Question

At T(K), the following equation is obtained for a first order reaction.
\[ \log \frac{k}{A} = -\frac{x}{T} \] The activation energy for this reaction is equal to (R = gas constant)

• A. \(2.303 \times R\)
• B. \(\frac{2.303 \, R}{x}\)
• C. \(\frac{x}{2.303 \, R}\)
• D. \(\frac{1}{2.303 \times R}\)

Hint:

Compare the given expression with the Arrhenius equation to identify activation energy. For logarithmic form, remember: \(\log(k/A) = -E_a/(2.303RT)\).

Answer 1

The Correct Option is A

Step 1: Compare with the Arrhenius equation. Arrhenius equation for a first-order reaction: \[ \log \left( \frac{k}{A} \right) = -\frac{E_a}{2.303 \, R \, T} \] Step 2: Match given equation. Given: \[ \log \left( \frac{k}{A} \right) = -\frac{x}{T} \] So comparing: \[ \frac{x}{T} = \frac{E_a}{2.303 \, R \, T} \Rightarrow x = \frac{E_a}{2.303 \, R} \Rightarrow E_a = 2.303 \, R \, x \] Hence, activation energy \(E_a = 2.303 \times R \times x\) So \(E_a/x = 2.303 \, R\)