Question

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature \(T(K)\): \[ \ln k = 14.34-\frac{1.25\times10^4}{T} \] Given \(R=1.987\text{ cal mol}^{-1}\text{K}^{-1}\). The activation energy of the reaction is:

• A. \(14.34\text{ kcal mol}^{-1}\)
• B. \(18.63\text{ kcal mol}^{-1}\)
• C. \(24.84\text{ kcal mol}^{-1}\)
• D. \(12.42\text{ kcal mol}^{-1}\)

Hint:

Compare the given equation with \(\ln k=\ln A-\frac{E_a}{RT}\). The coefficient of \(\frac{1}{T}\) gives \(\frac{E_a}{R}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the activation energy ($E_a$) of a chemical reaction. We are given an empirical logarithmic expression for the rate constant $k$ as a function of temperature $T$. We need to compare this expression with the theoretical Arrhenius equation to extract the value of $E_a$. Note that the final answer is required in $kcal/mol$, but the gas constant $R$ is given in $cal/mol \cdot K$.
Step 2: Key Formula or Approach:
The Arrhenius Equation in its logarithmic form is: \[ \ln k = \ln A - \frac{E_a}{RT} \] By comparing this with the provided equation: \[ \ln k = 14.34 - \frac{1.25 \times 10^4}{T} \] We can see that the term subtracted from $14.34$ must correspond to $E_a/RT$. Step 3: Detailed Explanation:

Equating terms: $\frac{E_a}{RT} = \frac{1.25 \times 10^4}{T}$.
The temperature $T$ cancels out from both sides, leaving: $\frac{E_a}{R} = 1.25 \times 10^4$.
Calculating $E_a$ in calories: $E_a = 1.25 \times 10^4 \times R$ $E_a = 12500 \times 1.987 = 24837.5 \text{ cal/mol}$.
Converting to kcal/mol: Since $1 \text{ kcal} = 1000 \text{ cal}$, we divide the result by $1000$. $E_a = \frac{24837.5}{1000} = 24.8375 \text{ kcal/mol}$.
This rounds perfectly to $24.84 \text{ kcal/mol}$, which is option (A).
Step 4: Final Answer:
The energy of activation is $24.84 kcal mol^{-1}$.