For the reaction A(g) + B(g) `hArr` C(g) at equilibrium the partial pressure of the species are `P_(A)=0.15"atm",P_(C)=P_(B)=0.30"atm"`. If the capacity of reaction vessel is reduced, the equilibrium is re-established. In the new situation partial pressure A and B become twice. What is the partial pressure of C?

Options

A. 0.30 atm
B. 0.60 atm
C. 1.20 atm
D. 1.80 atm

Answer

To solve the problem step by step, we will first establish the equilibrium constant \( K_p \) for the reaction and then use it to find the new partial pressure of \( C \) after the changes in the partial pressures of \( A \) and \( B \).

### Step 1: Write the expression for the equilibrium constant \( K_p \)

For the reaction:
\[ A(g) + B(g) \rightleftharpoons C(g) \]

The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_C}{P_A \cdot P_B} \]

### Step 2: Substitute the initial partial pressures into the \( K_p \) expression

From the problem, we know:
- \( P_A = 0.15 \, \text{atm} \)
- \( P_B = 0.30 \, \text{atm} \)
- \( P_C = 0.30 \, \text{atm} \)

Now substituting these values into the \( K_p \) expression:
\[ K_p = \frac{0.30}{0.15 \cdot 0.30} \]

### Step 3: Calculate \( K_p \)

Calculating the denominator:
\[ 0.15 \cdot 0.30 = 0.045 \]

Now substituting back into the equation for \( K_p \):
\[ K_p = \frac{0.30}{0.045} = \frac{30}{4.5} = \frac{300}{45} = \frac{20}{3} \approx 6.67 \]

### Step 4: Analyze the new conditions after the volume reduction

After the volume of the reaction vessel is reduced, the partial pressures of \( A \) and \( B \) become twice their initial values:
- New \( P_A = 2 \times 0.15 = 0.30 \, \text{atm} \)
- New \( P_B = 2 \times 0.30 = 0.60 \, \text{atm} \)

### Step 5: Set up the new \( K_p \) expression with the new pressures

Using the new partial pressures in the \( K_p \) expression:
\[ K_p = \frac{P_C}{P_A \cdot P_B} = \frac{P_C}{0.30 \cdot 0.60} \]

### Step 6: Equate the two expressions for \( K_p \)

Since \( K_p \) remains constant (assuming temperature is constant), we can set the two expressions equal to each other:
\[ \frac{20}{3} = \frac{P_C}{0.30 \cdot 0.60} \]

### Step 7: Calculate \( P_C \)

Calculating the denominator:
\[ 0.30 \cdot 0.60 = 0.18 \]

Now substituting back into the equation:
\[ \frac{20}{3} = \frac{P_C}{0.18} \]

Cross-multiplying to solve for \( P_C \):
\[ P_C = \frac{20}{3} \cdot 0.18 \]

Calculating \( P_C \):
\[ P_C = \frac{20 \cdot 0.18}{3} = \frac{3.6}{3} = 1.2 \, \text{atm} \]

### Final Answer
The partial pressure of \( C \) after the equilibrium is re-established is:
\[ \boxed{1.2 \, \text{atm}} \]

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