JEE Main 2026 — Chemical Kinetics Question with Solution

For Statement I:
According to the Arrhenius equation:
$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Given that the rate constant doubles, $k_2=2k_1$. The temperatures are $T_1=298$ K and $T_2=308$ K.
The gas constant $R=8.314$ J K${}^{-1}$ mol${}^{-1}=\frac{8.314}{4.2}$ cal K${}^{-1}$ mol${}^{-1}\approx 1.98$ cal K${}^{-1}$ mol${}^{-1}$.

Substituting the values:
$\log (2)=\frac{E_a}{2.303\times 1.98}\left(\frac{308-298}{298\times 308}\right)$

$0.301=\frac{E_a}{4.56}\left(\frac{10}{91784}\right)$

$E_a=\frac{0.301\times 4.56\times 91784}{10}\approx 12598$ cal/mol $\approx 12.6$ kcal/mol.
Thus, Statement I is true.

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