‹ Back to the paper
Answer the following.
The rate constant for a reaction is at and at . Calculate the activation energy for the reaction. (…
The rate constant for a reaction is $0\cdot05\text{ s}^{-1}$ at $300\text{ K}$ and $0\cdot5\text{ s}^{-1}$ at $310\text{ K}$.
Calculate the activation energy for the reaction. ($R = 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}$)
Answer
Answer
AIFormula used: Arrhenius equation at two temperatures: log10(k2/k1) = Ea/(2.303*R)*(T2 - T1)/(T1*T2) with R = 8.314 J K^-1 mol^-1, T1 = 300 K, T2 = 310 K, k1 = 0.05 s^-1, k2 = 0.5 s^-1, so Ea = 178000 J mol^-1
$\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2 - T_1}{T_1 T_2}\right)$
$\log\dfrac{0.5}{0.05} = 1 = \dfrac{E_a}{2.303 \times 8.314}\times\dfrac{10}{300 \times 310}$
$E_a = \dfrac{2.303 \times 8.314 \times 300 \times 310}{10} = 1.78 \times 10^{5}\ \mathrm{J\ mol^{-1}}$
Final answer: 178 kJ mol^-1
From ISC 2026 Improvement Chemistry Paper 1, question 17(ii).
Check your working with the Chemistry numericals calculator: solutions, electrochemistry and kinetics formulas: fill in what you know and get the rest.