For reaction $\mathrm{A} \rightarrow \mathrm{P}$, rate constant $\mathrm{k}=1.5 \times 10^{3} \mathrm{~s}^{-1}$ at $27^{\circ} \mathrm{C}$. If activation energy for the above reaction is $60 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then the temperature (in ${ }^{\circ} \mathrm{C}$) at which rate constant, $\mathrm{k}=4.5 \times 10^{3} \mathrm{~s}^{-1}$ is $\_\_\_\_$. (Nearest integer) Given: $\log 2=0.30, \log 3=0.48, \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, $\ln 10=2.3$

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Chemical Kinetics

For reaction

A \to P

, rate constant

k = 1.5 \times 1 0^{3} s^{- 1}

2 7^{\circ} C

. If activation energy for the above reaction is

60 kJ mol^{- 1}

, then the temperature (in

^{\circ} C

) at which rate constant,

k = 4.5 \times 1 0^{3} s^{- 1}

____

.
(Nearest integer)
Given:

lo g 2 = 0.30, lo g 3 = 0.48, R = 8.3 J K^{- 1} mol^{- 1}

ln 10 = 2.3

Answer: 41

From the two-temperature Arrhenius relation

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

with

\frac{k _{2}}{k _{1}} = 3

ln 3 = 2.3 \times 0.48 = 1.104

and

\frac{E _{a}}{R} = \frac{60000}{8.3} = 7228.9

. Then

\frac{1}{T _{1}} - \frac{1}{T _{2}} = \frac{1.104}{7228.9} = 1.527 \times 1 0^{- 4}

. With

T_{1} = 300 K

\frac{1}{T _{2}} = 3.333 \times 1 0^{- 3} - 1.527 \times 1 0^{- 4} = 3.181 \times 1 0^{- 3}

, so

T_{2} = 314.4 K = 41. 4^{\circ} C

, which to the nearest integer is

4 1^{\circ} C

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Integer answer