At a certain temperature and `2` atm pressure equilibrium constant `(K_(p))` is `25` for the reaction
`SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)`
Initially if we take `2` moles of each of the four gases and `2` moles of inert gas, what would be the equilibrium partial pressure of `NO_(2)`?

Options

A. (a) `1.33` atm
B. (b) `0.1665` atm
C. (c) `0.133` atm
D. (d) None of these

Answer

To solve the problem step-by-step, we will analyze the given reaction and the conditions provided.

### Step 1: Write the Reaction and Initial Conditions
The reaction is:
\[ \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \]

Initially, we have:
- 2 moles of SO₂
- 2 moles of NO₂
- 2 moles of SO₃
- 2 moles of NO
- 2 moles of inert gas

### Step 2: Define Changes at Equilibrium
Let \( x \) be the amount of SO₂ and NO₂ that reacts. At equilibrium, the moles of each gas will be:
- SO₂: \( 2 - x \)
- NO₂: \( 2 - x \)
- SO₃: \( 2 + x \)
- NO: \( 2 + x \)

### Step 3: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K_p \) is given as 25. The expression for \( K_p \) is:
\[ K_p = \frac{P_{\text{SO}_3} \cdot P_{\text{NO}}}{P_{\text{SO}_2} \cdot P_{\text{NO}_2}} \]
Substituting the equilibrium pressures in terms of \( x \):
\[ K_p = \frac{(2 + x)(2 + x)}{(2 - x)(2 - x)} = 25 \]

### Step 4: Simplify and Solve for \( x \)
This gives us:
\[ \frac{(2 + x)^2}{(2 - x)^2} = 25 \]
Taking the square root of both sides:
\[ \frac{2 + x}{2 - x} = 5 \]
Cross-multiplying:
\[ 2 + x = 5(2 - x) \]
Expanding:
\[ 2 + x = 10 - 5x \]
Rearranging gives:
\[ 6x = 8 \quad \Rightarrow \quad x = \frac{4}{3} \]

### Step 5: Calculate the Equilibrium Moles of NO₂
Now, substituting \( x \) back to find the equilibrium moles of NO₂:
\[ \text{Moles of NO}_2 = 2 - x = 2 - \frac{4}{3} = \frac{2}{3} \]

### Step 6: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = (2 - x) + (2 - x) + (2 + x) + (2 + x) + \text{(moles of inert gas)} \]
\[ = \left(2 - \frac{4}{3}\right) + \left(2 - \frac{4}{3}\right) + \left(2 + \frac{4}{3}\right) + \left(2 + \frac{4}{3}\right) + 2 \]
Calculating each term:
\[ = \frac{2}{3} + \frac{2}{3} + \frac{10}{3} + \frac{10}{3} + 2 = \frac{4}{3} + \frac{20}{3} + 2 = \frac{24}{3} + 2 = 8 + 2 = 10 \]

### Step 7: Calculate the Partial Pressure of NO₂
The partial pressure of NO₂ is given by:
\[ P_{\text{NO}_2} = \left(\frac{\text{Moles of NO}_2}{\text{Total moles}}\right) \times P_{\text{total}} \]
Where \( P_{\text{total}} = 2 \, \text{atm} \):
\[ P_{\text{NO}_2} = \left(\frac{\frac{2}{3}}{10}\right) \times 2 = \frac{2}{30} \times 2 = \frac{4}{30} = \frac{2}{15} \approx 0.133 \, \text{atm} \]

### Final Answer
The equilibrium partial pressure of NO₂ is approximately \( 0.133 \, \text{atm} \).

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