Question

Two chemical reactions of the same order have equal frequency factor value. Their activation energies differ by 26.8 kJ/mol. At 300 K if \(k_2 = xk_1\), find the value of \(x\).

• A. \(4.631 \times 10^4\)
• B. \(1.143 \times 10^3\)
• C. \(2.286 \times 10^3\)
• D. 4.665

Hint:

When frequency factors are equal, use \( \frac{k_2}{k_1} = e^{(E_{a1}-E_{a2})/RT} \). Always convert kJ to J before substitution.

Answer 1

The Correct Option is A

Step 1: Use Arrhenius equation.
\[ k = A e^{-E_a/RT} \]
Since both reactions have same frequency factor \(A\), we take ratio:
\[ \frac{k_2}{k_1} = \frac{Ae^{-E_{a2}/RT}}{Ae^{-E_{a1}/RT}} \]
\[ \frac{k_2}{k_1} = e^{-(E_{a2}-E_{a1})/RT} \]

Step 2: Define activation energy difference.
Given:
\[ E_{a1} - E_{a2} = 26.8\ \text{kJ/mol} \]
Convert to joules:
\[ 26.8\ \text{kJ/mol} = 26800\ \text{J/mol} \]
Thus,
\[ \frac{k_2}{k_1} = e^{(E_{a1}-E_{a2})/RT} \]
\[ x = e^{26800/(RT)} \]

Step 3: Substitute values.
\[ R = 8.314\ \text{J mol}^{-1}\text{K}^{-1},\quad T = 300\ K \]
\[ x = e^{26800/(8.314 \times 300)} \]

Step 4: Simplify denominator.
\[ 8.314 \times 300 \approx 2494.2 \]
\[ x = e^{26800/2494.2} \]
\[ x \approx e^{10.74} \]

Step 5: Evaluate exponential.
\[ e^{10.74} \approx 4.631 \times 10^4 \]

Step 6: Interpret result.
Since \(k_2 = xk_1\), this gives the required ratio.

Step 7: Final conclusion.
\[ \boxed{4.631 \times 10^4} \]