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The rate of the chemical reaction double...

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate `E_a`.

Text Solution

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According to Arrhenius equation,
`log"" (k_2)/(k_1) = (E_a)/( 2.303 R) [ (1)/( T_1 ) - (1)/( T_2) ]`
`(k_2)/( k_1) = 2, T_1 = 298 K, T_2 = 308 K`,
`R = 8.314" JK"^(-1)" mol"^(-1)`
`therefore log2 = (E_a)/( 2.303 xx (8.314" JK"^(-1)" mol"^(-1) ) ) [ (1)/( 298" K") - (1)/( 308" K" ) ]`
or `0.3010 = (E_a)/( 2.303 xx (8.314" JK"^(-1)" mol"^(-1) ) ) xx (10)/( 298 xx 308)`
`E_a = (0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/( 10)("J mol"^(-1) ) `
`= 52897.78" J mol"^(-1) = 52.898" kJ mol"^(-1)`
Activation energy for the reaction `=52.898" kJ mol"^(-1)`
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