The value of rate constant for a second order reaction is `6.7 xx 10^(-5) mol^(-) L s^(-)` at 298 K and `1.64 xx 10^(-4) mol^(-) L s^(-) ` at 313 K. Find the Arrhenius frequency factor A and activation energy of the reaction.

Answer

8649.48 collisions `s^(-)` at 298 K, `E_a = 46.2794.kJ mol^(-)`

Explanation

`k_1 = 6.7 xx 10^(-5) mol L^(-1) s^(-1) , T_1 = 298 K`
` k_2 = 1.64 xx 10^(-4) mol L^(-1) s^(-1) , T_2 =313 K`
`R= 8.314 JK^(-1) mol^(-1)`
` E_a = log (k_2)/(k_1) xx ( 2.303 R xx T_1 T_2)/( T_2 -T_1)`
`= ( 2.303 xx 8.314 xx 298 xx 313 )/((313 -298 )) xx log ( 1.64 xx 10^(-4))/( 6.7 xx 10^(-5))`
` = 119062.03 xx log 2.4477 = 119062.03 xx 0.3887 `
`= 46279.4 J = 46.2794 KJ`
Now` log K_1 = logA - (E_a)/(2.303 RT)`
or log `6.7 xx 10^(-5) = log A - ( 46279. 4 )/( 2.303 xx 8.314 xx 298 )`
` or log 6.7 + log 10^(-5) = log A - 8.111 `
`0.826 - 5 +8.111 = log A or log A or log A= 3.937 `
` A = ` antilog `3.937 = 8649.68 ` collisions `s^(-1)`