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, we need to analyze the given information about the equilibrium constants for the reaction \( \text{Br}_2 \rightleftharpoons 2\text{Br} \) at two different temperatures (500 K and 700 K). The equilibrium constants are given as \( K_{c1} = 1 \times 10^{-10} \) at 500 K and \( K_{c2} = 1 \times 10^{-5} \) at 700 K.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is \( \text{Br}_2 \rightleftharpoons 2\text{Br} \).

2. **Understand the Equilibrium Constant**:
The equilibrium constant \( K_c \) is a measure of the ratio of the concentration of products to reactants at equilibrium. For this reaction, it can be expressed as:
\[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \]

3. **Analyze the Given Values**:
- At 500 K, \( K_{c1} = 1 \times 10^{-10} \)
- At 700 K, \( K_{c2} = 1 \times 10^{-5} \)

We observe that as the temperature increases from 500 K to 700 K, the value of \( K_c \) increases significantly.

4. **Determine the Nature of the Reaction**:
According to Le Chatelier's principle, if the equilibrium constant increases with temperature, the reaction is endothermic. This is because an endothermic reaction absorbs heat, and increasing the temperature shifts the equilibrium to the right (towards the products).

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.

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