For the equilibrium in gaseous phase in 2 lit flask we start with 2 moles of `SO_(2)` and 1 mole of `O_(2)` at 3 atm, `2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g))`. When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant `K_(c) ` is :

Answer

To solve the problem step by step, we will follow the equilibrium reaction and apply the concepts of chemical equilibrium and the ideal gas law.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

### Step 2: Determine initial moles and total pressure
Initially, we have:
- Moles of \(\text{SO}_2 = 2\)
- Moles of \(\text{O}_2 = 1\)

The total initial moles \(n_i\) is:
\[ n_i = 2 + 1 = 3 \text{ moles} \]

The initial pressure \(P_i\) is given as \(3 \text{ atm}\).

### Step 3: Use the ideal gas law to find initial conditions
Using the ideal gas equation \(PV = nRT\), we can relate pressure to moles since the volume \(V\) is constant. Here, we can set up the relationship:
\[ \frac{P_i}{n_i} = \frac{P_f}{n_f} \]
where \(P_f\) is the final pressure at equilibrium, and \(n_f\) is the total moles at equilibrium.

### Step 4: Determine changes in moles at equilibrium
Let \(x\) be the amount of \(\text{SO}_2\) that reacts at equilibrium. The changes in moles will be:
- Moles of \(\text{SO}_2\) at equilibrium = \(2 - 2x\)
- Moles of \(\text{O}_2\) at equilibrium = \(1 - x\)
- Moles of \(\text{SO}_3\) produced = \(2x\)

Thus, the total moles at equilibrium \(n_f\) will be:
\[ n_f = (2 - 2x) + (1 - x) + 2x = 3 - x \]

### Step 5: Set up the pressure relationship
Given that the final pressure \(P_f\) is \(2.5 \text{ atm}\), we can use the relationship derived from the ideal gas law:
\[ \frac{3 \text{ atm}}{3 \text{ moles}} = \frac{2.5 \text{ atm}}{3 - x} \]

### Step 6: Solve for \(x\)
Cross-multiplying gives:
\[ 3 \cdot 2.5 = 3 \cdot (3 - x) \]
\[ 7.5 = 9 - 3x \]
\[ 3x = 9 - 7.5 \]
\[ 3x = 1.5 \]
\[ x = 0.5 \]

### Step 7: Calculate moles at equilibrium
Now substituting \(x\) back into the expressions for moles at equilibrium:
- Moles of \(\text{SO}_2\) = \(2 - 2(0.5) = 1\)
- Moles of \(\text{O}_2\) = \(1 - 0.5 = 0.5\)
- Moles of \(\text{SO}_3\) = \(2(0.5) = 1\)

### Step 8: Calculate concentrations
Since the volume of the flask is \(2 \text{ L}\):
- Concentration of \(\text{SO}_2 = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}\)
- Concentration of \(\text{O}_2 = \frac{0.5 \text{ mole}}{2 \text{ L}} = 0.25 \text{ M}\)
- Concentration of \(\text{SO}_3 = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}\)

### Step 9: Calculate the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) is given by:
\[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \]
Substituting the concentrations:
\[ K_c = \frac{(0.5)^2}{(0.5)^2 \cdot (0.25)} \]
\[ K_c = \frac{0.25}{0.25 \cdot 0.25} \]
\[ K_c = \frac{0.25}{0.0625} = 4 \]

### Final Answer
Thus, the equilibrium constant \(K_c\) is:
\[ K_c = 4 \]

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