A mixture of `SO_2 and O_2` at 1 atm in the mole ratio of 2:1 is passed through a catalyst at `1170^@C` at a rate sufficient for attainment of equilibrium. The existing gas, suddenly chilled and analysed, is found to contain 87% `SO_3` by volume. Calculate K, for the reaction: `SO_2+1/2 O_2 Leftrightarrow SO_3`

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction:

\[ SO_2 + \frac{1}{2} O_2 \leftrightarrow SO_3 \]

### Step-by-Step Solution:

1. **Determine Initial Moles:**
Given that the mole ratio of \( SO_2 \) to \( O_2 \) is 2:1, we can assume:
- Moles of \( SO_2 = 2 \) moles
- Moles of \( O_2 = 1 \) mole

Thus, the initial total moles \( n_{initial} = 2 + 1 = 3 \) moles.

2. **Define Change in Moles:**
Let \( x \) be the amount of \( SO_2 \) that reacts. According to the stoichiometry of the reaction:
- Change in moles of \( SO_2 = -x \)
- Change in moles of \( O_2 = -\frac{x}{2} \)
- Change in moles of \( SO_3 = +x \)

3. **Calculate Moles at Equilibrium:**
At equilibrium, the moles will be:
- Moles of \( SO_2 = 2 - x \)
- Moles of \( O_2 = 1 - \frac{x}{2} \)
- Moles of \( SO_3 = x \)

Therefore, the total moles at equilibrium:
\[ n_{total} = (2 - x) + \left(1 - \frac{x}{2}\right) + x = 3 - \frac{x}{2} \]

4. **Use Volume Percentage to Find \( x \):**
It is given that at equilibrium, the volume percentage of \( SO_3 \) is 87%. This means:
\[ \frac{x}{n_{total}} = 0.87 \]
Substituting \( n_{total} \):
\[ \frac{x}{3 - \frac{x}{2}} = 0.87 \]

5. **Solve for \( x \):**
Cross-multiplying gives:
\[ x = 0.87 \left(3 - \frac{x}{2}\right) \]
Expanding this:
\[ x = 2.61 - 0.435x \]
Rearranging:
\[ x + 0.435x = 2.61 \]
\[ 1.435x = 2.61 \]
\[ x = \frac{2.61}{1.435} \approx 1.82 \]

6. **Calculate Moles at Equilibrium:**
Now substituting \( x \) back to find the moles at equilibrium:
- Moles of \( SO_2 = 2 - 1.82 = 0.18 \)
- Moles of \( O_2 = 1 - \frac{1.82}{2} = 0.09 \)
- Moles of \( SO_3 = 1.82 \)

7. **Calculate Total Pressure:**
The total pressure is given as 1 atm. Thus, the partial pressures can be calculated using the mole fractions:
\[ P_{total} = 1 \text{ atm} \]
The total moles at equilibrium:
\[ n_{total} = 0.18 + 0.09 + 1.82 = 2.09 \]

8. **Calculate Partial Pressures:**
- \( P_{SO_2} = \frac{0.18}{2.09} \times 1 \text{ atm} \approx 0.086 \text{ atm} \)
- \( P_{O_2} = \frac{0.09}{2.09} \times 1 \text{ atm} \approx 0.0431 \text{ atm} \)
- \( P_{SO_3} = \frac{1.82}{2.09} \times 1 \text{ atm} \approx 0.871 \text{ atm} \)

9. **Calculate \( K_p \):**
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})(P_{O_2})^{1/2}} \]
Substituting the values:
\[ K_p = \frac{(0.871)^2}{(0.086)(0.0431)^{1/2}} \approx 58.60 \]

### Final Answer:
\[ K_p \approx 58.60 \]