For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?

Explanation

To determine the nature of the reaction \( \text{Br}_2 \rightleftharpoons 2\text{Br} \) based on the given equilibrium constants at two different temperatures, we can use the Van 't Hoff equation. Here's a step-by-step solution:

### Step 1: Write down the Van 't Hoff equation
The Van 't Hoff equation relates the change in the equilibrium constant \( K \) with temperature and is given by:

\[ \log \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{2.303 \cdot R} \cdot \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
...