At `440^(@)C`, the equilibrium constant (K) for the following reaction is `49.5, H_(2)(g)+I_(2)(g) hArr 2HI(g)`. If `0.2` mol of `H_(2)` and `0.2` mol of `I_(2)` are placed in a `10-L` vessel and permitted to react at this temperature, what will be the concentration of each substance at equilibrium?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]

### Step 2: Determine initial concentrations
We are given that:
- Moles of \( \text{H}_2 \) = 0.2 mol
- Moles of \( \text{I}_2 \) = 0.2 mol
- Volume of the vessel = 10 L

To find the initial concentrations, we use the formula:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]

Calculating the initial concentrations:
- Initial concentration of \( \text{H}_2 \):
\[ [\text{H}_2] = \frac{0.2 \text{ mol}}{10 \text{ L}} = 0.02 \text{ M} \]
- Initial concentration of \( \text{I}_2 \):
\[ [\text{I}_2] = \frac{0.2 \text{ mol}}{10 \text{ L}} = 0.02 \text{ M} \]
- Initial concentration of \( \text{HI} \):
\[ [\text{HI}] = 0 \text{ M} \quad (\text{since no HI is present initially}) \]

### Step 3: Set up the expression for equilibrium concentrations
Let \( x \) be the change in concentration of \( \text{H}_2 \) and \( \text{I}_2 \) that reacts to form \( \text{HI} \). At equilibrium, the concentrations will be:
- \( [\text{H}_2] = 0.02 - x \)
- \( [\text{I}_2] = 0.02 - x \)
- \( [\text{HI}] = 2x \)

### Step 4: Write the expression for the equilibrium constant \( K \)
The equilibrium constant expression for the reaction is:
\[ K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]
Substituting the equilibrium concentrations into the expression:
\[ 49.5 = \frac{(2x)^2}{(0.02 - x)(0.02 - x)} \]

### Step 5: Simplify and solve for \( x \)
Substituting \( (2x)^2 = 4x^2 \):
\[ 49.5 = \frac{4x^2}{(0.02 - x)^2} \]
Cross-multiplying gives:
\[ 49.5(0.02 - x)^2 = 4x^2 \]
Expanding and rearranging:
\[ 49.5(0.0004 - 0.04x + x^2) = 4x^2 \]
\[ 0.0198 - 1.98x + 49.5x^2 = 4x^2 \]
\[ 45.5x^2 - 1.98x + 0.0198 = 0 \]

### Step 6: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 45.5 \)
- \( b = -1.98 \)
- \( c = 0.0198 \)

Calculating the discriminant:
\[ D = (-1.98)^2 - 4 \cdot 45.5 \cdot 0.0198 \]
\[ D = 3.9204 - 3.6072 = 0.3132 \]
Calculating \( x \):
\[ x = \frac{1.98 \pm \sqrt{0.3132}}{2 \cdot 45.5} \]
Calculating the positive root:
\[ x \approx 0.1557 \]

### Step 7: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:
- \( [\text{H}_2] = 0.02 - x = 0.02 - 0.1557 = 0.00443 \text{ M} \)
- \( [\text{I}_2] = 0.02 - x = 0.00443 \text{ M} \)
- \( [\text{HI}] = 2x = 2 \cdot 0.1557 = 0.03144 \text{ M} \)

### Final Answer
The equilibrium concentrations are:
- \( [\text{H}_2] = 0.00443 \text{ M} \)
- \( [\text{I}_2] = 0.00443 \text{ M} \)
- \( [\text{HI}] = 0.03144 \text{ M} \)