આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Chemical Kinetics
આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Chemical Kinetics
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
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?
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