Twenty grams of HI is heated at `327^@C` in a bulb of 1-litre capacity. Calculate the volume percentage of `H_2, I_2,` and HI at equilibrium. Given that the mass law constant for the equation `2H1 Leftrightarrow H_2+I_2` is 0.0559 at `327^@C` when concentrations are expressed in moles/litre.

Answer

To solve the problem, we need to calculate the volume percentages of \( H_2 \), \( I_2 \), and \( HI \) at equilibrium when 20 grams of hydrogen iodide (HI) is heated. The equilibrium constant \( K_c \) is given as 0.0559 at \( 327^\circ C \).

### Step-by-Step Solution:

1. **Calculate the moles of HI:**
\[ \text{Molar mass of HI} = 128 \, \text{g/mol} \]
\[ \text{Moles of HI} = \frac{20 \, \text{g}}{128 \, \text{g/mol}} = 0.15625 \, \text{mol} \]

**Hint:** Use the formula for moles: \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \).

2. **Set up the reaction and initial conditions:**
The reaction is:
\[ 2 HI \rightleftharpoons H_2 + I_2 \]
- Initial moles of \( HI = 0.15625 \)
- Initial moles of \( H_2 = 0 \)
- Initial moles of \( I_2 = 0 \)

3. **Define the change in moles at equilibrium:**
Let \( \alpha \) be the degree of dissociation of \( HI \). At equilibrium:
- Moles of \( HI = 0.15625 - 2\alpha \)
- Moles of \( H_2 = \alpha \)
- Moles of \( I_2 = \alpha \)

4. **Write the expression for the equilibrium constant \( K_c \):**
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\alpha \cdot \alpha}{(0.15625 - 2\alpha)^2} = 0.0559 \]

5. **Solve for \( \alpha \):**
Rearranging gives:
\[ \alpha^2 = 0.0559 \cdot (0.15625 - 2\alpha)^2 \]
Expanding and simplifying leads to a quadratic equation in terms of \( \alpha \).

After solving the quadratic equation, we find:
\[ \alpha \approx 0.025 \]

**Hint:** When solving the quadratic equation, ensure to check for realistic values of \( \alpha \) (it should be less than \( \frac{0.15625}{2} \)).

6. **Calculate the equilibrium moles:**
- Moles of \( HI = 0.15625 - 2(0.025) = 0.10625 \)
- Moles of \( H_2 = 0.025 \)
- Moles of \( I_2 = 0.025 \)

7. **Calculate the total moles at equilibrium:**
\[ \text{Total moles} = 0.10625 + 0.025 + 0.025 = 0.15625 \]

8. **Calculate the volume percentages:**
- Volume percentage of \( H_2 \):
\[ \text{Volume percentage of } H_2 = \left( \frac{0.025}{0.15625} \right) \times 100 \approx 16\% \]
- Volume percentage of \( I_2 \):
\[ \text{Volume percentage of } I_2 = \left( \frac{0.025}{0.15625} \right) \times 100 \approx 16\% \]
- Volume percentage of \( HI \):
\[ \text{Volume percentage of } HI = \left( \frac{0.10625}{0.15625} \right) \times 100 \approx 68\% \]

### Final Results:
- Volume percentage of \( H_2 \): 16%
- Volume percentage of \( I_2 \): 16%
- Volume percentage of \( HI \): 68%