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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 8 April · Evening Shift · Chemistry · Chemical Kinetics

Given below are two statements: R=8.314 J K ^ -1 mol ^ -1 and 1 cal...

Given below are two statements:
R=8.314R=8.314 J K−1^{-1} mol−1^{-1} and 11 cal =4.2=4.2 J
Statement I: When Ea=12.6E_a=12.6 kcal/mol, the room temperature rate constant is doubled by a 10\,^\circC increase in temperature (298298 K to 308308 K)
Statement II: For a first order reactions A →\rightarrow B,
Given below are two statements: R=8.314 J K ^ -1 mol ^ -1 and 1 cal =4.2 J Statement I: When E_a=12.6 kcal/mol, the
Here [A]o[A]_o is the initial concentration of A and t1/2t_{1/2} is half life of reaction.
In the light of the above statements, choose the correct answer from the options given below:

Options

  1. ABoth Statement I and Statement II are true
  2. BBoth Statement I and Statement II are false
  3. CStatement I is true but Statement II is false
  4. DStatement I is false but Statement II is true

Correct answer

C. Statement I is true but Statement II is false

Step-by-step solution

For Statement I:
According to the Arrhenius equation:
log⁡(k2k1)=Ea2.303R(T2−T1T1T2)\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.303 R} \left( \dfrac{T_2 - T_1}{T_1 T_2} \right)

Given that the rate constant doubles, k2=2k1k_2 = 2k_1. The temperatures are T1=298T_1 = 298 K and T2=308T_2 = 308 K.
The gas constant R=8.314R = 8.314 J K−1^{-1} mol−1=8.3144.2^{-1} = \dfrac{8.314}{4.2} cal K−1^{-1} mol−1≈1.98^{-1} \approx 1.98 cal K−1^{-1} mol−1^{-1}.

Substituting the values:
log⁡(2)=Ea2.303×1.98(308−298298×308)\log(2) = \dfrac{E_a}{2.303 \times 1.98} \left( \dfrac{308 - 298}{298 \times 308} \right)

0.301=Ea4.56(1091784)0.301 = \dfrac{E_a}{4.56} \left( \dfrac{10}{91784} \right)

Ea=0.301×4.56×9178410≈12598E_a = \dfrac{0.301 \times 4.56 \times 91784}{10} \approx 12598 cal/mol ≈12.6\approx 12.6 kcal/mol.
Thus, Statement I is true.

For Statement II:
For a first order reaction, the half-life is given by:
t1/2=0.693kt_{1/2} = \dfrac{0.693}{k}
This indicates that the half-life of a first order reaction is independent of the initial concentration [A]o[A]_o. The graph of t1/2t_{1/2} versus [A]o[A]_o should be a horizontal line parallel to the concentration axis.
The given graph shows t1/2t_{1/2} directly proportional to [A]o[A]_o, which corresponds to a zero order reaction (t1/2=[A]o2kt_{1/2} = \dfrac{[A]_o}{2k}).
Thus, Statement II is false.

Therefore, Statement I is true but Statement II is false.

Answer: Statement I is true but Statement II is false
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