in the rate constant with temperature is:

$ \log \frac{K_{T_2}}{K_{T_1}} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) $

Where:

$ K_{T_1} $ is the rate constant at temperature $ T_1 $.

$ K_{T_2} $ is the rate constant at temperature $ T_2 $.

$ E_a $ is the activation energy.

$ R $ is the universal gas constant.

Given Values:

Initial rate constant, $ K_{T_1} = 3.46 \times 10^{-2} \, \mathrm{s}^{-1} $ at $ T_1 = 298 \, \mathrm{K} $

Activation energy, $ E_a = 50.1 \, \mathrm{kJ/mol} = 50.1 \times 10^3 \, \mathrm{J/mol} $

Universal gas constant, $ R = 8.314 \, \mathrm{J/K \cdot mol} $

Temperature, $ T_2 = 350 \, \mathrm{K} $

Calculation Steps:

Substitute the given values into the formula:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = \frac{50.1 \times 10^3}{2.303 \times 8.314} \times \left( \frac{350 - 298}{298 \times 350} \right) $

Evaluate the expression:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = 1.30 $

Solve for $ K_{T_2} $:

$ \frac{K_{T_2}}{3.46 \times 10^{-2}} = 10^{1.30} = 19.95 $

Calculate $ K_{T_2} $:

$ K_{T_2} = 3.46 \times 10^{-2} \times 19.95 = 0.6903 \, \mathrm{s}^{-1} $

Thus, the rate constant at 350 K is approximately $ 0.692 \, \mathrm{s}^{-1} $.

Chemical Kinetics

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