The rate coefficient (k) for a particular reactions is $$1.3 \times 10^{-4}$$ M$$^{-1}$$ s$$^{-1}$$ at 100°C, and $$1.3 \times 10^{-3}$$ M$$^{-1}$$ s$$^{-1}$$ at 150°C. What is the energy of activation (E$$_A$$) (in kJ) for this reaction? (R = molar gas constant = 8.314 JK$$^{-1}$$ mol$$^{-1}$$)

The activation energy ($$E_A$$) can be found using the Arrhenius equation, which relates the rate constant ($$k$$) to temperature ($$T$$) and activation energy. The formula for two different temperatures is:

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

where:

• $$k_1$$ and $$k_2$$ are the rate constants at temperatures $$T_1$$ and $$T_2$$ respectively,
• $$R$$ is the gas constant (8.314 J K⁻¹ mol⁻¹),
• $$E_A$$ is the activation energy in J/mol.

First, convert the temperatures from Celsius to Kelvin:

$$T_1 = 100^\circ \text{C} + 273 = 373 \text{K}$$

$$T_2 = 150^\circ \text{C} + 273 = 423 \text{K}$$

Given:

$$k_1 = 1.3 \times 10^{-4} \text{M}^{-1} \text{s}^{-1}$$

$$k_2 = 1.3 \times 10^{-3} \text{M}^{-1} \text{s}^{-1}$$

Calculate the ratio $$\frac{k_2}{k_1}$$:

$$\frac{k_2}{k_1} = \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} = 10$$

Take the natural logarithm of the ratio:

$$\ln\left(\frac{k_2}{k_1}\right) = \ln(10) \approx 2.302585$$

Now, compute the difference in the reciprocal temperatures:

$$\frac{1}{T_1} = \frac{1}{373} \approx 0.002681$$

$$\frac{1}{T_2} = \frac{1}{423} \approx 0.002364$$

$$\frac{1}{T_1} - \frac{1}{T_2} = 0.002681 - 0.002364 = 0.000317$$

Alternatively, use the exact fraction:

$$\frac{1}{T_1} - \frac{1}{T_2} = \frac{T_2 - T_1}{T_1 T_2} = \frac{423 - 373}{373 \times 423} = \frac{50}{373 \times 423}$$

Calculate $$373 \times 423$$:

$$373 \times 400 = 149200$$

$$373 \times 23 = 373 \times 20 + 373 \times 3 = 7460 + 1119 = 8579$$

$$373 \times 423 = 149200 + 8579 = 157779$$

So,

$$\frac{1}{T_1} - \frac{1}{T_2} = \frac{50}{157779} \approx 0.000317$$

Now rearrange the Arrhenius equation to solve for $$E_A$$:

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

Substitute the values:

$$E_A = \frac{8.314 \times 2.302585}{0.000317}$$

First, compute the numerator:

$$8.314 \times 2.302585 = 19.14369169$$

Now divide by the denominator:

$$E_A = \frac{19.14369169}{0.000317} \approx 60402.77 \text{J/mol}$$

Alternatively, using the exact fraction:

$$E_A = \frac{8.314 \times 2.302585 \times 157779}{50}$$

First, compute $$8.314 \times 157779 = 1311774.606$$

Then, $$1311774.606 \times 2.302585 \approx 3020890.06$$

Then, divide by 50:

$$\frac{3020890.06}{50} = 60417.8012 \text{J/mol}$$

Both methods give approximately 60400 J/mol. Convert to kJ/mol by dividing by 1000:

$$E_A \approx 60.4 \text{kJ/mol}$$

Rounding to the nearest whole number, $$E_A \approx 60 \text{kJ/mol}$$.

Comparing with the options:

• A. 16
• B. 60
• C. 99
• D. 132

Hence, the correct answer is Option B.