
The half life (t_1/2) of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration.
(i) t_1/2 for a zero order reaction = \( \frac{[R]_0}{2k} \) where [R]₀ = initial concentration, k = rate constant
(ii) t_1/2 for a first order reaction = \( \frac{0.693}{k} \)

Teacher's Note:
1. Zero order half-life depends on [R]₀; first order half-life does not – a favourite one-line exam contrast.

Question 30. A first order reaction is 50% complete in 25 minutes. Calculate the time for 80% completion of the reaction. [CBSE 2019 (56/3/2)]
Answer:
t_1/2 = \( \frac{0.693}{k} \) ⇒ k = \( \frac{0.693}{t_{1/2}} \)
k = \( \frac{0.693}{25\ min} \) = 2.772 × 10⁻² min⁻¹
[R] = [R]₀ − 80% of [R]₀ or [R] = [R]₀ − \( \frac{80\times[R]_0}{100} \) or [R] = 0.2[R]₀
Substituting the value of k and [R] in the expression
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \), we get
t = \( \frac{2.303}{2.772\times10^{-2}} \) × log \( \frac{[R]_0}{0.2[R]_0} \)
t = \( \frac{2.303}{2.772\times10^{-2}} \) log 5
t = \( \frac{2.303\times0.699}{2.772\times10^{-2}} \) or t = 58.07 min

Teacher's Note:
1. 80% completion means 20% remains, so [R]₀/[R] = 5, not 0.8.

Question 31. ²³⁸₉₂U changes to ²⁰⁶₈₂Pb by successive radioactive decay. A sample of uranium ore was analysed and found to contain 1.0 g of ²³⁸U and 0.1 g of ²⁰⁶Pb had accumulated due to decay of ²³⁸U, find out the age of the ore. (Half-life of ²³⁸U = 4.5 × 10⁹ years) [HOTS]
Answer:
[A]₀ = Initial amount of ²³⁸U = amount of ²³⁸U left at time t + amount of ²³⁸U decayed
[A]₀ = 1.0 + amount of ²³⁸U decayed
Now, amount of ²³⁸U decayed = \( \frac{0.1\times238}{206} \) g = 0.1155 g
∴ [A]₀ = 1.0 g + 0.1155 g = 1.1155 g
Determination of k: k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{4.5\times10^9} \) = 0.154 × 10⁻⁹ year⁻¹
Determination of time: t = \( \frac{2.303}{k} \) log \( \frac{[A]_0}{[A]} \)
Substituting the values of [A]₀ = 1.1155 g and k = 0.154 × 10⁻⁹ year⁻¹
t = \( \frac{2.303}{0.154\times10^{-9}} \) log \( \frac{1.1155}{1} \)
= 0.7099 × 10⁹ year
= 7.099 × 10⁸ year

Teacher's Note:
1. Convert the mass of Pb formed into the mass of U decayed using the mole ratio (1 U atom gives 1 Pb atom), i.e., multiply by 238/206.

Teacher's Note:
1. Here 2 mol gas give 3 mol gas, so p_A = 1.5 − 2p_t – not 2p₀ − p_t, which is only for A → B + C.

Question 33. The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.2 − \( \frac{1.0\times10^4}{T} \) K
Calculate E_a for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K⁻¹ mol⁻¹) [CBSE Delhi 2016]
Answer:
Comparing the equation, log k = 14.2 − \( \frac{1.0\times10^4}{T} \) K with the equation, log k = log A − \( \frac{E_a}{2.303RT} \), we get
\( \frac{E_a}{2.303RT} \) = \( \frac{1.0\times10^4\ K}{T} \) or E_a = 1.0 × 10⁴ K × 2.303 × R
E_a = 1.0 × 10⁴ K × 2.303 × 8.314 J K⁻¹ mol⁻¹
= 19.1471 × 10⁴ J mol⁻¹ = 191.47 kJ mol⁻¹
For a first order reaction, t_1/2 = \( \frac{0.693}{k} \) or k = \( \frac{0.693}{t_{1/2}} \)
∴ k = \( \frac{0.693}{200\ min} \) = 3.465 × 10⁻³ min⁻¹

Teacher's Note:
1. Match the given equation term by term with log k = log A − E_a/2.303RT; the coefficient of 1/T equals E_a/2.303R.

Question 34. The rate constant of a first order reaction increases from 2 × 10⁻² to 4 × 10⁻² when the temperature changes from 300 K to 310 K. Calculate the energy of activation (E_a).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) [CBSE Patna 2015]
Answer:
Substituting k₁ = 2 × 10⁻², k₂ = 4 × 10⁻², T₁ = 300 K, T₂ = 310 K, R = 8.314 J K⁻¹ mol⁻¹ in the expression
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303\times R} \)\( \left(\frac{T_2 - T_1}{T_1 T_2}\right) \), we get
log \( \frac{4\times10^{-2}}{2\times10^{-2}} \) = \( \frac{E_a}{2.303\times8.314} \)\( \left(\frac{310 - 300}{300\times310}\right) \)
log 2 = \( \frac{E_a}{19.147} \) × \( \frac{10}{300\times310} \)
E_a = 0.3010 × 19.147 × 300 × 31 = 53598 J mol⁻¹
E_a = 53.598 kJ mol⁻¹

Teacher's Note:
1. Keep R in J K⁻¹ mol⁻¹ so E_a comes out in J mol⁻¹; convert to kJ only at the end.

Question 35. 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]
Answer:
Given T₁ = 280 K, k₁ = 1.6 × 10⁶ s⁻¹, k₂ = ?, E_a = 0, T₂ = 300 K.
By Arrhenius equation,
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \)\( \left[\frac{T_2 - T_1}{T_1 T_2}\right] \)
As, E_a = 0
∴ log \( \frac{k_2}{k_1} \) = 0
or \( \frac{k_2}{k_1} \) = 1 or k₂ = k₁
Thus, the rate constant at 300 K is 1.6 × 10⁶ s⁻¹.

Teacher's Note:
1. When E_a = 0, k = A at every temperature, so no calculation with R is actually needed.

Question 36. A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is again 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.
[R = 8.314 J K⁻¹ mol⁻¹; log 4 = 0.602] [HOTS]
Answer:
For a first order reaction, k = \( \frac{0.693}{t_{1/2}} \)
T₁ = 300 K, k₁ = \( \frac{0.693}{20} \) = 3.465 × 10⁻² min⁻¹
T₂ = 350 K, k₂ = \( \frac{0.693}{5} \) = 1.386 × 10⁻¹ min⁻¹
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \)\( \left[\frac{T_2 - T_1}{T_1 T_2}\right] \)
log \( \frac{1.386\times10^{-1}}{3.465\times10^{-2}} \) = \( \frac{E_a}{2.303\times8.314} \)\( \left(\frac{350 - 300}{350\times300}\right) \)
log 4 = \( \frac{E_a}{2.303\times8.314} \)\( \left(\frac{50}{350\times300}\right) \)
0.602 = \( \frac{E_a}{19.147} \)\( \left(\frac{50}{350\times300}\right) \)
E_a = \( \frac{0.602\times19.147\times350\times300}{50} \)
= 24205.63 J mol⁻¹
= 24.206 kJ mol⁻¹

Teacher's Note:
1. Since k ∝ 1/t_1/2, k₂/k₁ = 20/5 = 4 directly; there is no need to compute each k separately.

Long Answer Questions (5 Marks)

Question 37. Consider the reaction R \( \xrightarrow{k} \) P. The change in concentration of R with time is shown in the following plot:
(i) Predict the order of the reaction.
(ii) Derive the expression for the time required for the completion of the reaction.
(iii) What does the slope of the above line indicate? [CBSE 2019 (56/4/1)]

Answer:
(i) The reaction R → P is a zero order reaction.
(ii) For the reaction R \( \xrightarrow{k} \) P
rate = \( \frac{-d[R]}{dt} \) = k
d[R] = −k dt
Integrating both sides,
[R] = −kt + C, where C = constant of integration   …(i)
At t = 0, [R] = [R]₀
Substituting this in equation (i), C = [R]₀
Substituting the value of C in equation (i)
[R] = −kt + [R]₀   …(ii)
kt = [R]₀ − [R]
⇒ t = \( \frac{[R]_0 - [R]}{k} \)
On completion of reactions, [R] = 0
∴ t = \( \frac{[R]_0}{k} \)
(iii) From equation (ii), we have slope of curve
Slope = \( \frac{d[R]}{dt} \) = −k

Teacher's Note:
1. A zero order reaction is the only kind that actually reaches [R] = 0 in a finite time; first order reactions approach it only gradually.

Question 38. For a certain chemical reaction variation in the concentration ln [R] vs. time plot is given alongside. For this reaction
(i) what is the order of the reaction?
(ii) what are the units of rate constant k?
(iii) give the relationship between k and t_1/2 (half-life period).
(iv) what is the slope of the curve?
(v) draw the plot log [R]₀/[R] vs time t(s). [HOTS]

Answer:
(i) First order
(ii) time⁻¹ (s⁻¹)
(iii) k = \( \frac{0.693}{t_{1/2}} \)
(iv) slope = −k (rate constant)
(v) The plot of log [R]₀/[R] vs time is a straight line passing through the origin:

Teacher's Note:
1. For first order, ln[R] vs t has slope −k, while log([R]₀/[R]) vs t rises from the origin with slope k/2.303.

Teacher's Note:
1. You can check the first order assumption by computing k from another pair (e.g., 40 min): it should give the same value.

Question 40. (i) Write the rate law for a first order reaction. Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant.
(ii) The activation energy of a reaction is 75.2 kJ mol⁻¹ in the absence of a catalyst and it lowers to 50.14 kJ mol⁻¹ with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C?
Answer:
(i) Consider the first order reaction, R → P
For this reaction, rate law which relates the rate of reaction to the concentration of reactants can be given as
Rate = \( \frac{-d[R]}{dt} \) = k[R]
For a first order reaction,
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \), where [R]₀ = initial concentration, [R] = concentration at time t.
At t_1/2, [R] = [R]₀/2
So, the above equation becomes
t_1/2 = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]_0/2} \)
t_1/2 = \( \frac{2.303}{k} \) log 2 or t_1/2 = \( \frac{2.303}{k} \) × 0.3010
t_1/2 = \( \frac{0.693}{k} \)
This shows that half life of a first order reaction is independent of the initial concentration of the reactant.
(ii) According to Arrhenius equation,
log k = log A − \( \frac{E_a}{2.303RT} \)
For uncatalysed reaction, log k₁ = log A − \( \frac{E_{a_1}}{2.303RT} \)   …(i)
For catalysed reaction, log k₂ = log A − \( \frac{E_{a_2}}{2.303RT} \)   …(ii)
A is equal for both reactions.
Subtracting equation (i) from equation (ii),
log \( \frac{k_2}{k_1} \) = \( \frac{E_{a_1} - E_{a_2}}{2.303RT} \) = \( \frac{(75.2 - 50.14)\ kJ\ mol^{-1}}{2.303\times8.314\ J K^{-1} mol^{-1}\times298\ K} \) = 4.39
\( \frac{k_2}{k_1} \) = antilog (4.39) = 2.45 × 10⁴
Rate of reaction increases by 2.45 × 10⁴ times.

Teacher's Note:
1. Convert the E_a difference to J mol⁻¹ (25060 J mol⁻¹) before dividing by R in J K⁻¹ mol⁻¹, and use T in kelvin (298 K).