Multiple Choice
Using the Arrhenius Equation, calculate the activation energy for the decomposition of acetaldehyde, CH3CHO, given that the rate constant is 1.1 × 10⁻² L mol⁻¹ s⁻¹ at 703 K and 4.95 L mol⁻¹ s⁻¹ at 865 K.
15. Chemical Kinetics
Arrhenius Equation
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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.
A
1.50 kJ/mol
B
15.0 kJ/mol
C
150 kJ/mol
D
75.0 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 \( E_a \): \( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \). Here, \( k_1 = 2.57 \text{ M}^{-1}\text{s}^{-1} \) at \( T_1 = 701 \text{ K} \) and \( k_2 = 567 \text{ M}^{-1}\text{s}^{-1} \) at \( T_2 = 895 \text{ K} \).
Calculate the natural logarithm of the ratio of the rate constants: \( \ln\left(\frac{567}{2.57}\right) \).
Calculate the difference in the inverse of the temperatures: \( \frac{1}{701} - \frac{1}{895} \).
Rearrange the equation to solve for \( E_a \): \( E_a = \frac{R \cdot \ln\left(\frac{567}{2.57}\right)}{\frac{1}{701} - \frac{1}{895}} \). Substitute the values and solve for \( E_a \).
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