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Solve the following.
A first order reaction is 50% completed in 30 minutes at 300 K and in 10 minutes at 320 K…
A first order reaction is $50\%$ completed in $30\text{ minutes}$ at $300\text{ K}$ and in $10\text{ minutes}$ at $320\text{ K}$. Calculate the activation energy of the reaction. ($R = 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}$)
Answer
Answer
Formulas used: Half-life, first order: t_half = 0.693/k with t_half = 30 min, so k = 0.0231 min^-1 Half-life, first order: t_half = 0.693/k with t_half = 10 min, so k = 0.0693 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 = 0.0231 min^-1, k2 = 0.0693 min^-1, so Ea = 43.85 kJ mol^-1
For a first order reaction $k = \dfrac{0\cdot693}{t_{1/2}}$:
$k_1 = \dfrac{0\cdot693}{30} = 0\cdot0231\text{ min}^{-1}$ at $300\text{ K}$; $k_2 = \dfrac{0\cdot693}{10} = 0\cdot0693\text{ min}^{-1}$ at $320\text{ K}$
$\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2\cdot303\,R}\left(\dfrac{T_2-T_1}{T_1T_2}\right)$
$\log 3 = \dfrac{E_a}{2\cdot303\times8\cdot314}\times\dfrac{20}{300\times320}$
$E_a = 43850\text{ J mol}^{-1}$
Final answer: 43.85 kJ mol^-1