Question

What is the activation energy for the decomposition of $N_{2}O_{5}$ as,

$N_{2}O_{5}\rightleftharpoons2N O_{2}+\frac{1}{2}O_{2}$

If the values of rate constant are $3.45\times 10^{- 5} \, $ at $27℃$ and $=6.9\times 10^{- 3} \, at \, 67℃?$

• 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,

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

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

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

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

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

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

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

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