Doubling the absolute temperature of a reaction with Ea = 50 kJ/mol from 300 K to 600 K changes the rate constant by approximately

A factor of about 2.7 \times 10^4 k₂/k₁ = exp[Ea/R · (1/T₁ - 1/T₂)] = exp[50000/8.314 · (1/300 - 1/600)] = exp[6014 · 0.001667] = exp(10.02) ≈ 2.25e4.

A factor of about 2.7 \times 10^4

Collision Model

AP Chemistry· difficulty 4/5

E_a = 50

A
$Decreases by half$
B
$2.7 \times 10^4$
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C
$A factor of 2$
D
$No change$

Explanation

k₂/k₁ = exp[Ea/R \cdot (1/T₁ - 1/T₂)] = exp[50000/8.314 \cdot (1/300 - 1/600)] = exp[6014 \cdot 0.001667] = exp(10.02) \approx 2.25e4.

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