For a first-order reaction, the plot of ₁₀ k against 1 T is a straight line with a slope of -1.0 10^4 K . What is the activation energy of the reaction? (Given R = 8.314 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

According to the Arrhenius equation: k = A e^ - E_a RT Taking logarithm to the base 10 on both sides: ₁₀ k = ₁₀ A - E_a 2.303 R T Comparing this with the equation of a straight line y = mx + c , the plot of ₁₀ k vs 1 T gives a straight line with: Slope ( m ) = - E_a 2.303 R Given slope = -1.0 10^4 K - E_a 2.303 8.314 = -1.0 10^4 E_a = 1.0 10^4 2.303 8.314 J mol ⁻¹ E_a = 191471.4 J mol ⁻¹ E_a 191.5 kJ mol ⁻¹ Using - E_a R instead of - E_a 2.303 R leads to the incorrect answer of 83.1 kJ mol ⁻¹ . Answer: 191.5 kJ mol