JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

For reaction $A \to P$ , rate constant $k = 1.5 \times 1 0^{3}$ s $^{- 1}$ at $2 7^{\circ}$ 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 $^{- 1}$ is _____. (Nearest integer)

Given : $lo g 2 = 0.30$ , $lo g 3 = 0.48$ , $R = 8.3$ J K $^{- 1}$ mol $^{- 1}$ , $ln 10 = 2.3$

Your answer

Solution

Apply the integrated Arrhenius equation:

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

Substituting $k_{2} / k_{1} = 3$ , $E_{a} = 60, 000$ J/mol, $R = 8.3$ J/K/mol, $T_{1} = 300$ K:

$ln 3 = 2.3 \times 0.48 \approx 1.104$

$1.104 = \frac{60000}{8.3} \cdot \frac{T _{2} - 300}{300 T _{2}}$

Solving: $\frac{T _{2} - 300}{T _{2}} \approx \frac{1.104 \times 8.3 \times 300}{60000} \approx 0.0458$

$T_{2} \approx \frac{300}{1 - 0.0458} \approx 314.4$ K $\approx 41. 4^{\circ}$ C

Nearest integer: $4 1^{\circ}$ C.

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