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Solve the following.
A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K…
A first order reaction is $50\%$ completed in $40\text{ minutes}$ at $300\text{ K}$ and in $20\text{ minutes}$ at $320\text{ K}$. Calculate the activation energy of the reaction.
(Given: $\log 2 = 0\cdot3010$, $\log 4 = 0\cdot6021$, $R = 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}$)
Answer
Answer
Formulas used: Half-life, first order: t_half = 0.693/k with t_half = 40 min, so k = 0.01733 min^-1 Half-life, first order: t_half = 0.693/k with t_half = 20 min, so k = 0.03465 min^-1 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 = 320 K, k1 = 0.01733 min^-1, k2 = 0.03465 min^-1, so Ea = 27.66 kJ mol^-1
For a first order reaction $t_{1/2}=\dfrac{0.693}{k}$, so $k_1=\dfrac{0.693}{40}=0.01733\text{ min}^{-1}$ at 300 K and $k_2=\dfrac{0.693}{20}=0.03465\text{ min}^{-1}$ at 320 K, hence $\dfrac{k_2}{k_1}=2$.
$\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303\,R}\left(\dfrac{T_2-T_1}{T_1T_2}\right)$
$0.3010=\dfrac{E_a}{2.303\times8.314}\times\dfrac{20}{300\times320}$
$E_a=\dfrac{2.303\times8.314\times0.3010\times300\times320}{20}=27664\text{ J mol}^{-1}$
Final answer: 27.66 kJ mol^-1