Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1
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)
T1=300 KT_{1}=300 \mathrm{~K}T1=300 K (Rate constant)
K2=2 K1\mathrm{K}_{2}=2 \mathrm{~K}_{1}K2=2 K1, on increase temperature by 9 K9 \mathrm{~K}9 K
T2=309 K\mathrm{T}_{2}=309 \mathrm{~K}T2=309 K
Ea=?\mathrm{Ea}=?Ea=?
log K2 K1=Ea2.3 R[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 K1K2=2.3 REa[ T2 ⋅ T1T2−T1]
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]log 2=2.3 × 8.3Ea[309 × 3009]
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}Ea=90.3 × 309 × 300 × 2.3 × 8.3
=58988.1 J / mole=58988.1 \mathrm{~J} / \mathrm{mole}=58988.1 J / mole
≃ 59 kJ / mole\simeq 59 \mathrm{~kJ} / \mathrm{mole}≃ 59 kJ / mole
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