객관식
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?
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15. Chemical Kinetics
Arrhenius Equation
객관식
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
검증된 단계별 안내1
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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