- Keneitsino Lydia
- March 4, 2026
Arrhenius Equation Solved MCQs Class 12 : Master Chemical Kinetics with Confidences
Arrhenius Equation Solved MCQs Class 12 is one of the most important preparation themes in chemical kinetics for board exams and competitive examinations like NEET and JEE. Understanding the temperature dependence of reaction rates is not just about memorizing formulas, but about deeply connecting activation energy, rate constant, and molecular collisions. That is why Arrhenius Equation Solved MCQs Class 12 becomes a powerful revision tool before exams.
The Arrhenius equation is written as:
k = A e^(-Ea/RT)
Here, k is the rate constant, A is the frequency factor, Ea is activation energy, R is the gas constant, and T is temperature in Kelvin. When aspirants practice Arrhenius Equation Solved MCQs Class 12, they repeatedly apply this formula to calculate activation energy, rate constant at different temperatures, and temperature coefficients. This repetition strengthens conceptual clarity.
In Class 12 chemistry, many numerical problems involve comparing rate constants at two different temperatures. The logarithmic form of the Arrhenius equation is especially useful:
log k₂/k₁ = (Ea / 2.303R) (1/T₁ − 1/T₂)
Through Arrhenius Equation Solved MCQs Class 12, aspirants learn how to manipulate this expression efficiently without confusion. Board exam questions often test whether an aspirant can identify how much the rate changes when temperature increases by 10 K or 20 K.
One major concept reinforced through Arrhenius Equation Solved MCQs Class 12 is the idea of activation energy. Activation energy represents the minimum energy barrier that reacting molecules must overcome. Higher activation energy means slower reaction at a given temperature. When solving Arrhenius Equation Solved MCQs Class 12 , learners see how even a small increase in temperature significantly increases the fraction of molecules having energy greater than Ea.
Another important area where Arrhenius Equation Solved MCQs Class 12 helps is graphical interpretation. The plot of ln k versus 1/T gives a straight line with slope equal to −Ea/R. Many aspirants initially struggle with identifying slope relationships, but repeated exposure through Arrhenius Equation Solved MCQs Class 12 builds confidence in extracting activation energy from graphs.
In competitive exams, conceptual twists are common. For example, questions may ask how the rate changes if activation energy decreases due to a catalyst. Arrhenius Equation Solved MCQs Class 12 trains aspirants to quickly conclude that lowering Ea increases k exponentially, not linearly. This exponential dependence is what makes catalysis so effective.
Temperature coefficient problems are another common application. If the rate doubles for every 10°C rise, students must connect this observation with the Arrhenius equation. Practicing Arrhenius Equation Solved MCQs Class 12 ensures that such reasoning becomes automatic rather than mechanical.
A strong benefit of studying Arrhenius Equation Solved MCQs Class 12 is improved numerical speed. Since exams are time-bound, aspirants need to quickly substitute values, convert units properly, and avoid calculation errors. Repeated numerical exposure makes logarithmic calculations and exponential comparisons much easier.
Another advantage of Arrhenius Equation Solved MCQs Class 12 is conceptual linking with collision theory. The frequency factor A reflects the number of effective collisions, while Ea determines the energy barrier. Aspirants understand that temperature affects both kinetic energy and the fraction of molecules exceeding activation energy. This holistic understanding grows stronger when practicing Arrhenius Equation Solved MCQs Class 12 regularly.
Many Class 12 aspirants also prepare for entrance exams simultaneously. Arrhenius Equation Solved MCQs Class 12 bridges board-level clarity with competitive-level application. Problems involving comparison of two reactions, determination of unknown activation energy, and estimation-based reasoning become easier after consistent practice.
To master Arrhenius Equation Solved MCQs Class 12 , aspirants should remember three core ideas: activation energy controls sensitivity to temperature, rate constant increases exponentially with temperature, and graphical interpretation simplifies numerical work. When these principles are internalized, solving Arrhenius-based problems becomes systematic and less intimidating.
30 Arrhenius Equation Solved MCQs Class 12:
1.
For the reaction, A + B ⇌ 2C
The value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x × 10⁻¹ M. The value of x is ….. (Nearest integer)
A. 25
B. 5
C. 6
D. 36
Answer: A
2.
PCl₅(g) → PCl₃(g) + Cl₂(g)
In the above first order reaction, the concentration of PCl₅ reduces from 50 mol L⁻¹ to 10 mol L⁻¹ in 120 minutes at 300 K. The rate constant is x × 10⁻² min⁻¹.
(Given log 5 = 0.6989)
A. 1.34
B. 1
C. 2
D. 2.34
Answer: A
3.
The reaction 2A + B₂ → 2AB is an elementary reaction. If the volume is reduced by a factor of 3, the rate increases by a factor of:
A. 27
B. 25
C. 20
D. 30
Answer: A
4.
For the reaction
[PtCl₄]²⁻ + H₂O ⇌ [Pt(H₂O)Cl₃]⁻ + Cl⁻
Given rate expressions, the equilibrium constant Kc is:
A. 50
B. 5
C. 25
D. 30
Answer: A
5.
If temperature increases, the value of rate constant (k) will:
A. Decrease
B. Increase
C. Remain Constant
D. Become Zero
Answer: B
6.
For reaction A → B,
log₁₀ k = 20.35 − (2.47 × 10³)/T
Energy of activation (kJ mol⁻¹):
A. 50
B. 47
C. 400
D. 500
Answer: B
7.
Virus inactivation follows first order kinetics. 10% is inactivated initially. Rate constant is:
A. 106 × 10⁻³ min⁻¹
B. 100 × 10⁻³ min⁻¹
C. 110 × 10⁻³ min⁻¹
D. 120 × 10⁻³ min⁻¹
Answer: A
8.
Formic acid decomposition (first order).
k₃₀₀ = 1 × 10⁻³ s⁻¹
Ea = 11.488 kJ mol⁻¹
Rate constant at 200 K is:
A. 10 × 10⁻⁵
B. 4 × 10⁻⁵
C. 6 × 10⁻⁵
D. 8 × 10⁻⁵
Answer: A
9.
Activation energy = 80.9 kJ mol⁻¹ at 700 K. Fraction of molecules = e⁻ˣ.
x =
A. 1
B. 2
C. 10
D. 14
Answer: D
10.
For first order reaction k = 2.4 × 10⁻⁴ s⁻¹.
Ratio t₉₉.₉ / t₅₀ =
A. 1
B. 5
C. 10
D. 15
Answer: C
11.
For first order reaction t₁/₂ = 1200 s.
k =
A. 5.8 × 10⁻⁴ s⁻¹
B. 5.8 × 10⁻² s⁻¹
C. 0.58 × 10⁻⁵ s⁻¹
D. 0.58 × 10⁻² s⁻¹
Answer: A
12.
k = 4.606 × 10⁻³ s⁻¹.
Time for 400 g → 50 g:
A. 75.2 min
B. 7.52 min
C. 46.06 min
D. 15.05 min
Answer: B
13.
Temperature coefficient = 2.
Rate constant at 15°C:
A. 2 × 10⁻¹
B. 2 × 10⁻²
C. 5 × 10⁻⁴
D. 4 × 10⁻³
Answer: C
14.
K = 4.5 × 10⁻⁴ L mol⁻¹ s⁻¹.
Order of reaction:
A. Zero
B. Second
C. First
D. Third
Answer: B
15.
Rate doubles from 300K to 600K.
Activation energy:
A. 3.45 kJ mol⁻¹
B. 69.0 kJ mol⁻¹
C. 96.8 kJ mol⁻¹
D. 19.6 kJ mol⁻¹
Answer: A
16.
First order decomposition. t₁/₂ = 40 min.
k =
A. 1.73×10⁻⁴ s⁻¹
B. 2.88×10⁻⁴ s⁻¹
C. 2.88×10⁻² s⁻¹
D. 1.73×10⁻² s⁻¹
Answer: B
17.
Activation energy order from plot:
A. E_b > E_a > E_d > E_c
B. E_c > E_d > E_b > E_a
C. E_c > E_a > E_b > E_d
D. E_c > E_a > E_d > E_b
Answer: D
18.
2A + 3B → 4C + D
If rate of formation of C = 2.8 × 10⁻³ mol L⁻¹ s⁻¹
Rate of disappearance of B:
A. (4/3)(2.8×10⁻³)
B. (3/4)(2.8×10⁻³)
C. 2.8×10⁻¹
D. (1/4)(2.8×10⁻³)
Answer: B
19.
k = 231 × 10⁻⁵ s⁻¹.
Time for 4g → 2g:
A. 310 s
B. 300 s
C. 210 s
D. 230.3 s
Answer: B
20.
k = 4.606 × 10⁻³ s⁻¹.
Time for 2g → 0.2g:
A. 100 s
B. 200 s
C. 500 s
D. 1000 s
Answer: C
21.
2A + B → P
Unit of rate constant:
A. mol⁻¹ L s⁻¹
B. L mol⁻¹ s
C. mol L⁻¹ s⁻¹
D. L² mol⁻² s⁻¹
Answer: A
22.
Incorrect statement (enthalpy diagram):
A. D is kinetically stable
B. Reverse has highest activation enthalpy
C. C is thermodynamically stable
D. ΔH‡ difference is 5 kJ mol⁻¹
Answer: D
23.
From given table, rate law is:
A. k[A]²[B]
B. k[A][B]²
C. k[A][B]
D. k[A]²[B]²
Answer: B
24.
If t₁/₂ vs [R]₀ is horizontal line, unit of k:
A. mol L⁻¹ s⁻¹
B. L mol⁻¹ s⁻¹
C. mol s⁻¹
D. s⁻¹
Answer: D
25.
rate = k[A]²[B].
If [A] doubled:
A. ×8
B. ×4
C. ×3
D. ×2
Answer: B
26.
For N₂ + 3H₂ → 2NH₃
If Δ[NH₃]/Δt = 2×10⁻⁴
Then −Δ[H₂]/Δt =
A. 1×10⁻⁴
B. 3×10⁻⁴
C. 4×10⁻⁴
D. 6×10⁻⁴
Answer: B
27.
N₂O₂ → 2NO (first order)
[NO] expression:
A. 2[N₂O₂]₀(1−e⁻ᵏᵗ)
B. [N₂O₂]₀(1−e⁻ᵏᵗ)
C. 2[N₂O₂]₀e⁻ᵏᵗ
D. [N₂O₂]₀e⁻ᵏᵗ
Answer: B
28.
From Arrhenius plot, rate constant at 500 K:
A. 4×10⁻⁴ s⁻¹
B. 10⁻⁴ s⁻¹
C. 10⁻⁵ s⁻¹
D. 2×10⁻⁴ s⁻¹
Answer: B
29.
k = 3.0×10⁻⁴ s⁻¹
Rate = 2.4×10⁻⁵ mol L⁻¹ s⁻¹
Concentration of N₂O₅:
A. 4
B. 1.2
C. 0.02
D. 0.08
Answer: D
30.
At 27°C, time for 75% completion = 20 s.
Rate constant:
A. 0.693 s⁻¹ mol⁻¹ L
B. 0.0693 s⁻¹
C. 0.693 s⁻¹
D. 0.0693 s⁻¹ mol⁻¹ L
Answer: B
Conclusion on Arrhenius Equation Solved MCQs Class 12
Ultimately, Arrhenius Equation Solved MCQs Class 12 is not just about solving numbers; it is about building analytical thinking. With consistent practice, aspirants develop speed, accuracy, and conceptual strength. This makes Arrhenius Equation Solved MCQs Class 12 a highly valuable component of Class 12 chemistry preparation and competitive exam success.