In a 7.0 Levacuated chamber, 0.50 mol `H_(2)` and 0.50 mol `I_(2)` react at `427^(@)` C.
`H_(2) (g) + I_(2) (g) hArr HI (g) ` + Heat , At the given temperature, `K_(c)` = 49 for the reaction.
What is the total pressure (atm) in the chamber ?

Options

A. 83.14
B. 831.4
C. 8.21
D. None of these

Answer

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial concentrations
Initially, we have:
- Moles of \( \text{H}_2 = 0.50 \) mol
- Moles of \( \text{I}_2 = 0.50 \) mol
- Moles of \( \text{HI} = 0 \) mol

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the moles of \( \text{H}_2 \) and \( \text{I}_2 \) that react at equilibrium. Thus, at equilibrium:
- Moles of \( \text{H}_2 = 0.50 - x \)
- Moles of \( \text{I}_2 = 0.50 - x \)
- Moles of \( \text{HI} = 2x \)

### Step 4: 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 equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(0.50 - x)(0.50 - x)} = 49 \]

### Step 5: Solve for \( x \)
Expanding the equation:
\[ \frac{4x^2}{(0.50 - x)^2} = 49 \]
Cross-multiplying gives:
\[ 4x^2 = 49(0.50 - x)^2 \]
Expanding the right side:
\[ 4x^2 = 49(0.25 - x + x^2) \]
\[ 4x^2 = 12.25 - 49x + 49x^2 \]
Rearranging the equation:
\[ 0 = 45x^2 - 49x + 12.25 \]

### Step 6: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 45 \)
- \( b = -49 \)
- \( c = 12.25 \)

Calculating the discriminant:
\[ D = (-49)^2 - 4 \times 45 \times 12.25 = 2401 - 2205 = 196 \]
Thus:
\[ x = \frac{49 \pm \sqrt{196}}{2 \times 45} = \frac{49 \pm 14}{90} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{63}{90} = 0.7 \) (not possible since it exceeds initial moles)
2. \( x = \frac{35}{90} \approx 0.39 \)

### Step 7: Calculate equilibrium concentrations
Substituting \( x \) back:
- Moles of \( \text{H}_2 = 0.50 - 0.39 = 0.11 \)
- Moles of \( \text{I}_2 = 0.50 - 0.39 = 0.11 \)
- Moles of \( \text{HI} = 2(0.39) = 0.78 \)

### Step 8: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ n_{total} = 0.11 + 0.11 + 0.78 = 1.0 \text{ mol} \]

### Step 9: Calculate total pressure using the ideal gas law
Using the ideal gas law \( PV = nRT \):
- \( P = \frac{nRT}{V} \)
- \( R = 0.0821 \, \text{atm L/(mol K)} \)
- \( T = 427 + 273 = 700 \, K \)
- \( V = 7.0 \, L \)

Calculating pressure:
\[ P = \frac{(1.0 \, \text{mol})(0.0821 \, \text{atm L/(mol K)})(700 \, K)}{7.0 \, L} \]
\[ P = \frac{57.47}{7.0} \approx 8.21 \, \text{atm} \]

### Step 10: Conclusion
The total pressure in the chamber at equilibrium is approximately:
\[ \text{Total Pressure} \approx 8.21 \, \text{atm} \]