The equilibrium constants of a reaction at `27^@`C and `127^@`C are `1.6 xx 10^(-3)` and `7.6 xx 10^(-2)` respectively. Is the reaction exothermic or endothermic ?

Explanation

To determine whether the reaction is exothermic or endothermic based on the given equilibrium constants at two different temperatures, we can use the Van 't Hoff equation, which relates the change in the equilibrium constant with temperature to the enthalpy change of the reaction.

### Step-by-Step Solution:

1. **Identify the given data:**
- Equilibrium constant at \( T_1 = 27^\circ C = 300 K \): \( K_1 = 1.6 \times 10^{-3} \)
- Equilibrium constant at \( T_2 = 127^\circ C = 400 K \): \( K_2 = 7.6 \times 10^{-2} \)

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