The equilibrium constant, `K_(p)` for the reaction
`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`
is `44.0atm^(-1) "at" 1000 K`. What would be the partial pressure of `O_(2)` if at equilibrium the amound of `SO_(2)` and `SO_(3)` is the same?

Options

A. (A) `16.0 atm`
B. (B) `0.023 atm`
C. (C) `1atm`
D. (D) `0.75atm`

Answer

To solve the problem, we will follow these steps:

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

The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2 \cdot (P_{\text{O}_2})} \]

### Step 2: Understand the condition given in the problem.
We are told that at equilibrium, the amounts of \( \text{SO}_2 \) and \( \text{SO}_3 \) are the same. This implies:
\[ P_{\text{SO}_2} = P_{\text{SO}_3} \]

### Step 3: Substitute the equality into the \( K_p \) expression.
Since \( P_{\text{SO}_2} = P_{\text{SO}_3} \), we can denote this common value as \( P \). Therefore, we can rewrite the \( K_p \) expression as:
\[ K_p = \frac{(P)^2}{(P)^2 \cdot (P_{\text{O}_2})} \]

### Step 4: Simplify the expression.
The \( (P)^2 \) terms in the numerator and denominator cancel out:
\[ K_p = \frac{1}{P_{\text{O}_2}} \]

### Step 5: Substitute the known value of \( K_p \).
We know that \( K_p = 44.0 \, \text{atm}^{-1} \):
\[ 44 = \frac{1}{P_{\text{O}_2}} \]

### Step 6: Solve for \( P_{\text{O}_2} \).
To find \( P_{\text{O}_2} \), we take the reciprocal of \( K_p \):
\[ P_{\text{O}_2} = \frac{1}{44} \]

### Step 7: Calculate the value.
Calculating this gives:
\[ P_{\text{O}_2} \approx 0.02273 \, \text{atm} \]

### Step 8: Round to three significant figures.
Rounding to three significant figures, we have:
\[ P_{\text{O}_2} \approx 0.023 \, \text{atm} \]

### Final Answer:
Thus, the partial pressure of \( O_2 \) at equilibrium is approximately:
\[ P_{\text{O}_2} = 0.023 \, \text{atm} \]

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