Question

What is the activation energy for the decomposition of $N_2{O}_5$ as,

$N_2{O}_5\rightleftharpoons 2N{O}_2+\frac{1}{2}{O}_2$

If the values of rate constant are $3.45\times 10^{-5}$ at $27℃$ and $=6.9\times 10^{-3}at67℃?$

• A. $\text{112.3 }\text{kJ}{\text{mol}}^{-\text{1}}$
• B. $\text{200}\text{.5}\text{kJ}{\text{mol}}^{-\text{1}}$
• C. $\text{149}\text{.5}\text{kJ}{\text{mol}}^{-\text{1}}$
• D. $\text{11}\text{.25}\text{kJ}{\text{mol}}^{-\text{1}}$

Answer 1

Answer: (A)

To be solved with the help of formula,

$\text{ log}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{2}\text{.303 R}}\left[\frac{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right]$

${\text{T}}_{\text{1}}=273+27=300\text{K}$

${\text{T}}_{\text{2}}=273+67=340\text{K}$

$\text{log}\frac{6.9\times 10^{-3}}{3.45\times 10^{-5}}=\frac{{\text{E}}_{\text{a}}}{2.303\times 8.31}\left[\frac{340-300}{340\times 300}\right]$

$\text{log}200=\frac{{\text{E}}_{\text{a}}}{19.1379}\times \frac{40}{102000}$

$2.3010=\frac{{\text{E}}_{\text{a}}}{19.14}\times \frac{4}{10200}$

${\text{E}}_{\text{a}}=\frac{19.14\times 10200\times 2.3010}{4}$

$\text{=}\text{112304}\text{.907 J}{\text{mol}}^{-\text{1}}\text{=}\text{112}\text{.3 kJ}{\text{mol}}^{-\text{1}}$