This MCQ module is based on: NCERT Exercises and Solutions: Chemical Kinetics
This assessment will be based on: NCERT Exercises and Solutions: Chemical Kinetics
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Chemical kinetics is the quantitative study of the rates and mechanisms of chemical reactions. The key learnings of this chapter are summarized below.
\(r = -\dfrac{1}{a}\dfrac{d[A]}{dt} = +\dfrac{1}{c}\dfrac{d[C]}{dt}\). Units: mol L⁻¹ s⁻¹ (or atm s⁻¹ for gases).
Rate = k [A]ˣ [B]ʸ. Powers x, y are determined experimentally; the sum is the order.
Order = experimental (can be 0, fractional, integer); molecularity = number of species in elementary step (positive integer).
[R] = [R]₀ − kt; t₁/₂ = [R]₀/(2k); units of k = mol L⁻¹ s⁻¹.
k = (2.303/t) log([R]₀/[R]); t₁/₂ = 0.693/k; units of k = s⁻¹. Independent of starting concentration.
True 2nd-order reactions appear 1st-order when one reactant is in vast excess (e.g. ester hydrolysis, sucrose inversion).
k = A e^(−Eₐ/RT); log(k₂/k₁) = (Eₐ/2.303R)(T₂−T₁)/(T₁T₂).
Lowers Eₐ, alters rate without changing equilibrium constant or ΔH.
Rate = P Z exp(−Eₐ/RT). Requires sufficient energy AND proper orientation.
| Quantity | Formula |
|---|---|
| Average rate | \(-\Delta[R]/\Delta t\) |
| Instantaneous rate | \(-d[R]/dt\) |
| Rate law (general) | \(\text{Rate} = k[A]^x[B]^y\) |
| Zero order integrated | \([R] = [R]_0 - kt\) |
| Zero order t₁/₂ | \([R]_0/(2k)\) |
| First order integrated | \(k = (2.303/t)\log([R]_0/[R])\) |
| First order t₁/₂ | \(0.693/k\) |
| Arrhenius | \(k = A\,e^{-E_a/RT}\) |
| Two-point Arrhenius | \(\log(k_2/k_1) = (E_a/2.303R)(T_2-T_1)/(T_1 T_2)\) |
| Temperature coefficient | \(\mu = k_{T+10}/k_T \approx 2-3\) |
For the reaction \(R \to P\), the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate using units of time in both minutes and seconds.
\(r_{\text{av}} = -\dfrac{\Delta[R]}{\Delta t} = -\dfrac{0.02 - 0.03}{25}\)
= 0.01/25 = 4 × 10⁻⁴ M min⁻¹ = 6.66 × 10⁻⁶ M s⁻¹.
In a reaction \(2A \to \text{Products}\), the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval.
\(\text{Rate} = -\dfrac{1}{2}\dfrac{\Delta[A]}{\Delta t} = -\dfrac{1}{2}\dfrac{0.4 - 0.5}{10}\)
= 0.05/10 = 0.005 mol L⁻¹ min⁻¹ = 5 × 10⁻³ M min⁻¹.
For a reaction A + B → Products, the rate law is given by Rate = k[A]^(1/2)[B]². What is the order of reaction?
Order = ½ + 2 = 2.5.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?
For second order: Rate = k[X]². If [X] becomes 3[X], new Rate = k(3[X])² = 9k[X]².
Rate becomes 9 times the original.
A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
\(t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{[R]} = \dfrac{2.303}{1.15 \times 10^{-3}}\log\dfrac{5}{3}\)
= 2002.6 × log(1.667) = 2002.6 × 0.2218 = 444.2 s ≈ 7.4 min.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant.
For first order, \(k = 0.693/t_{1/2} = 0.693/60\) = 1.155 × 10⁻² min⁻¹ = 1.925 × 10⁻⁴ s⁻¹.
What will be the effect of temperature on rate constant?
The rate constant increases with temperature. According to Arrhenius equation \(k = A\,e^{-E_a/RT}\); a 10 K rise typically doubles or triples k. Quantitatively: \(\log(k_2/k_1) = (E_a/2.303R)(T_2 - T_1)/(T_1 T_2)\).
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Eₐ.
\(\log 2 = \dfrac{E_a}{2.303 \times 8.314}\dfrac{10}{298 \times 308}\)
\(0.301 = \dfrac{E_a}{19.147} \cdot \dfrac{10}{91784} = \dfrac{E_a}{19.147} \times 1.089 \times 10^{-4}\)
\(E_a = \dfrac{0.301 \times 19.147}{1.089 \times 10^{-4}} = 5.29 \times 10^{4}\) J/mol = 52.9 kJ/mol.
The activation energy for a reaction is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Fraction = \(e^{-E_a/RT} = e^{-209500/(8.314 \times 581)} = e^{-43.36}\)
\(\log(\text{fraction}) = -43.36/2.303 = -18.83\)
Fraction = 1.471 × 10⁻¹⁹.
From the rate expression for the reaction below, determine the order and dimensions of the rate constant:
(i) 3 NO(g) → N₂O(g) Rate = k[NO]²
(ii) H₂O₂(aq) + 3 I⁻(aq) + 2 H⁺ → 2 H₂O(l) + I₃⁻ Rate = k[H₂O₂][I⁻]
(iii) CH₃CHO(g) → CH₄(g) + CO(g) Rate = k[CH₃CHO]^(3/2)
(iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g) Rate = k[C₂H₅Cl]
(i) Order = 2; units = (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹.
(ii) Order = 1+1 = 2; units = L mol⁻¹ s⁻¹.
(iii) Order = 3/2; units = (mol L⁻¹)^(−1/2) s⁻¹ = L^(1/2) mol^(−1/2) s⁻¹.
(iv) Order = 1; units = s⁻¹.
For the reaction 2A + B → A₂B, the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.
Initial rate = 2.0 × 10⁻⁶ × (0.1)(0.2)² = 2.0 × 10⁻⁶ × 0.1 × 0.04 = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹.
When [A] = 0.06, A consumed = 0.04 mol L⁻¹. By stoichiometry 2A:B, B consumed = 0.02; so [B] = 0.20 − 0.02 = 0.18 mol L⁻¹.
Rate = 2.0 × 10⁻⁶ × (0.06)(0.18)² = 2.0 × 10⁻⁶ × 0.06 × 0.0324 = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹.
The decomposition of NH₃ on platinum surface is zero order. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹?
Reaction: \(2NH_3 \to N_2 + 3H_2\). Rate = −(1/2)(d[NH₃]/dt) = d[N₂]/dt = (1/3)(d[H₂]/dt).
Since rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ (zero order):
d[N₂]/dt = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
d[H₂]/dt = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
The decomposition of dimethyl ether leads to formation of CH₄, H₂ and CO and the rate of reaction is given by Rate = k[(CH₃)₂O]^(3/2). The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of partial pressure. What are the units of rate and rate constant?
Rate (in pressure units) = bar s⁻¹ (or atm s⁻¹).
k = Rate / (pressure)^(3/2) = bar s⁻¹ / bar^(3/2) = bar^(−1/2) s⁻¹.
Mention the factors that affect the rate of a chemical reaction.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Rate = k[A]². (i) If [A] → 2[A]: new rate = k(2[A])² = 4 × old rate.
(ii) If [A] → ½[A]: new rate = k(½[A])² = ¼ × old rate (one-fourth).
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
The rate constant increases with temperature; usually doubles for every 10 K rise. Quantitatively given by Arrhenius equation \(k = A\,e^{-E_a/RT}\) where Eₐ is activation energy. Two-temperature form: \(\log(k_2/k_1) = (E_a/2.303R)\bigl[(T_2 - T_1)/(T_1 T_2)\bigr]\).
In a pseudo first order hydrolysis of ester in water the following results were obtained:
| t/s | 0 | 30 | 60 | 90 |
|---|---|---|---|---|
| [Ester]/M | 0.55 | 0.31 | 0.17 | 0.085 |
(i) Calculate average rate between 30 and 60 s.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
(i) Avg rate (30→60s) = −(0.17 − 0.31)/30 = 0.14/30 = 4.67 × 10⁻³ M s⁻¹.
(ii) For first order: \(k = (2.303/t)\log([R]_0/[R])\). Use t = 30 s:
k = (2.303/30) log(0.55/0.31) = 0.0768 × log(1.774) = 0.0768 × 0.249 = 0.0191 s⁻¹ at 30 s.
At 60 s: k = (2.303/60) log(0.55/0.17) = 0.0384 × 0.510 = 0.0196 s⁻¹.
At 90 s: k = (2.303/90) log(0.55/0.085) = 0.0256 × 0.811 = 0.0208 s⁻¹.
Average k ≈ 1.98 × 10⁻² s⁻¹.
A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when concentrations of both A and B are doubled?
(i) Rate = k[A][B]².
(ii) [B] → 3[B]: Rate = k[A](3[B])² = 9 × original.
(iii) [A]→2[A], [B]→2[B]: Rate = k(2[A])(2[B])² = 8k[A][B]² = 8 × original.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations:
| Run | A/M | B/M | r₀/M s⁻¹ |
|---|---|---|---|
| 1 | 0.20 | 0.30 | 5.07 × 10⁻⁵ |
| 2 | 0.20 | 0.10 | 5.07 × 10⁻⁵ |
| 3 | 0.40 | 0.05 | 1.43 × 10⁻⁴ |
Determine the rate law and rate constant.
Comparing runs 1 and 2 ([A] same, [B] tripled, rate same) → order in B = 0.
Comparing runs 2 and 3 ([B] doesn't matter; [A] doubled from 0.20 to 0.40, rate became 1.43e−4/5.07e−5 = 2.82 ≈ 2^(1.5)).
Order in A = 1.5. So Rate = k[A]^(3/2). From run 1: k = 5.07 × 10⁻⁵ / (0.20)^(3/2) = 5.07e−5 / 0.0894 = 5.67 × 10⁻⁴ M^(−1/2) s⁻¹.
The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
\(t = (2.303/k)\log([R]_0/[R]) = (2.303/60)\log 16\)
= 0.03838 × 1.204 = 0.0462 s ≈ 4.62 × 10⁻² s.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 μg of ⁹⁰Sr was absorbed in the bones of a newly born baby, how much of it will remain after 10 years and 60 years?
k = 0.693/28.1 = 0.02466 yr⁻¹.
After 10 yr: \(t = (2.303/k)\log([R]_0/[R])\) → \(\log([R]_0/[R]) = (k\,t)/2.303 = (0.02466 \times 10)/2.303 = 0.1071\) → [R]₀/[R] = 1.279 → [R] = 1/1.279 = 0.782 μg.
After 60 yr: log([R]₀/[R]) = (0.02466 × 60)/2.303 = 0.6425 → [R]₀/[R] = 4.39 → [R] = 0.228 μg.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
For 99%: \(t_{99} = (2.303/k)\log(100/1) = (2.303/k)(2)\).
For 90%: \(t_{90} = (2.303/k)\log(100/10) = (2.303/k)(1)\).
So \(t_{99}/t_{90} = 2\), i.e. \(t_{99} = 2 t_{90}\). Q.E.D.
A first order reaction takes 40 min for 30% decomposition. Calculate t₁/₂.
For 30% decomposition, [R]/[R]₀ = 0.70.
k = (2.303/40) log(1/0.70) = 0.0576 × 0.1549 = 8.92 × 10⁻³ min⁻¹.
t₁/₂ = 0.693/k = 0.693/(8.92 × 10⁻³) = 77.7 min.
For the decomposition of azoisopropane to hexane and N₂ at 543 K, the following data were obtained:
| t/s | P (mm Hg) |
|---|---|
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
Reaction: A(g) → B(g) + N₂(g). At time t: \(p_t = p_0 + p\), where p is decrease in azoisopropane. So \(p_A = p_0 - p = 2p_0 - p_t\).
At t = 360 s: \(p_A = 2(35) - 54 = 16\) mm Hg.
k = (2.303/360) log(35/16) = 0.0064 × log(2.1875) = 0.0064 × 0.3398 = 2.18 × 10⁻³ s⁻¹.
At t = 720 s: \(p_A = 70 - 63 = 7\) mm Hg. k = (2.303/720) log(35/7) = 0.0032 × 0.6990 = 2.24 × 10⁻³ s⁻¹.
Average k ≈ 2.21 × 10⁻³ s⁻¹.
The rate constant for the decomposition of N₂O₅ at various temperatures is given:
| T/°C | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| 10⁵ × k / s⁻¹ | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph of log k vs 1/T and calculate Eₐ and A.
Compute log k and 1/T (in K) for each entry; plotting yields a straight line of slope = −Eₐ/(2.303R).
Using endpoints T₁ = 273 K, k₁ = 7.87 × 10⁻⁷ and T₂ = 353 K, k₂ = 2.140 × 10⁻²:
log(k₂/k₁) = log(2.140e−2 / 7.87e−7) = log(27192) = 4.434.
(T₂ − T₁)/(T₁ T₂) = 80 / (273 × 353) = 8.30 × 10⁻⁴ K⁻¹.
Eₐ = (4.434 × 2.303 × 8.314) / (8.30 × 10⁻⁴) = 84.9 / 8.30 × 10⁻⁴ × ... ≈ 1.022 × 10⁵ J/mol = 102.3 kJ/mol.
Using k = A e^(−Eₐ/RT) at T = 273 K: A = 7.87 × 10⁻⁷ / e^(−102300/(8.314 × 273)) = 7.87 × 10⁻⁷ / e^(−45.05) = 7.87 × 10⁻⁷ × 3.4 × 10¹⁹ ≈ 2.7 × 10¹³ s⁻¹.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
\(k = A\,e^{-E_a/RT}\) → \(A = k\,e^{E_a/RT}\)
\(E_a/RT = 179900/(8.314 \times 546) = 39.62\)
\(A = 2.418 \times 10^{-5} \times e^{39.62} = 2.418 \times 10^{-5} \times 1.62 \times 10^{17}\)
\(A = \mathbf{3.91 \times 10^{12} \text{ s}^{-1}}\).
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation, assuming it does not change with temperature.
k₂/k₁ = 4, T₁ = 293 K, T₂ = 313 K.
\(\log 4 = (E_a/2.303R)\,(T_2 - T_1)/(T_1 T_2)\)
0.6021 = (Eₐ/19.147) × (20/91709) = (Eₐ/19.147) × 2.181 × 10⁻⁴
Eₐ = (0.6021 × 19.147)/(2.181 × 10⁻⁴) = 5.29 × 10⁴ J/mol = 52.9 kJ/mol.
The rate constant for first order decomposition of H₂O₂ is given by log k = 14.34 − (1.25 × 10⁴ K)/T. Calculate Eₐ. At what temperature will t₁/₂ be 256 minutes?
Comparing with log k = log A − Eₐ/(2.303RT): Eₐ/(2.303R) = 1.25 × 10⁴.
Eₐ = 1.25 × 10⁴ × 2.303 × 8.314 = 239340 J/mol ≈ 239.3 kJ/mol.
For t₁/₂ = 256 min = 256 × 60 = 15360 s: k = 0.693/15360 = 4.51 × 10⁻⁵ s⁻¹.
log k = log(4.51 × 10⁻⁵) = −4.346.
−4.346 = 14.34 − (1.25 × 10⁴/T) → 1.25 × 10⁴/T = 18.69 → T = 12500/18.69 = 668.9 K.
The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10 °C and Eₐ = 60 kJ/mol. At what temperature would k be 1.5 × 10⁴ s⁻¹?
T₁ = 283 K, k₁ = 4.5 × 10³ s⁻¹, k₂ = 1.5 × 10⁴ s⁻¹, Eₐ = 60000 J/mol.
log(k₂/k₁) = (Eₐ/2.303R)(T₂ − T₁)/(T₁ T₂)
log(1.5e4 / 4.5e3) = log(3.333) = 0.5229.
0.5229 = (60000/19.147)(T₂ − 283)/(283 T₂) = 3133.5 × (T₂ − 283)/(283 T₂).
(T₂ − 283)/(283 T₂) = 0.5229/3133.5 = 1.668 × 10⁻⁴.
(T₂ − 283) = 1.668 × 10⁻⁴ × 283 T₂ = 0.04721 T₂.
T₂(1 − 0.04721) = 283 → T₂ = 283/0.9528 = 297 K (≈ 24 °C).
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s⁻¹, calculate k at 318 K and Eₐ.
For 10% at 298 K: t = (2.303/k₁) log(100/90) = (2.303/k₁)(0.0458).
For 25% at 308 K: t = (2.303/k₂) log(100/75) = (2.303/k₂)(0.1249).
Setting equal: (0.0458/k₁) = (0.1249/k₂) → k₂/k₁ = 0.1249/0.0458 = 2.727.
log(k₂/k₁) = log(2.727) = 0.4357.
0.4357 = (Eₐ/19.147)(10/(298 × 308)) = (Eₐ/19.147)(1.089 × 10⁻⁴).
Eₐ = (0.4357 × 19.147)/(1.089 × 10⁻⁴) = 76630 J/mol ≈ 76.6 kJ/mol.
At 318 K: k = A e^(−Eₐ/RT) = 4 × 10¹⁰ × e^(−76630/(8.314 × 318)) = 4 × 10¹⁰ × e^(−28.99)
= 4 × 10¹⁰ × 2.59 × 10⁻¹³ = 1.036 × 10⁻² s⁻¹.
Setup: Pick any everyday process that goes faster as it gets warmer (food spoiling, milk turning, banana ripening, ice melting).
Banana ripens roughly 3× as fast at 25 °C than at 15 °C. So k₂/k₁ ≈ 3, ΔT = 10 K, T₁ = 288 K, T₂ = 298 K.
log 3 = (Eₐ/19.147)(10/(288 × 298)) → Eₐ = (0.477 × 19.147)/(1.166 × 10⁻⁴) = 78300 J/mol ≈ 78 kJ/mol.
If we drop another 10 K (5 °C, fridge): rate becomes 1/3 of 15 °C rate, so shelf life triples again. From 25 °C to 5 °C = 9× longer shelf life. This is the chemistry of refrigeration!
Give two experiments with [A] doubled. Find the order in A from the rate ratio.
Order in A = log₂(ratio) = 1.00
Rate doubles → first order.
Q1. Which graph correctly represents a first order reaction? L4
Q2. The half-life period of a radioactive substance is 5 years. The fraction left after 15 years is: L3
Q3. (Short answer) The rate constant of a reaction at 280 K is 9 × 10⁻³ s⁻¹ and at 320 K is 0.36 s⁻¹. Estimate Eₐ in kJ/mol. L4
Q4. (Fill in the blank) The Arrhenius factor A is also called the _________ factor. L1
Q5. (Long answer) A drug has a recommended shelf life of 24 months at 25 °C. If its decomposition follows first order kinetics with Eₐ = 60 kJ/mol, what shelf life would you expect at 30 °C? L5
(A) Both true & R explains A. (B) Both true but R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion: The number of half-lives elapsed for first order decay to 12.5% of original concentration is 3.
Reason: 12.5% = (1/2)³, so 3 half-lives have elapsed.
Assertion: Photochemical reactions are independent of temperature.
Reason: The energy required for photochemical reactions comes from absorbed photons, not thermal collisions.
Assertion: Rate of reaction is always positive.
Reason: Reactant concentration decreases with time so a negative sign is included in its rate expression.
For the reaction R → P, the concentration of reactant changes from 0.03 M to 0.02 M in 25 min. Calculate average rate of reaction using units of time both in min and s.
In a reaction, 2A → Products, conc. of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 min. Find rate during this interval.
For a reaction A + B → Product, the rate law is given by r = k[A]^(½)[B]². What is the order of the reaction?
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?
Sit a full paper on what you have been studying, marked question by question.