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Question:

For the reaction of H₂ and I₂, the rate constant is 2.5 × 10⁻⁴ dm³ mol⁻¹ s⁻¹ at 327 °C and 1.0 dm³ mol⁻¹ s⁻¹ at 527 °C. The activation energy for the reaction, in kJ mol⁻¹ is: (R = 8.314 J K⁻¹ mol⁻¹)

150

72

166

59

Solution:

Correct option is B. 166
H₂(g) + I₂(g) → 2HI(g)
Apply Arrhenius equation
log₁₀(K₂/K₁) = Ea/2.303R(1/T₁ - 1/T₂)
log₁₀(1/2.5 × 10⁻⁴) = Ea/2.303 × 8.314(1/600 - 1/800)
∴ Ea ≈ 166 kJ/mol