The equilibrium constant for the reaction Br₂ ⇌ 2Br at 500 K and 700 K are 1 × 10^–10 and 1 × 10^–5 respectively. The reaction is;

The correct answer is: Endothermic

We are given the equilibrium constant ($K$) for the dissociation reaction:

${\mathrm{Br}}_2\rightleftharpoons 2\mathrm{Br}$

at two different temperatures:

At $T=500\text{K}$, $K=1\times {10}^{-10}$
At $T=700\text{K}$, $K=1\times {10}^{-5}$

Step 1: Observe the trend between temperature and K

As the temperature increases from 500 K to 700 K, the equilibrium constant increases from $1\times {10}^{-10}$ to $1\times {10}^{-5}$. This is an increase by a factor of ${10}^5$.

Step 2: Apply Le Chatelier's Principle

According to Le Chatelier's Principle, when the temperature of a system at equilibrium is increased:

• If the reaction is endothermic (absorbs heat), the equilibrium shifts to the right (toward products), causing $K$ to increase.
• If the reaction is exothermic (releases heat), the equilibrium shifts to the left (toward reactants), causing $K$ to decrease.

Step 3: Apply the Van 't Hoff equation reasoning

The Van 't Hoff equation relates the change in equilibrium constant with temperature:

$\ln \frac{K_2}{K_1}=\frac{\Delta H_r}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

Here, $T_2=700\text{K}>T_1=500\text{K}$, so $\frac{1}{T_1}-\frac{1}{T_2}>0$ (a positive value).

Also, $K_2>K_1$, so $\ln \frac{K_2}{K_1}>0$ (also positive).

Since both sides of the equation are positive, $\Delta H_r$ must be positive.

Step 4: Conclusion

A positive $\Delta H$ means the reaction absorbs heat from the surroundings. Therefore, the dissociation of ${\mathrm{Br}}_2$ into bromine atoms is an endothermic reaction. This makes physical sense as well — breaking the Br–Br bond requires energy input.

Hence, the reaction is Endothermic.