1.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400 K ?(R = 8.314 JK^(-1)mol^(-1))

Answer»

`234.65 kJ mol^(-1) K^(-1)`
`434.65 kJ mol^(-1) K^(-1)`
`434.65 J mol^(-1) K^(-1)`
`334.65 J mol^(-1) K^(-1)`

Solution :`T_(1) = 200K, k = k_(1) , T_(2) = 400K, k= k_(2) = 2k_(1)`
log `((k_(2))/(k_(1))) = (2.303 E_(a))/(R ) ((T_(2) - T_(1))/(T_(1) T_(2)))`
`log ((2k_(2))/(k_(1))) = (2.303 E_(a))/(8.314J K^(-1) mol^(-1)) ((400K - 200K)/(200 K xx 400 K))`
`E_(a) = (0.3010 xx 8.314J mol^(-1) xx 200 xx 400)/(2.303 xx 200)`
`E_(a) = 434.65 J mol^(-1)`


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