1.

The rate of most reactions becomes double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.

Answer»

Solution :`T_(1) = 298 K, T_(2) = 308 K`.
`R = 8.314 J mol^(-1) K^(-1)`
Activation energy
`K_(2) = 2K_(1)`
`log.(K_(2))/(K_(1)) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) XX T_(2))]`
`log 2 = (E_(a))/(2.303 xx 8.314)[(308 - 298)/(308 xx 298)]`
`0.3010 = (E_(a))/(19.147)[(10)/(91784)]`
`0.3010 = (E_(a))/(19.147) xx 1.08 xx 10^(-4)`
`E_(a) = (0.3010 xx 19.147)/(1.089 xx 10^(-4))`
`= (5.763)/(1.089 xx 10^(-4))`
`= 52922.37 J mol^(-1)`


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