- iamal
- February 17, 2026
Best Order of Reaction Numerical MCQs Class 12 Chemistry to Improve Your Score
Understanding reaction order is one of the most important topics in chemical kinetics for board exams and competitive tests. When aspirants practice order of reaction numerical mcqs class 12 chemistry, they are essentially learning how to interpret rate data, compare concentration changes, and determine how reaction rate depends on reactant concentration. Mastery of order of reaction numerical mcqs class 12 chemistry helps aspirants build clarity in rate laws and integrated rate equations.
In chemical kinetics, the order of reaction is defined as the sum of the powers of the concentration terms in the rate law. For example, if the rate law is written as rate = k[A]²[B], then the reaction is second order with respect to A, first order with respect to B, and third order overall. Many aspirants encounter such expressions while solving order of reaction numerical mcqs class 12 chemistry and must carefully interpret experimental data.
When concentration data and initial rates are provided, comparison methods are used. If doubling the concentration of A doubles the rate, the reaction is first order with respect to A. If doubling A increases the rate four times, the order with respect to A is two. These concepts form the backbone of order of reaction numerical mcqs class 12 chemistry. Logical comparison and proportional reasoning are essential skills here.
Another important area in order of reaction numerical mcqs class 12 chemistry involves zero-order reactions. In zero-order reactions, rate = k and is independent of concentration. The integrated rate equation becomes:
[A] = [A]₀ − kt
This leads to a linear plot of concentration versus time. Many numerical problems ask aspirants to determine k from slope values or concentration-time data, which is common in order of reaction numerical mcqs class 12 chemistry practice sets.
First-order reactions are equally significant in order of reaction numerical mcqs class 12 chemistry. The integrated equation is:
ln[A] = ln[A]₀ − kt
or
k = (2.303/t) log([A]₀/[A])
The half-life expression is:
t₁/₂ = 0.693/k
Aspirants often calculate k using half-life or determine remaining concentration after a given time. These standard formulas appear repeatedly in order of reaction numerical mcqs class 12 chemistry exercises.
Second-order reactions also play a central role in order of reaction numerical mcqs class 12 chemistry. For a second-order reaction involving a single reactant, the integrated form is:
1/[A] = 1/[A]₀ + kt
In such numericals, aspirants may be required to compute k from concentration changes over time or compare half-life values at different concentrations. Recognizing that half-life depends on initial concentration in second-order reactions is a key concept tested in order of reaction numerical mcqs class 12 chemistry.
Graphical interpretation is another major component of order of reaction numerical mcqs class 12 chemistry. A straight line for concentration vs time indicates zero order, ln[A] vs time indicates first order, and 1/[A] vs time indicates second order. Many conceptual numericals require identifying order from graph patterns, reinforcing understanding through order of reaction numerical mcqs class 12 chemistry practice.
Temperature does not affect order but influences rate constant. However, in order of reaction numerical mcqs class 12 chemistry, aspirants must distinguish between order and molecularity. Molecularity is theoretical and applies to elementary reactions, whereas order is determined experimentally. Confusing these two concepts is a common mistake addressed in order of reaction numerical mcqs class 12 chemistry preparation.
Another frequently tested idea in order of reaction numerical mcqs class 12 chemistry is fractional order. In complex reactions, the rate law may include fractional powers, such as rate = k[A]¹/². Such cases arise in chain reactions or adsorption mechanisms. Aspirants must remember that order need not match stoichiometric coefficients in order of reaction numerical mcqs class 12 chemistry.
Pseudo-first-order reactions are also important in order of reaction numerical mcqs class 12 chemistry. When one reactant is present in large excess, its concentration remains effectively constant, and the reaction behaves as first order. This simplification often appears in competitive-level order of reaction numerical mcqs class 12 chemistry problems.
Order of Reaction Numerical MCQs Class 12 Chemistry
1.
For the reaction A + B → products, it is observed that:
(i) On doubling the initial concentration of A only, the rate of reaction is doubled.
(ii) On doubling the initial concentration of both A and B, the rate increases by a factor of 8.
The rate law is:
A. Rate = k[A][B]²
B. Rate = k[A]²[B]
C. Rate = k[A][B]
D. Rate = k[A]²[B]²
Answer: A
2.
Bromination of acetone in acidic medium:
CH₃COCH₃ + Br₂ → CH₃COCH₂Br + H⁺ + Br⁻
From experimental data, the rate equation is:
A. Rate = k[CH₃COCH₃][Br₂][H⁺]
B. Rate = k[CH₃COCH₃][H⁺]
C. Rate = k[CH₃COCH₃][Br₂]
D. Rate = k[CH₃COCH₃][Br₂][H⁺]²
Answer: B
3.
For the reaction:
N₂ + 3H₂ → 2NH₃
If
d[NH3]dt=2×10−4 mol L−1s−1\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}
Then
−d[H2]dt=-\frac{d[H_2]}{dt} =
A. 4 × 10⁻⁴
B. 6 × 10⁻⁴
C. 1 × 10⁻⁴
D. 3 × 10⁻⁴
Answer: D
4.
For the reaction:
BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O
The relation between rates is:
A.
d[Br2]dt=−53d[Br−]dt\frac{d[Br_2]}{dt} = -\frac{5}{3}\frac{d[Br^-]}{dt}
B.
d[Br2]dt=53d[Br−]dt\frac{d[Br_2]}{dt} = \frac{5}{3}\frac{d[Br^-]}{dt}
C.
d[Br2]dt=33d[Br−]dt\frac{d[Br_2]}{dt} = \frac{3}{3}\frac{d[Br^-]}{dt}
D.
d[Br2]dt=35d[Br−]dt\frac{d[Br_2]}{dt} = \frac{3}{5}\frac{d[Br^-]}{dt}
Answer: D
5.
For reaction: 2A + B → 3C + D
Which does not express the rate?
A.
−12d[A]dt-\frac{1}{2}\frac{d[A]}{dt}
B.
−13d[C]dt-\frac{1}{3}\frac{d[C]}{dt}
C.
−d[B]dt-\frac{d[B]}{dt}
D.
d[D]dt\frac{d[D]}{dt}
Answer: B
6.
For reaction: 2NO + O₂ → 2NO₂
If volume becomes 1/3rd, rate increases by:
A. 3 times
B. 9 times
C. 27 times
D. 36 times
Answer: C
7.
For reaction: 2NO₂ + F₂ → 2NO₂F
Correct rate expressions:
A.
−12dp[NO2]dt-\frac{1}{2}\frac{dp[NO_2]}{dt}
B.
12dp[NO2]dt\frac{1}{2}\frac{dp[NO_2]}{dt}
C.
12dp[NO2F]dt\frac{1}{2}\frac{dp[NO_2F]}{dt}
D. Both I and III
Answer: D
8.
40% HI decomposes at 300K. ΔG° ≈
A. 2735 J mol⁻¹
B. 2300
C. 2700
D. 2730
Answer: A
9.
Half-life of A = 100 s
Half-life of B = 50 s
Time when [A] = 4[B]:
A. 100 s
B. 200 s
C. 50 s
D. 400 s
Answer: B
10.
Rate constants at 500K and 700K: 0.02 and 0.2 s⁻¹
Activation energy:
A. 66.9
B. 33.45
C. 22.3
D. 44.45
Answer: B
11.
2N₂O₅ → 4NO₂ + O₂
If
−d[N2O5]dt=1.2×10−5-\frac{d[N_2O_5]}{dt} = 1.2 \times 10^{-5}
Then
d[NO2]dt=\frac{d[NO_2]}{dt} =
A. 1.2 × 10⁻²
B. 3.6 × 10⁻⁵
C. 2.4 × 10⁻⁵
D. 4.8 × 10⁻⁵
Answer: C
12.
First-order reaction, t½ = 60 min
k =
A. 1.92 × 10⁻²
B. 1.15 × 10⁻²
C. 1.25 × 10⁻⁴
D. 1.92 × 10⁻⁴
Answer: D
13.
Difference in activation energies when kA = 2kB:
A. RT ln 2
B. −2.303 RT
C. −RT ln 2
D. 0
Answer: C
14.
Unit of rate constant (zero order):
A. mol L⁻¹ s⁻¹
B. L mol⁻¹ s⁻¹
C. L⁻¹ mol s
D. s⁻¹
Answer: A
15.
For
2NO2⇌N2O42NO_2 \rightleftharpoons N_2O_4
Rate of disappearance of NO₂:
A.
2k1k2[NO2]2\frac{2k_1}{k_2}[NO_2]^2
B. 2k₁[N₂O₄] − 2k₂[NO₂]
C. 2k₁[NO₂]² − k₂[N₂O₄]
D. (2k₁ − k₂)[NO₂]
Answer: C
16.
For equilibrium:
H₂O + CO → H₂ + CO₂
K = 81
If kf = 162
kb =
A. 132
B. 2
C. 261
D. 243
Answer: B
17.
2N₂O₅ → 4NO₂ + O₂
Rate = 1.02 × 10⁻⁴
k = 3.4 × 10⁻⁵
Concentration =
A. 1.3 M
B. 3 M
C. 3.4 × 10⁻⁴
D. 10.2 × 10⁻⁴
Answer: B
18.
From experimental data:
Rate law =
A. k[A]
B. k[B]³
C. k[A][B]
D. k[A]²[B]
Answer: B
19.
Doubling both reactants increases rate 8 times; doubling B doubles rate.
Rate law:
A. k[A]¹/²[B]¹/²
B. k[A]²[B]
C. k[A][B]²
D. k[A][B]
Answer: B
20.
New rate / old rate when A doubled, B halved:
A. 1/2ⁿ⁺ᵐ
B. m+n
C. n−m
D. 2ⁿ⁻ᵐ
Answer: D
21.
Rate constant depends on:
A. Temperature
B. Initial concentration
C. Extent
D. Time
Answer: A
22.
Temperature coefficient = 2
Increase 30°C → 90°C
Rate increases:
A. 60
B. 64
C. 150
D. 400
Answer: B
23.
First-order decomposition data
k =
A. 0.0693
B. 0.693
C. 6.93
D. 6.93 × 10⁻⁴
Answer: A
24.
For A + 2B → Products
Unit of k:
A. s⁻¹
B. mol L⁻¹ s⁻¹
C. L mol⁻¹ s⁻¹
D. atm⁻¹
Answer: A
25.
From data table
Rate law =
A. k[A]²[B]
B. k[A][B]
C. k[A][B]²
D. k[B]³
Answer: D
26.
Time for 99% completion (first order):
A.
t=4.606kt = \frac{4.606}{k}
B.
t=2.303kt = \frac{2.303}{k}
C.
t=0.693kt = \frac{0.693}{k}
D.
t=6.909kt = \frac{6.909}{k}
Answer: A
27.
Based on kinetic data:
A. k[A]²[B]
B. k[A][B]
C. k[A][B]²
D. k[A]²[B]²
Answer: C
28.
Temperature for k = 10⁻⁴:
A. 526 K
B. 600 K
C. 700 K
D. 800 K
Answer: A
29.
32% reactant left
k =
A. 2 × 10⁻³
B. 1 × 10⁻³
C. 3 × 10⁻³
D. 4 × 10⁻³
Answer: A
30.
For first-order reaction:
N₂O₅ concentration change
k =
A. 7 × 10⁻³
B. 8 × 10⁻³
C. 9 × 10⁻³
D. 5 × 10⁻³
Answer: A
Conclusion on Order of Reaction Numerical MCQs Class 12 Chemistry
Overall, consistent practice of order of reaction numerical mcqs class 12 chemistry builds analytical thinking and formula application skills. By understanding rate laws, integrated equations, graphical methods, and proportional reasoning, aspirants can confidently solve order of reaction numerical mcqs class 12 chemistry and strengthen their performance in board and entrance examinations.