For `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g),K_(p)=3.4`. At an instant, the partial pressures of `SO_(2), O_(2) and SO_(3)` were found to be 0.41 atm, 0.16 atm and 0.57 atm respectively. Comment on the status of equilibrium process.

Answer

To solve the problem, we need to analyze the given reaction and the provided partial pressures to determine the status of the equilibrium process.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The balanced chemical equation is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

2. **Identify the Given Values:**
- \( K_p = 3.4 \)
- Partial pressures at a certain instant:
- \( P_{SO_2} = 0.41 \, \text{atm} \)
- \( P_{O_2} = 0.16 \, \text{atm} \)
- \( P_{SO_3} = 0.57 \, \text{atm} \)

3. **Calculate the Reaction Quotient \( Q \):**
The reaction quotient \( Q \) is calculated using the formula:
\[ Q = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]
Substituting the values:
\[ Q = \frac{(0.57)^2}{(0.41)^2 \cdot (0.16)} \]

- Calculate \( (0.57)^2 = 0.3249 \)
- Calculate \( (0.41)^2 = 0.1681 \)
- Now substitute:
\[ Q = \frac{0.3249}{0.1681 \cdot 0.16} \]
- Calculate \( 0.1681 \cdot 0.16 = 0.026896 \)
- Finally, calculate \( Q \):
\[ Q = \frac{0.3249}{0.026896} \approx 12.07 \]

4. **Compare \( Q \) with \( K_p \):**
We have:
- \( Q \approx 12.07 \)
- \( K_p = 3.4 \)

Since \( Q > K_p \), we can conclude that the reaction quotient is greater than the equilibrium constant.

5. **Determine the Direction of the Shift:**
According to Le Chatelier's principle:
- If \( Q > K_p \), the reaction will shift to the left (toward the reactants) to reach equilibrium.

6. **Conclusion:**
The equilibrium process is currently not at equilibrium and will shift in the backward direction (toward the formation of \( SO_2 \) and \( O_2 \)).

### Summary:
The reaction is not at equilibrium because \( Q (12.07) > K_p (3.4) \). Therefore, the equilibrium will shift to the left.