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. `1 xx 10^(-3) M `
B. `0.5 xx 10^(-3)` M
C. `2 xx 10^(-3)` M
D. `4xx10^(-3)` M

Answer

To find the equilibrium concentration of \( H_2 \) for the reaction:

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

given that the equilibrium constant \( K \) is 32, the equilibrium concentration of \( I_2 \) is \( 0.5 \times 10^{-3} \, M \), and the equilibrium concentration of \( HI \) is \( 8 \times 10^{-3} \, M \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K \)

The equilibrium constant expression for the reaction is given by:

\[ K = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 2: Substitute the known values into the equation

We know that:
- \( K = 32 \)
- \( [HI] = 8 \times 10^{-3} \, M \)
- \( [I_2] = 0.5 \times 10^{-3} \, M \)

Substituting these values into the equilibrium expression gives:

\[ 32 = \frac{(8 \times 10^{-3})^2}{[H_2] \times (0.5 \times 10^{-3})} \]

### Step 3: Simplify the equation

Calculating \( (8 \times 10^{-3})^2 \):

\[ (8 \times 10^{-3})^2 = 64 \times 10^{-6} = 6.4 \times 10^{-5} \]

Now substituting this back into the equation:

\[ 32 = \frac{6.4 \times 10^{-5}}{[H_2] \times (0.5 \times 10^{-3})} \]

### Step 4: Rearranging the equation to find \( [H_2] \)

We can rearrange the equation to solve for \( [H_2] \):

\[ [H_2] = \frac{6.4 \times 10^{-5}}{32 \times (0.5 \times 10^{-3})} \]

Calculating the denominator:

\[ 32 \times (0.5 \times 10^{-3}) = 16 \times 10^{-3} = 0.016 \]

Now substituting this back:

\[ [H_2] = \frac{6.4 \times 10^{-5}}{0.016} \]

### Step 5: Calculate \( [H_2] \)

Now performing the division:

\[ [H_2] = 4.0 \times 10^{-3} \, M \]

### Conclusion

Thus, the equilibrium concentration of \( H_2 \) is:

\[ [H_2] = 4.0 \times 10^{-3} \, M \]

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