Saurabh Koranglekar
Last Activity: 4 Years ago
To determine the activation energy (Ea) of the reaction, we use the Arrhenius equation in its logarithmic form:
k₂/k₁ = e^[(Ea/R) * (1/T₁ - 1/T₂)]
Taking the natural logarithm on both sides:
ln(k₂/k₁) = (Ea/R) * (1/T₁ - 1/T₂)
Given that the rate doubles when temperature increases from 300 K to 310 K, we have:
k₂/k₁ = 2
Taking the logarithm (base 10) on both sides:
log 2 = (Ea / 2.303R) * (1/T₁ - 1/T₂)
Substituting the given values:
0.301 = (Ea / (2.303 * 8.314)) * (1/300 - 1/310)
First, calculate (1/300 - 1/310):
1/300 = 0.003333
1/310 = 0.003226
(1/300 - 1/310) = 0.000107
Now, calculate 2.303 * 8.314:
2.303 * 8.314 = 19.14
Now solve for Ea:
0.301 = (Ea / 19.14) * 0.000107
Ea = (0.301 * 19.14) / 0.000107
Ea = 53858.88 J/mol
Ea ≈ 53.86 kJ/mol
Final Answer: Activation energy (Ea) ≈ 53.86 kJ/mol