A) 72
B) 166
C) 150
D) 59
Correct Answer: B
Solution :
\[{{H}_{2}}(g)+{{I}_{2}}(g)\to 2HI(g)\] Apply Arrhenius equation \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{1}{600}-\frac{1}{800} \right)\] \[\log \frac{1}{2.5\times {{10}^{-4}}}=\frac{{{E}_{a}}}{2.303\times 8.31}\left( \frac{200}{600\times 800} \right)\] \[\therefore \]\[{{\text{E}}_{\text{a}}}\approx 166\text{kJ/mol}\]You need to login to perform this action.
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