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 step by step, we need to analyze the given equilibrium reaction and the information provided.

### Step 1: Write the equilibrium expression for the reaction.

The reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

### Step 2: Define the variables based on the given conditions.

We are told that equal moles of \( SO_2 \) and \( SO_3 \) are present at equilibrium. Let's denote the partial pressures as follows:
- Let \( P_{SO_2} = P \)
- Let \( P_{SO_3} = P \)
- Let \( P_{O_2} = x \)

Since the moles of \( SO_2 \) and \( SO_3 \) are equal, we can express their partial pressures as \( P \).

### Step 3: Substitute the values into the equilibrium expression.

Now, substituting the values into the equilibrium expression:
\[ K_p = \frac{(P)^2}{(P)^2 \cdot (x)} \]

Since \( P_{SO_2} = P \) and \( P_{SO_3} = P \), we can simplify:
\[ K_p = \frac{P^2}{P^2 \cdot x} = \frac{1}{x} \]

### Step 4: Set the equilibrium constant equal to the given value.

We know from the problem that \( K_p = 5 \). Therefore, we can set up the equation:
\[ \frac{1}{x} = 5 \]

### Step 5: Solve for \( x \).

To find \( x \), we take the reciprocal:
\[ x = \frac{1}{5} \]

### Step 6: Interpret the result.

The value of \( x \) represents the partial pressure of \( O_2 \) at equilibrium. Thus:
\[ P_{O_2} = 0.2 \, \text{atm} \]

### Final Answer:
The partial pressure of \( O_2 \) at equilibrium is \( 0.2 \, \text{atm} \).

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