Question

For a reaction at $300 \text{ K}$, on addition of catalyst, activation energy of reaction lowered by $10 \text{ kJ}$. Then calculate the value of $\log \frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}$

• A. $1.74$
• B. $0.174$
• C. $17.4$
• D. $3.48$

Hint:

A catalyst increases the rate constant exponentially with respect to the decrease in activation energy, as defined by the Arrhenius equation derivation: $\log(K_{\text{cat}}/K_{\text{uncat}}) = \Delta E_a/(2.303RT)$.

Answer 1

The Correct Option is A

The question requires calculating the change in the rate constant when a catalyst lowers the activation energy of a reaction. For this, we need to use the Arrhenius equation, which describes how the rate constant \( K \) is affected by temperature and activation energy:

\(K = A \cdot e^{-E_a / (R \cdot T)}\)

where:

• \( K \) = rate constant
• \( A \) = pre-exponential factor
• \( E_a \) = activation energy
• \( R \) = universal gas constant = 8.314 J/mol·K
• \( T \) = temperature in Kelvin

The problem states that the temperature \( T \) = 300 K and the catalyst lowers the activation energy by 10 kJ/mol, which we convert to joules: \( 10 \times 10^3 \, \text{J/mol} \).

Let's find the change in the natural log of the rate constants for catalyzed and uncatalyzed reactions:

\(\log \left(\frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}\right) = \frac{E_{\text{uncatalysed}} - E_{\text{catalysed}}}{2.303 \cdot R \cdot T}\)

Given:

• \( E_{\text{uncatalysed}} - E_{\text{catalysed}} = 10,000 \, \text{J/mol} \)

Substitute these values into the equation:

\(\log \left(\frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}\right) = \frac{10,000}{2.303 \cdot 8.314 \cdot 300}\)

Calculating the denominator:

\(2.303 \cdot 8.314 \cdot 300 = 5748.462 \, \text{J/mol}\)

Now, calculate the log value:

\(\log \left(\frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}\right) = \frac{10,000}{5748.462} \approx 1.74\)

Therefore, the value of \(\log \frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}\) is approximately 1.74. Hence, the correct option is 1.74.