Question
The slope of the line for the graph of $\log k$ vs $1 / \mathrm{T}$ for the reaction,$$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$$is $-5000 .$ Calculate the energy of activation of the reaction:(a) $95.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$(b) $9.57 \mathrm{~kJ} \mathrm{~mol}^{-1}$(c) $957 \mathrm{~kJ} \mathrm{~mol}^{-1}$(d) $0.957 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Step 1
Step 1: The Arrhenius equation is given by: $$ k = A e^{-E_a/RT} $$ where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature. Show more…
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