Question

The rate constant of a first-order reaction is \( 3.46 \times 10^3 \, {s}^{-1} \) at 298K. What is the rate constant of the reaction at 350 K if its activation energy is 50.1 kJ mol\(^{-1}\) (R = 8.314 J K\(^{-1}\) mol\(^{-1}\))?

• A. 0.592 s\(^{-1}\)
• B. 0.692 s\(^{-1}\)
• C. 0.792 s\(^{-1}\)
• D. 0.892 s\(^{-1}\)

Hint:

Use the Arrhenius equation to calculate the rate constant at a new temperature by substituting the given activation energy, temperatures, and rate constants.

Answer 1

The Correct Option is B

We can calculate the rate constant at 350 K using the Arrhenius equation: \[ k_2 = k_1 \exp \left( \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \right) \] where: - \( k_1 \) is the rate constant at temperature \( T_1 \) (298 K),
- \( k_2 \) is the rate constant at temperature \( T_2 \) (350 K),
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.
Step 1:
Given: - \( k_1 = 3.46 \times 10^3 \, {s}^{-1} \),
- \( E_a = 50.1 \, {kJ/mol} = 50100 \, {J/mol} \),
- \( T_1 = 298 \, {K} \),
- \( T_2 = 350 \, {K} \),
- \( R = 8.314 \, {J/mol·K} \).
Step 2:
Substitute the values into the Arrhenius equation: \[ k_2 = 3.46 \times 10^3 \exp \left( \frac{-50100}{8.314} \left( \frac{1}{350} - \frac{1}{298} \right) \right). \] Step 3:
Calculate the exponent: \[ \frac{1}{350} - \frac{1}{298} = -0.000231, \] \[ \frac{-50100}{8.314} = -6026.66, \] \[ k_2 = 3.46 \times 10^3 \exp \left( 6026.66 \times 0.000231 \right) = 3.46 \times 10^3 \times \exp(1.39). \] Step 4:
Now, calculate the final rate constant: \[ k_2 = 3.46 \times 10^3 \times 4.01 = 0.692 \, {s}^{-1}. \] Thus, the rate constant at 350 K is \( 0.692 \, {s}^{-1} \).

Answer 2

To find the rate constant at 350 K for a reaction with activation energy 50.1 kJ mol\(^{-1}\), we use the Arrhenius equation:

\( k = A e^{-\frac{E_a}{RT}} \)

In this expression:

• \(k\) is the rate constant.
• \(A\) is the pre-exponential factor.
• \(E_a\) is the activation energy.
• \(R\) is the universal gas constant (\(8.314 \text{ J K}^{-1} \text{ mol}^{-1}\)).
• \(T\) is the temperature in Kelvin.

For comparing rate constants at two different temperatures (\(T_1\) and \(T_2\)), the equation is:

\( \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \)

Given:

• \(k_1 = 3.46 \times 10^3 \, \text{s}^{-1}\) at \(T_1 = 298 \text{ K}\)
• \(E_a = 50.1 \times 10^3 \text{ J mol}^{-1}\)
• \(T_2 = 350 \text{ K}\)

Substitute these values into the equation:

\( \ln \left( \frac{k_2}{3.46 \times 10^3} \right) = \frac{-50.1 \times 10^3}{8.314} \left( \frac{1}{350} - \frac{1}{298} \right) \)

Calculate the terms:

\( \ln \left( \frac{k_2}{3.46 \times 10^3} \right) = \frac{-50.1 \times 10^3}{8.314} \times (-0.000476) \approx 2.013 \)

Solve for \(k_2\):

\( \frac{k_2}{3.46 \times 10^3} \approx e^{2.013} \approx 7.49 \)

Thus,

\( k_2 \approx 7.49 \times 3.46 \times 10^3 \approx 0.692 \, \text{s}^{-1} \)

The rate constant at 350 K is 0.692 s\(^{-1}\).