CBSE Class 12 Chemistry 2018 Solved Paper

Question : 13 of 26

Marks: +1, -0

A first order reaction is

50 \%

completed in

{4 0}

minutes at

300 {\K}

and in 20 minutes at

320 {\K}

. Calculate the activation energy of the reaction.

\;\;\text " (Given : "\; \log 2=0.3010, \log 4=0.6021,

R=8.314 \; {\JK}^{-1} {\mol}^{-1} \;\text " ) "\;

Solution:

\;k_2=0.693 \/ 20 \;\text ", "\;

\;k_1=0.693 \/ 40

\;\log \;{k_2}/{k_1}=\;{E_a}/{2.303 R} [\;{1}/{T_1}-\;{1}/{T_2}]

\;k_2 \/ k_1=2

\;\log 2=\;{E_a}/{2.303 × 8.314} [\;{320-300}/{320 × 300}]

\;E_a=27663.8 {\J} \/ {\mol} \;\text " or "\; 27.66 {\kJ} \/ {\mol}

\;

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