The equilibrium constant for the reaction `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 32 at a given temperature. The equilibrium concentration of `I_(2)` and `HI` are `0.5xx10^(-3)` and `8xx10^(-3)M` respectively. The equilibrium concentration of `H_(2)` is

Options

A. `1xx10^(-3)M`
B. `0.5xx10^(-3)M`
C. `2xx10^(-3)M`
D. `4xx10^(-3)M`

Answer

To solve the problem, we need to find the equilibrium concentration of \( H_2 \) given the equilibrium constant \( K \) and the concentrations of \( I_2 \) and \( HI \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

2. **Write the expression for the equilibrium constant \( K \)**:
The equilibrium constant \( K \) is defined as:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the equilibrium concentrations of hydrogen iodide, hydrogen, and iodine, respectively.

3. **Substitute the known values into the equilibrium expression**:
We know:
- \( K = 32 \)
- \( [I_2] = 0.5 \times 10^{-3} \, M \)
- \( [HI] = 8 \times 10^{-3} \, M \)

Substituting these values into the equation gives:
\[ 32 = \frac{(8 \times 10^{-3})^2}{[H_2] \times (0.5 \times 10^{-3})} \]

4. **Calculate \( (8 \times 10^{-3})^2 \)**:
\[ (8 \times 10^{-3})^2 = 64 \times 10^{-6} = 6.4 \times 10^{-5} \]

5. **Rearrange the equation to solve for \( [H_2] \)**:
\[ 32 = \frac{6.4 \times 10^{-5}}{[H_2] \times (0.5 \times 10^{-3})} \]
Multiplying both sides by \( [H_2] \times (0.5 \times 10^{-3}) \):
\[ 32 \times [H_2] \times (0.5 \times 10^{-3}) = 6.4 \times 10^{-5} \]

6. **Solve for \( [H_2] \)**:
\[ [H_2] = \frac{6.4 \times 10^{-5}}{32 \times (0.5 \times 10^{-3})} \]
\[ = \frac{6.4 \times 10^{-5}}{16 \times 10^{-3}} = \frac{6.4}{16} \times 10^{-2} = 0.4 \times 10^{-2} = 4 \times 10^{-3} \, M \]

### Final Answer:
The equilibrium concentration of \( H_2 \) is \( 4 \times 10^{-3} \, M \).

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