In a 7.0 L evacuated chamber, 0.50 mol `H_(2)` and 0.50 mol `I_(2)` react at `427^(@)` C.
`H_(2) (g) + I_(2) (g) hArr HI (g) ` + Heat , At the given temperature, `K_(c)` = 49 for the reaction.
The `K_(c)` for above reaction at `500^(@)` C is

Options

A. less than 49
B. more than 49
C. equal to 49
D. None of the above

Answer

To determine the value of \( K_c \) for the reaction at \( 500^\circ C \), we need to analyze the effect of temperature on the equilibrium constant for an exothermic reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
The reaction given is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) + \text{Heat} \]
This indicates that the reaction is exothermic because heat is released.

2. **Understand the Relationship Between Temperature and \( K_c \)**:
For exothermic reactions, the equilibrium constant \( K_c \) decreases as the temperature increases. This is due to Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in temperature, the equilibrium will shift to counteract that change.

3. **Given Data**:
- At \( 427^\circ C \), \( K_c = 49 \).
- We need to find \( K_c \) at \( 500^\circ C \).

4. **Analyze the Temperature Change**:
Since \( 500^\circ C \) is higher than \( 427^\circ C \), we expect the value of \( K_c \) to decrease.

5. **Conclusion**:
Therefore, since the temperature is increased, the \( K_c \) at \( 500^\circ C \) will be less than \( 49 \).

### Final Answer:
The \( K_c \) for the reaction at \( 500^\circ C \) is **less than 49**.

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