Scelta multipla
Using the Arrhenius Equation, calculate the activation energy (Ea) for the reaction between nitrogen dioxide and carbon monoxide given the rate constants at 701 K and 895 K are 2.57 M⁻¹s⁻¹ and 567 M⁻¹s⁻¹, respectively.
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15. Chemical Kinetics
Arrhenius Equation
Scelta multipla
Given the rate constants k_1 = 2.5 \(\times\) 10^{-3} \(\text{ s}\)^{-1} at 300 \(\text{ K}\) and k_2 = 1.2 \(\times\) 10^{-2} \(\text{ s}\)^{-1} at 320 \(\text{ K}\), what is the activation energy (E_a) of the reaction? (R = 8.314 \(\text{ J mol}\)^{-1} \(\text{K}\)^{-1})
A
54.2 \(\text{ kJ mol}\)^{-1}
B
34.7 \(\text{ kJ mol}\)^{-1}
C
21.6 \(\text{ kJ mol}\)^{-1}
D
8.3 \(\text{ kJ mol}\)^{-1}
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Identify the given data: rate constants \(k_1 = 2.5 \times 10^{-3} \text{ s}^{-1}\) at temperature \(T_1 = 300 \text{ K}\), and \(k_2 = 1.2 \times 10^{-2} \text{ s}^{-1}\) at temperature \(T_2 = 320 \text{ K}\). The gas constant is \(R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}\).
Recall the Arrhenius equation in its two-point form to find the activation energy \(E_a\):
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
Rearrange the equation to solve for \(E_a\):
\(E_a = -R \times \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\)
Calculate the natural logarithm of the ratio of rate constants: \(\ln\left(\frac{k_2}{k_1}\right)\), and compute the difference in the reciprocals of the temperatures: \(\frac{1}{T_2} - \frac{1}{T_1}\).
Substitute the values of \(R\), \(\ln\left(\frac{k_2}{k_1}\right)\), and \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) into the rearranged formula to find \(E_a\). Remember to convert \(E_a\) from joules per mole to kilojoules per mole by dividing by 1000.
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