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 determine the status of the equilibrium process for the reaction

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

with a given equilibrium constant \( K_p = 3.4 \) and the partial pressures of the gases at a certain instant as \( P_{SO_2} = 0.41 \, \text{atm} \), \( P_{O_2} = 0.16 \, \text{atm} \), and \( P_{SO_3} = 0.57 \, \text{atm} \), we will follow these steps:

### Step 1: Calculate the Reaction Quotient \( Q_p \)

The reaction quotient \( Q_p \) is calculated using the formula:

\[ Q_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

Substituting the given partial pressures:

\[ Q_p = \frac{(0.57)^2}{(0.41)^2 \cdot (0.16)} \]

Calculating each term:

- \( (0.57)^2 = 0.3249 \)
- \( (0.41)^2 = 0.1681 \)
- Therefore, \( Q_p = \frac{0.3249}{0.1681 \cdot 0.16} \)

Calculating the denominator:

\[ 0.1681 \cdot 0.16 = 0.026896 \]

Now substituting back into the equation for \( Q_p \):

\[ Q_p = \frac{0.3249}{0.026896} \approx 12.06 \]

### Step 2: Compare \( Q_p \) with \( K_p \)

Now we compare the calculated \( Q_p \) with the given \( K_p \):

- \( K_p = 3.4 \)
- \( Q_p \approx 12.06 \)

### Step 3: Determine the Direction of the Shift

Since \( Q_p > K_p \):

- This indicates that the concentration of products is higher than at equilibrium.
- According to Le Chatelier's principle, if the concentration of products is too high, the equilibrium will shift to the left (toward the reactants) to reduce the concentration of products.

### Conclusion

Thus, the reaction is not at equilibrium and will shift in the backward direction (toward the reactants) to reach equilibrium.

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