Answer:- 

Given:

Rate constants; k₂ = 2k₁,

Temperatures: T₁ = 303 K, T₂ = 313 K

To find:

Activation energy of the reaction (E_a)

Formula:

log₁₀ (k₂/k₁) = E_a / (2.303R) × ((T₂ - T₁) / (T₂T₁))

Calculation:

log₁₀ (2k₁/k₁) = E_a / (2.303 × 8.314 J K^-1 mol^-1) × ((313 K - 303 K) / (313 K × 303 K))

log₁₀ 2 = E_a / (2.303 × 8.314 J mol^-1 K^-1) × (10 / (313 × 303))

0.3010 = E_a / (19.147 J mol^-1) × 1.0544 × 10^-4

E_a = (0.3010 × 19.147) / (1.0544 × 10^-4) J mol^-1

E_a = 54659 J mol^-1 = 54.66 kJ mol^-1

The energy of activation of the reaction is 54.66 kJ mol^-1.