Question

The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation. 

Given: R = 8. 314 J K⁻¹ mol⁻¹, log 4 = 0.6021

विकल्प

• 3.80 kJ/mol

• 3804 kJ/mol

• 38.04 kJ/mol

• 380.4 kJ/mol

Answer 1

38.04 kJ/mol

Explanation:

Given: Initial temperature (T₁) = 27°C = 300 K

Final temperature (T₂) = 57°C = 330 K

R = 8. 314 J K⁻¹ mol⁻¹

log 4 = 0.6021

By using the Arrhenius equation:

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

log 4 = `E_a/(2.303 xx 8.314) ((330 - 300)/(300 xx 330))`    ...[Since the rate quadruples, the ratio `k_2/k_1` = 4]

0.6021 = `E_a/19.147 (30/99000)`

E_a = `(0.6021 xx 19.147 xx 99000)/30`

= 38040 J mol⁻¹

= 38.04 kJ mol⁻¹

∴ The energy of activation for the reaction is 38.04 kJ mol⁻¹.