Multiple Choice
The rate constant for a reaction is 1 × 10⁻³ M sec⁻¹ at 27°C and the Arrhenius frequency factor is 3500 sec⁻¹. What is the activation energy?
15. Chemical Kinetics
Arrhenius Equation
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Using the Arrhenius Equation, what is the activation energy (Ea) for the reaction between nitrogen dioxide and carbon monoxide, given that the rate constant at 701 K is 2.57 M⁻¹·s⁻¹ and at 895 K is 567 M⁻¹·s⁻¹?
A
125 kJ/mol
B
200 kJ/mol
C
150 kJ/mol
D
175 kJ/mol
Verified step by step guidance1
Identify the Arrhenius Equation: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.
Use the two-point form of the Arrhenius Equation to solve for \( E_a \): \( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \), where \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively.
Substitute the given values into the equation: \( k_1 = 2.57 \text{ M}^{-1}\cdot\text{s}^{-1} \), \( k_2 = 567 \text{ M}^{-1}\cdot\text{s}^{-1} \), \( T_1 = 701 \text{ K} \), and \( T_2 = 895 \text{ K} \).
Calculate the natural logarithm of the ratio of the rate constants: \( \ln\left(\frac{567}{2.57}\right) \).
Solve for \( E_a \) by rearranging the equation: \( E_a = R \cdot \ln\left(\frac{567}{2.57}\right) \cdot \left(\frac{1}{701} - \frac{1}{895}\right)^{-1} \).
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