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

### Step 2: Set up the initial conditions
We are given:
- Initial moles of \( H_2 = 0.500 \, \text{mol} \)
- Initial moles of \( I_2 = 0.500 \, \text{mol} \)
- Volume of the vessel = 5.25 L

### Step 3: Calculate initial concentrations
To find the initial concentrations, we use the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
Thus, the initial concentrations are:
- \( [H_2] = \frac{0.500}{5.25} \approx 0.0952 \, \text{mol/L} \)
- \( [I_2] = \frac{0.500}{5.25} \approx 0.0952 \, \text{mol/L} \)
- \( [HI] = 0 \, \text{mol/L} \) (since no product is formed initially)

### Step 4: Define the change in concentration at equilibrium
Let \( x \) be the change in moles of \( H_2 \) and \( I_2 \) that react to form \( HI \). Therefore, at equilibrium:
- Moles of \( H_2 = 0.500 - x \)
- Moles of \( I_2 = 0.500 - x \)
- Moles of \( HI = 2x \)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x/5.25)^2}{((0.500 - x)/5.25)((0.500 - x)/5.25)} \]

### Step 6: Substitute the given \( K_c \) value
We know \( K_c = 54.3 \), so we can set up the equation:
\[ 54.3 = \frac{(2x/5.25)^2}{((0.500 - x)/5.25)^2} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 54.3 = \frac{4x^2}{(0.500 - x)^2} \]

### Step 8: Cross-multiply and rearrange
Cross-multiplying gives:
\[ 54.3(0.500 - x)^2 = 4x^2 \]

Expanding and rearranging leads to:
\[ 54.3(0.25 - x + x^2) = 4x^2 \]
\[ 13.575 - 54.3x + 54.3x^2 = 4x^2 \]
\[ 50.3x^2 - 54.3x + 13.575 = 0 \]

### Step 9: Solve the quadratic equation
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:
\[ D = (-54.3)^2 - 4(50.3)(13.575) \]
\[ D = 2948.49 - 2724.51 = 223.98 \]

Now, substituting into the quadratic formula:
\[ x = \frac{54.3 \pm \sqrt{223.98}}{2 \times 50.3} \]
Calculating \( \sqrt{223.98} \approx 15 \):
\[ x \approx \frac{54.3 \pm 15}{100.6} \]

Calculating the two possible values for \( x \):
1. \( x \approx \frac{69.3}{100.6} \approx 0.688 \) (not possible since it exceeds initial moles)
2. \( x \approx \frac{39.3}{100.6} \approx 0.390 \)

### Step 10: Calculate equilibrium moles
Now we can find the equilibrium moles:
- Moles of \( H_2 = 0.500 - 0.390 \approx 0.110 \, \text{mol} \)
- Moles of \( I_2 = 0.500 - 0.390 \approx 0.110 \, \text{mol} \)
- Moles of \( HI = 2 \times 0.390 \approx 0.780 \, \text{mol} \)

### Final Answer
At equilibrium:
- Moles of \( H_2 \approx 0.110 \, \text{mol} \)
- Moles of \( I_2 \approx 0.110 \, \text{mol} \)
- Moles of \( HI \approx 0.780 \, \text{mol} \)