At 500 K, the equilibrium costant for the reaction `H_(2(g))+I_(2(g))hArr2HI_((g))" is "24.8" If "(1)/(2)mol//L` of HI is present at equilibrium, what are the concentrations of `H_(2)andI_(2)`, assuming that we started by taking HI and reached the equilibrium at 500 K ?

Options

A. `0.068"mol L"^(-1)`
B. `1.020"mol L"^(-1)`
C. `0.10"mol L"^(-1)`
D. `1.20"mol L"^(-1)`

Explanation

To solve the problem, we need to find the equilibrium concentrations of \( H_2 \) and \( I_2 \) given that the equilibrium constant \( K_c \) for the reaction

\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

is 24.8 at 500 K, and that the concentration of \( HI \) at equilibrium is \( 0.5 \, \text{mol/L} \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the reaction**:
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the equilibrium concentrations of the respective species.

2. **Reverse the reaction**:
Since we are starting with \( HI \) and want to find the concentrations of \( H_2 \) and \( I_2 \), we reverse the reaction:
\[ 2 HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
The new equilibrium constant \( K'_c \) for this reversed reaction is:
\[ K'_c = \frac{1}{K_c} = \frac{1}{24.8} \]

3. **Set up the new equilibrium expression**:
For the reversed reaction, the equilibrium expression becomes:
\[ K'_c = \frac{[H_2][I_2]}{[HI]^2} \]

4. **Substitute the known values**:
We know that at equilibrium, \( [HI] = 0.5 \, \text{mol/L} \). Substitute this into the equilibrium expression:
\[ \frac{1}{24.8} = \frac{[H_2][I_2]}{(0.5)^2} \]
This simplifies to:
\[ \frac{1}{24.8} = \frac{[H_2][I_2]}{0.25} \]

5. **Rearrange to find \( [H_2][I_2] \)**:
Multiply both sides by \( 0.25 \):
\[ [H_2][I_2] = \frac{0.25}{24.8} \]

6. **Calculate \( [H_2][I_2] \)**:
\[ [H_2][I_2] = 0.01008 \, \text{mol}^2/\text{L}^2 \]

7. **Assume \( [H_2] = [I_2] \)**:
Let \( [H_2] = [I_2] = x \). Then:
\[ x^2 = 0.01008 \]

8. **Solve for \( x \)**:
\[ x = \sqrt{0.01008} \approx 0.1 \, \text{mol/L} \]

9. **Conclusion**:
Therefore, the concentrations of \( H_2 \) and \( I_2 \) at equilibrium are both \( 0.1 \, \text{mol/L} \).

### Final Answer:
\[ [H_2] = [I_2] = 0.1 \, \text{mol/L} \]