The rate constants of a reaction at 500 K and 700 K are `0.02 s^(-1)` and `0.07 s^(-1)` respectively. Calculate the values of `E_a` and A.

Answer

`E_a = 18.231 KJ ; A = 1.604 `

Explanation

`K_1 = 0.02 S^(-1) .,T_1 = 500 K , k_2 = 0.07 s^(-1) ,T_2 = 700 K`
According to Arrhenius equation
`log (k_2)/(k_1) = (E_a)/( 2.303 R) [(T_2 -T_1)/(T_1 T_2)]`
`or log (0.07 )/( 0.02 )= ( E_a)/( 8.314 xx 2.303 JK^(-1) mol^(-1) ) [ (7000-500)/( 700 xx 500 )]`
` or [ log 0.07 - log 0.02 ]`
` = (E_a)/(2.303 xx 8.314 JK^(-1) mol^(-1)) xx(200 )/( 350000)`
` or [-1.1549 - ( -1.6990 )]`
` = (E_a)/(2.303 xx 8.314 JK^(-1) mol^(-1)) xx( 200 )/( 350000 )`
` or ( 0.5441 xx 8.314 xx 350000 )/( 200 ) xx 2.303 = E_a`
` or E_a = 18231.4 J = 18.231 KJ mol^(-1)`
Now ` log K_1 = log A- (E_a)/(2.303 RT_1)`
Now ` log k_1 = loga - (E_a)/(2.303 RT_1)`
` or log 0.02 = logA - ( 18231.4 )/( 2.303 xx 8.314 xx 500 )`
` - 1.6990 = log A - 1.9043 `
` or log A=- 1.6990 + 1.9043 `
` or log A= 0.2053`
` or log A = 0.2053`
` or A=A ` antilog `0.2053 = 1.6043`