For a certain chemical reaction, the plot of k versus 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 the natural logarithm on both sides yields: k = A - E_a RT This represents the equation of a straight line ( y = mx + c ), where y = k and x = 1 T . The slope ( m ) of this line is - E_a R . Equating the given slope to the theoretical expression: - E_a R = -1.0 10^4 K E_a = 1.0 10^4 8.314 J mol ⁻¹ E_a = 83140 J mol ⁻¹ = 83.14 kJ mol ⁻¹ Using the factor 2.303 is a common mistake that occurs when confusing the k plot with a k plot. Answer: 83.14 kJ mol ⁻¹