QuestionJuly 9, 2025

Given the following data for the reaction Aarrow B determine the activation energy, E_(a) of the reaction. k(s^-1) & T(mathrm(~K)) 3.10 times 10^-4 & 275.0 7.77 times 10^-3 & 375.0 a) 2.68kJ/mol b) 15.1kJ/mol c) 27.6kJ/mol d) 95.1kJ/mol e) 18.5kJ/mol

Given the following data for the reaction Aarrow B determine the activation energy, E_(a) of the reaction. k(s^-1) & T(mathrm(~K)) 3.10 times 10^-4 & 275.0 7.77 times 10^-3 & 375.0 a) 2.68kJ/mol b) 15.1kJ/mol c) 27.6kJ/mol d) 95.1kJ/mol e) 18.5kJ/mol
Given the following data for the reaction Aarrow B determine the activation energy, E_(a) of the reaction. k(s^-1) & T(mathrm(~K)) 3.10 times 10^-4 & 275.0 7.77 times 10^-3 & 375.0 a) 2.68kJ/mol b) 15.1kJ/mol c) 27.6kJ/mol d) 95.1kJ/mol e) 18.5kJ/mol

Solution
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Answer

c) 27.6 \, \text{kJ/mol} Explanation 1. Use the Arrhenius Equation The Arrhenius equation is k = A e^{-\frac{E_a}{RT}}. Taking the natural logarithm of both sides gives \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}. 2. Calculate Slope from Two Points Using the linear form \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A, calculate the slope m using two points (\frac{1}{T_1}, \ln k_1) and (\frac{1}{T_2}, \ln k_2). \ln(3.10 \times 10^{-4}) = -8.077, \ln(7.77 \times 10^{-3}) = -4.857. \frac{1}{275} = 0.003636, \frac{1}{375} = 0.002667. Slope m = \frac{-4.857 - (-8.077)}{0.002667 - 0.003636} = \frac{3.22}{-0.000969} = -3323.52. 3. Calculate Activation Energy **E_a = -m \cdot R**, where R = 8.314 \, \text{J/mol K}. E_a = 3323.52 \times 8.314 = 27628.5 \, \text{J/mol} = 27.6 \, \text{kJ/mol}.

Explanation

1. Use the Arrhenius Equation<br /> The Arrhenius equation is $k = A e^{-\frac{E_a}{RT}}$. Taking the natural logarithm of both sides gives $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$.<br /><br />2. Calculate Slope from Two Points<br /> Using the linear form $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$, calculate the slope $m$ using two points $(\frac{1}{T_1}, \ln k_1)$ and $(\frac{1}{T_2}, \ln k_2)$. <br /> $\ln(3.10 \times 10^{-4}) = -8.077$, $\ln(7.77 \times 10^{-3}) = -4.857$.<br /> $\frac{1}{275} = 0.003636$, $\frac{1}{375} = 0.002667$.<br /> Slope $m = \frac{-4.857 - (-8.077)}{0.002667 - 0.003636} = \frac{3.22}{-0.000969} = -3323.52$.<br /><br />3. Calculate Activation Energy<br /> **$E_a = -m \cdot R$**, where $R = 8.314 \, \text{J/mol K}$.<br /> $E_a = 3323.52 \times 8.314 = 27628.5 \, \text{J/mol} = 27.6 \, \text{kJ/mol}$.
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