Verified Solution • 2025–26 Session moderate

QJEEM22Chedf8: JEE Main 2022 Chemistry - Class 12 JEEM22Chedf8 Solution: It has been found that for a chemical reaction with rise in…

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol −- 1. [nearest integer]

(Given ln10 = 2.3, R = 8.3 J K −- 1 mol −- 1, log 2 = 0.30)

Step-by-Step Verified Solution

T1=300 KT_{1}=300 \mathrm{~K}

(Rate constant)

K2=2 K1\mathrm{K}_{2}=2 \mathrm{~K}_{1} , on increase temperature by 9 K9 \mathrm{~K}

T2=309 K\mathrm{T}_{2}=309 \mathrm{~K}

Ea=?\mathrm{Ea}=?

log⁡K2 K1=Ea2.3R[T2−T1 T2⋅T1]\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \cdot \mathrm{T}_{1}}\right]

log⁡2=Ea2.3×8.3[9309×300]\log 2=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{9}{309 \times 300}\right]

Ea=0.3×309×300×2.3×8.39\mathrm{Ea}=\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}

=58988.1 J/mole=58988.1 \mathrm{~J} / \mathrm{mole}

≃59 kJ/mole\simeq 59 \mathrm{~kJ} / \mathrm{mole}

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