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Question

At $27^\circ\text{C}$ in presence of a catalyst, activation energy of a reaction is lowered by $10 \text{ kJ mol}^{-1}$. The logarithm of ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is....
(Consider that the frequency factor for both the reactions is same)

The correct answer is
1.741

Catalyst Impact on Reaction Rate Constant

The effect of a catalyst on a reaction rate is typically explained using the Arrhenius equation. A catalyst increases the reaction rate by lowering the activation energy ($E_a$) without changing the frequency factor ($A$).

Applying the Arrhenius Equation

The Arrhenius equation relates the rate constant ($k$) to temperature ($T$) and activation energy ($E_a$):

$k = A e^{-E_a / (RT)}$

Taking the natural logarithm:

$\ln k = \ln A - \frac{E_a}{RT}$

Where:

  • $k$ is the rate constant
  • $A$ is the frequency factor
  • $E_a$ is the activation energy
  • $R$ is the ideal gas constant ($8.314 \text{ J mol}^{-1} \text{K}^{-1}$)
  • $T$ is the absolute temperature in Kelvin

Calculating the Log Ratio of Rate Constants

Let $k_u$ and $E_{a,u}$ be the rate constant and activation energy for the uncatalysed reaction, and $k_c$ and $E_{a,c}$ be for the catalysed reaction. We are given:

  • Temperature, $T = 27^\circ\text{C} = 27 + 273 = 300 \text{ K}$
  • Change in activation energy, $E_{a,u} - E_{a,c} = 10 \text{ kJ mol}^{-1} = 10000 \text{ J mol}^{-1}$
  • Frequency factor, $A$ is the same for both.
  • Gas constant, $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$

For the uncatalysed reaction:

$\ln k_u = \ln A - \frac{E_{a,u}}{RT}$

For the catalysed reaction:

$\ln k_c = \ln A - \frac{E_{a,c}}{RT}$

Subtracting the first equation from the second gives the change in the logarithm of the rate constant:

$\ln k_c - \ln k_u = \left(\ln A - \frac{E_{a,c}}{RT}\right) - \left(\ln A - \frac{E_{a,u}}{RT}\right)$ $\ln \left(\frac{k_c}{k_u}\right) = \frac{E_{a,u} - E_{a,c}}{RT}$

Substitute the given values:

$\ln \left(\frac{k_c}{k_u}\right) = \frac{10000 \text{ J mol}^{-1}}{(8.314 \text{ J mol}^{-1} \text{K}^{-1})(300 \text{ K})}$ $\ln \left(\frac{k_c}{k_u}\right) = \frac{10000}{2494.2} \approx 4.009$

The question asks for the logarithm of the ratio, which typically refers to the base-10 logarithm ($\log_{10}$). To convert from natural logarithm ($\ln$) to base-10 logarithm:

$\log_{10} x = \frac{\ln x}{\ln 10}$

Using $\ln 10 \approx 2.303$:

$\log_{10} \left(\frac{k_c}{k_u}\right) = \frac{4.009}{2.303} \approx 1.741$

Therefore, the logarithm of the ratio $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is approximately $1.741$.

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