Question

The slope of Arrhenius Plot (\(\ln k\) v/s \(\frac{1}{T}\)) of the first-order reaction is \(-5 \times 10^3 \, K\). The value of \(E_a\) of the reaction is. Choose the correct option for your answer. [Given \(R = 8.314 \, JK^{-1}mol^{-1}\)]

• A. \(166 \, kJ \, mol^{-1}\)
• B. \(-83 \, kJ \, mol^{-1}\)
• C. \(41.5 \, kJ \, mol^{-1}\)
• D. \(83.0 \, kJ \, mol^{-1}\)

Hint:

For Arrhenius plots: \[ \ln k \text{ vs } \frac{1}{T} \Rightarrow \text{Slope} = -\frac{E_a}{R} \] while \[ \log k \text{ vs } \frac{1}{T} \Rightarrow \text{Slope} = -\frac{E_a}{2.303R} \] Always check whether the logarithm is natural log or common log.

Answer 1

The Correct Option is C

Concept: The Arrhenius equation explains the dependence of rate constant on temperature. It is expressed as: \[ k = A e^{-\frac{E_a}{RT}} \] where:
• \(k\) = rate constant
• \(A\) = Arrhenius frequency factor
• \(E_a\) = activation energy
• \(R\) = gas constant
• \(T\) = absolute temperature Taking natural logarithm on both sides converts the equation into linear form: \[ \ln k = \ln A - \frac{E_a}{RT} \] This equation represents the equation of a straight line: \[ y = mx + c \] Hence, for a graph of \(\ln k\) versus \(\frac{1}{T}\): \[ \text{Slope} = -\frac{E_a}{R} \]

Step 1: Writing the relation between slope and activation energy.
For Arrhenius plot: \[ \text{Slope} = -\frac{E_a}{R} \] Given: \[ \text{Slope} = -5 \times 10^3 \, K \] Substituting into the equation: \[ -\frac{E_a}{R} = -5 \times 10^3 \] Removing negative signs: \[ \frac{E_a}{R} = 5 \times 10^3 \] Therefore: \[ E_a = 5 \times 10^3 \times R \]

Step 2: Substituting the value of gas constant.
Given: \[ R = 8.314 \, JK^{-1}mol^{-1} \] Thus: \[ E_a = 5 \times 10^3 \times 8.314 \] \[ E_a = 41570 \, Jmol^{-1} \]

Step 3: Converting Joule into kilojoule.
Since: \[ 1 \, kJ = 1000 \, J \] Therefore: \[ E_a = \frac{41570}{1000} \] \[ E_a = 41.57 \, kJmol^{-1} \] Approximating: \[ E_a \approx 41.5 \, kJmol^{-1} \] Final Answer: \[ \boxed{41.5 \, kJmol^{-1}} \]