At 27 °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)

We need to find the logarithm of the ratio of rate constants for catalysed and uncatalysed reactions.

Temperature: $$T = 27°C = 300$$ K

Reduction in activation energy: $$\Delta E_a = 10$$ kJ/mol = 10000 J/mol

Frequency factor is the same for both reactions.

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

For the ratio:

$$\frac{k_{cat}}{k_{uncat}} = \frac{A e^{-E_{a,cat}/RT}}{A e^{-E_{a,uncat}/RT}} = e^{(E_{a,uncat} - E_{a,cat})/RT} = e^{\Delta E_a/RT}$$

$$\log\left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{\Delta E_a}{2.303 \times RT}$$

$$\log\left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{10000}{2.303 \times 8.314 \times 300}$$

$$= \frac{10000}{2.303 \times 2494.2} = \frac{10000}{5744.14} \approx 1.741$$

Therefore, the logarithm of the ratio is Option 4: 1.741.