The value of `K_p` for the reaction,
`2SO_2(g)+O_2(g) hArr 2SO_3(g)` is 5
what will be the partial pressure of `O_2` at equilibrium when equal moles of `SO_2 and SO_3` are present at equilibrium ?

Options

A. 0.5
B. 0.3
C. 0.2
D. 0.1

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced equation and the expression for \( K_p \)
The balanced equation for the reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]
where \( P_{SO_3} \), \( P_{SO_2} \), and \( P_{O_2} \) are the partial pressures of the respective gases.

### Step 2: Define the initial conditions and changes
Let’s assume that initially, we have:
- \( P_{SO_2} = 1 \) atm
- \( P_{O_2} = 1 \) atm
- \( P_{SO_3} = 0 \) atm

At equilibrium, let \( \alpha \) be the change in pressure for \( SO_2 \) and \( SO_3 \):
- \( P_{SO_2} = 1 - \alpha \)
- \( P_{O_2} = 1 - \frac{\alpha}{2} \)
- \( P_{SO_3} = 2\alpha \)

### Step 3: Substitute into the \( K_p \) expression
Substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(2\alpha)^2}{(1 - \alpha)^2 \cdot (1 - \frac{\alpha}{2})} \]

Given that \( K_p = 5 \), we can set up the equation:
\[ 5 = \frac{(2\alpha)^2}{(1 - \alpha)^2 \cdot (1 - \frac{\alpha}{2})} \]

### Step 4: Solve for \( \alpha \)
Expanding the equation:
\[ 5 = \frac{4\alpha^2}{(1 - \alpha)^2 \cdot (1 - \frac{\alpha}{2})} \]

Cross-multiplying gives:
\[ 5(1 - \alpha)^2(1 - \frac{\alpha}{2}) = 4\alpha^2 \]

Expanding the left side:
\[ 5(1 - 2\alpha + \alpha^2)(1 - \frac{\alpha}{2}) = 4\alpha^2 \]
\[ 5(1 - 2\alpha + \alpha^2 - \frac{\alpha}{2} + \alpha^2 \cdot \frac{\alpha}{2}) = 4\alpha^2 \]

This simplifies to a quadratic equation in terms of \( \alpha \). Solving this quadratic will yield the value of \( \alpha \).

### Step 5: Find the partial pressure of \( O_2 \)
Once we have \( \alpha \), we can find the partial pressure of \( O_2 \) at equilibrium:
\[ P_{O_2} = 1 - \frac{\alpha}{2} \]

### Step 6: Calculate the final value
Substituting the value of \( \alpha \) back into the equation will give us the final partial pressure of \( O_2 \).