Question

For a first-order reaction, a plot of \( \ln k \) (y-axis) and \( \frac{1}{T} \) (x-axis) gave a straight line with a slope equal to \( -10^4 \) and an intercept equal to 2.303 (on the y-axis). What is the activation energy \( E_a \) (in kJ/mol) of the reaction? (Given \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))

• A. 8.314 kJ/mol
• B. 2.303 kJ/mol
• C. 2303 kJ/mol
• D. 83.14 kJ/mol

Hint:

The slope of the \( \ln k \) vs. \( \frac{1}{T} \) plot is used to calculate the activation energy of a reaction using the Arrhenius equation.

Answer 1

The Correct Option is A

The Arrhenius equation is given by:
$$k = Ae^{-\frac{E_a}{RT}}$$
Taking the natural logarithm of both sides, we get:
$$\ln k = \ln A - \frac{E_a}{RT}$$
Comparing this equation with the equation of a straight line, $y = mx + c$, where $y = \ln k$ and $x = \frac{1}{T}$, we can see that:
Slope (m) = $-\frac{E_a}{R}$
Intercept (c) = $\ln A$
Given:
Slope = $-10^3$ K
Intercept = 2.303
R = 8.314 J mol$^{-1}$ K$^{-1}$
We have:
$$-\frac{E_a}{R} = -10^3$$
$$E_a = R \times 10^3$$
$$E_a = 8.314 \text{ J mol}^{-1} \text{K}^{-1} \times 10^3 \text{ K}$$
$$E_a = 8314 \text{ J mol}^{-1}$$
To convert J mol$^{-1}$ to kJ mol$^{-1}$, we divide by 1000:
$$E_a = \frac{8314}{1000} \text{ kJ mol}^{-1}$$
$$E_a = 8.314 \text{ kJ mol}^{-1}$$
Therefore, the activation energy ($E_a$) of the reaction is 8.314 kJ mol$^{-1}$. Final Answer: The final answer is $\boxed{8.314}$