Question

For a first order decomposition of a certain reaction, rate constant is given by the equation 
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\)) 
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.

• A. 161.1
• B. 171.1
• C. 181.1
• D. 191.1

Hint:

When you see a rate constant equation in the form \(\log k = C - B/T\), immediately identify the slope term `B` with \(E_a/(2.303R)\). This allows you to quickly set up the equation \(E_a = B \times 2.303 \times R\) to find the activation energy.

Answer 1

The Correct Option is D

Step 1: Arrhenius Equation form: The standard Arrhenius equation is: \[ k = A e^{-E_a/RT} \] Taking log on both sides: \[ \ln k = \ln A - \frac{E_a}{RT} \] Convert to base 10: \[ 2.303 \log k = 2.303 \log A - \frac{E_a}{RT} \] \[ \log k = \log A - \frac{E_a}{2.303 R T} \]
Step 2: Comparing with given equation: Given: \( \log k = 7.14 - \frac{1 \times 10^4}{T} \) Comparing coefficients of \( \frac{1}{T} \): \[ \frac{E_a}{2.303 R} = 1 \times 10^4 \]
Step 3: Calculating \( E_a \): \[ E_a = 10^4 \times 2.303 \times R \] Given \( R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1} \). \[ E_a = 10000 \times 2.303 \times 8.3 \] \[ E_a = 10000 \times 19.1149 \] \[ E_a = 191149 \, \text{J mol}^{-1} \] \[ E_a \approx 191.1 \, \text{kJ mol}^{-1} \]