The equilibrium constants for the reaction `Br_(2)hArr 2Br`
at 500 K and 700 K are `1xx10^(-10)` and `1xx10^(-5)` respectively. The reaction is:

Options

A. Endothermic
B. Exothermic
C. Fast
D. Slow

Answer

To solve the problem regarding the equilibrium constants for the reaction \( \text{Br}_2 \rightleftharpoons 2\text{Br} \) at different temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is \( \text{Br}_2 \rightleftharpoons 2\text{Br} \). This indicates that one mole of bromine gas dissociates into two moles of bromine atoms.

2. **Understand the Given Data**:
We have two equilibrium constants:
- At 500 K, \( K_c = 1 \times 10^{-10} \)
- At 700 K, \( K_c = 1 \times 10^{-5} \)

3. **Analyze the Change in Equilibrium Constant**:
As the temperature increases from 500 K to 700 K, the equilibrium constant \( K_c \) increases from \( 1 \times 10^{-10} \) to \( 1 \times 10^{-5} \). This suggests that the reaction favors the formation of products (Br atoms) at higher temperatures.

4. **Determine the Nature of the Reaction**:
The increase in \( K_c \) with temperature indicates that the reaction is endothermic. In an endothermic reaction, heat is absorbed, and increasing the temperature shifts the equilibrium towards the products, thereby increasing \( K_c \).

5. **Conclusion**:
Since the equilibrium constant increases with temperature, we conclude that the reaction \( \text{Br}_2 \rightleftharpoons 2\text{Br} \) is endothermic.

### Final Answer:
The reaction is endothermic.