Question

For a reaction, the graph of \( \ln k \) (on y-axis) and \( 1/T \) (on x-axis) is a straight line with a slope \( -2 \times 10^4 \text{ K} \). The activation energy of the reaction (in kJ mol\(^{-1}\)) is \((R = 8.3~\text{J K}^{-1} \text{mol}^{-1})\)

• A. 332
• B. 432
• C. 166
• D. 216

Hint:

In Arrhenius plots, the slope equals \(-E_a/R\). Multiply the slope by \( -R \) to find activation energy. Convert to kJ if needed.

Answer 1

The Correct Option is C

Step 1: Use Arrhenius equation in linear form

\[ \ln k = \ln A - \frac{E_a}{R} . \frac{1}{T} \]

Comparing this with the straight-line equation \( y = mx + c \), the slope \( m = -\frac{E_a}{R} \)

Step 2: Plug in the known values

\[ \text{Slope} = -\frac{E_a}{R} = -2 \times 10^4 \]

\[ \Rightarrow \frac{E_a}{R} = 2 \times 10^4 \Rightarrow E_a = R . 2 \times 10^4 = 8.3 . 2 \times 10^4 = 1.66 \times 10^5~\text{J mol}^{-1} \]

\[ = \frac{1.66 \times 10^5}{1000} = 166~\text{kJ mol}^{-1} \]

Final Answer:

166 kJ mol\(^{-1}\)