Question

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be ______. (R = 8.314 JK⁻¹ mol⁻¹ and log 2 = 0.301)

पर्याय

• 53.6 kJ mol⁻¹

• 48.6 kJ mol⁻¹

• 58.5 kJ mol⁻¹

• 60.5 kJ mol⁻¹

Answer 1

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be 53.6 kJ mol⁻¹.

Explanation:

Given: `k_2/k_1 = 2` (rate doubles)

T₁​ = 300 K

T₂ = 310 K

R = 8.314 JK⁻¹ mol⁻¹

log 2 = 0.301

By using the Arrhenius equation in the logarithmic form:

`log (k_2/k_1) = E_a/2.303 R ((T_2 - T_1)/(T_1T_2))`

`log (2) = E_a/(2.303 xx 8.314) ((310 - 300)/(300 xx 310))`

`0.301 = E_a/19.147 xx 10/93000`

`0.301 = (E_a xx 10)/(19.147 xx 93000)`

`0.301 = (10 E_a)/1782671`

`E_a = (0.301 xx 1782671)/10`

= `536884/10`

= 53688 J/mol

∴ E_a = 53.7 kJ/mol