Consider the following reaction
`H_(2)(g)+I_(2)(g) hArr 2HI(g), K_(c)=54.3` at 698K
If we start with 0.500 mol `H_(2)` and 0.500 mol `I_(2)(g)` in a 5.25-L vessel at 698 K, how many moles of each gas will be present at equilibrium?

Options

A. `n_(H_(2))=n_(I_(2))=0.107, n_(HI)=0.788`
B. `n_(H_(2))=n_(I_(2))=0.788, n_(HI)=0.107`
C. `n_(H_(2))=n_(HI)=0.788,n_(I_(2))=0.107`
D. `n_(H_(2))=n_(HI)=0.107, n_(I_(2))=0.788`

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: Set up the initial conditions
We start with:
- Moles of \( \text{H}_2 = 0.500 \, \text{mol} \)
- Moles of \( \text{I}_2 = 0.500 \, \text{mol} \)
- Moles of \( \text{HI} = 0 \, \text{mol} \)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of \( \text{H}_2 \) and \( \text{I}_2 \) that react at equilibrium. Therefore, at equilibrium, we have:
- Moles of \( \text{H}_2 = 0.500 - x \)
- Moles of \( \text{I}_2 = 0.500 - x \)
- Moles of \( \text{HI} = 2x \)

### Step 4: Calculate concentrations
The volume of the vessel is 5.25 L. The concentrations at equilibrium will be:
- \[ [\text{H}_2] = \frac{0.500 - x}{5.25} \]
- \[ [\text{I}_2] = \frac{0.500 - x}{5.25} \]
- \[ [\text{HI}] = \frac{2x}{5.25} \]

### Step 5: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]
Substituting the concentrations into the equation:
\[ 54.3 = \frac{\left(\frac{2x}{5.25}\right)^2}{\left(\frac{0.500 - x}{5.25}\right)\left(\frac{0.500 - x}{5.25}\right)} \]

### Step 6: Simplify the equation
This simplifies to:
\[ 54.3 = \frac{(2x)^2}{(0.500 - x)^2} \]
\[ 54.3 = \frac{4x^2}{(0.500 - x)^2} \]

### Step 7: Cross-multiply and solve for \( x \)
Cross-multiplying gives:
\[ 54.3(0.500 - x)^2 = 4x^2 \]
Expanding and rearranging:
\[ 54.3(0.25 - x + x^2) = 4x^2 \]
\[ 13.575 - 54.3x + 54.3x^2 = 4x^2 \]
\[ 50.3x^2 - 54.3x + 13.575 = 0 \]

### Step 8: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 50.3 \)
- \( b = -54.3 \)
- \( c = 13.575 \)

Calculating the discriminant:
\[ b^2 - 4ac = (-54.3)^2 - 4(50.3)(13.575) \]
\[ = 2947.69 - 2734.77 = 212.92 \]

Finding \( x \):
\[ x = \frac{54.3 \pm \sqrt{212.92}}{2 \times 50.3} \]
Calculating \( \sqrt{212.92} \approx 14.59 \):
\[ x = \frac{54.3 \pm 14.59}{100.6} \]

Calculating the two possible values:
1. \( x = \frac{68.89}{100.6} \approx 0.684 \) (not possible since it exceeds initial moles)
2. \( x = \frac{39.71}{100.6} \approx 0.394 \)

### Step 9: Calculate equilibrium moles
Now substituting \( x \) back to find the moles at equilibrium:
- Moles of \( \text{H}_2 = 0.500 - 0.394 = 0.106 \)
- Moles of \( \text{I}_2 = 0.500 - 0.394 = 0.106 \)
- Moles of \( \text{HI} = 2 \times 0.394 = 0.788 \)

### Final Answer
At equilibrium:
- Moles of \( \text{H}_2 = 0.106 \)
- Moles of \( \text{I}_2 = 0.106 \)
- Moles of \( \text{HI} = 0.788 \)