A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK-1 mol-1 )

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

k_{50}=40min\Rightarrow 40=\frac{0.693}{k_{1}}

at T = 300K

k_{50}=20min at 320K \Rightarrow \frac{0.693}{k_{2}}

R = 8·314 JK-1 mol-1 

E_{a}=?

1st order reaction

\frac{k_{2}}{k_{1}}=\frac{\frac{0.693}{20}}{\frac{0.693}{40}}=\frac{40}{20}=2

\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303R}\left ( \frac{T_{2}-T_{1}}{T_{1}T_{2}} \right )

\log 2=\frac{E_{a}}{2.303\times 8.314}\left ( \frac{320-300}{320\times 300} \right )

E_{a}=\frac{0.3010\times 2.303\times 8.314\times 320\times 300}{20}

E_{a}=27764J

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