`N_(2)O_(3)` is an unstable oxide of nitrogen and it decomposes into NO (g) and `NO_(2)(g)`where `NO_(2)(g)` is further dimerise dimerise into `N_(2)O_(4)` as
`N_(2)O_(3)(g)hArrNO_(2)(g)+NO(g)" ",K_(p_(1)=2.5` bar
`2NO_(2)(g)hArrN_(2)O_(4)(g)" ": K_(P2)`
A flask is initially filled with pure `N_(2)O_(3)(g)` having pressure `2` bar and equilibria was established.
At equilibrium partial pressure of NO (g) was found to be `1.5` ber.
The value of `K_(P2)` is

Options

A. `0.16 "bar"^(-1)`
B. `0.32 "bar"^(-1)`
C. `0.48 "bar"^(-1)`
D. `0.64 "bar"^(-1)`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the reactions and their equilibrium constants
1. The first reaction is the decomposition of \(N_2O_3\):
\[ N_2O_3(g) \rightleftharpoons NO_2(g) + NO(g) \quad K_{p1} = 2.5 \text{ bar} \]
2. The second reaction is the dimerization of \(NO_2\):
\[ 2NO_2(g) \rightleftharpoons N_2O_4(g) \quad K_{p2} \]

### Step 2: Set up the initial conditions and changes
- Initial pressure of \(N_2O_3\) = 2 bar, and at equilibrium, the partial pressure of \(NO\) is given as 1.5 bar.
- Let \(x\) be the change in pressure of \(NO\) formed from \(N_2O_3\). Thus, \(x = 1.5\) bar.

### Step 3: Calculate the changes in pressures
- The change in pressure for \(N_2O_3\) will be \(2 - x\) (since it decomposes).
- The change in pressure for \(NO_2\) will be \(x\) (formed from \(N_2O_3\)).
- Therefore, at equilibrium:
- Pressure of \(N_2O_3\) = \(2 - x\)
- Pressure of \(NO\) = \(x = 1.5\) bar
- Pressure of \(NO_2\) = \(x - y\) (where \(y\) is the amount of \(NO_2\) that dimerizes)

### Step 4: Use \(K_{p1}\) to find \(y\)
Using the expression for \(K_{p1}\):
\[ K_{p1} = \frac{P_{NO_2} \cdot P_{NO}}{P_{N_2O_3}} = 2.5 \]
Substituting the values:
\[ 2.5 = \frac{(x - y) \cdot x}{(2 - x)} \]
Substituting \(x = 1.5\):
\[ 2.5 = \frac{(1.5 - y) \cdot 1.5}{(2 - 1.5)} \]
This simplifies to:
\[ 2.5 = \frac{(1.5 - y) \cdot 1.5}{0.5} \]
Multiplying both sides by 0.5:
\[ 1.25 = (1.5 - y) \cdot 1.5 \]
Dividing both sides by 1.5:
\[ \frac{1.25}{1.5} = 1.5 - y \]
Calculating:
\[ \frac{1.25}{1.5} = \frac{5}{6} \implies 1.5 - y = \frac{5}{6} \]
Thus:
\[ y = 1.5 - \frac{5}{6} = \frac{9}{6} - \frac{5}{6} = \frac{4}{6} = \frac{2}{3} \text{ bar} \]

### Step 5: Calculate the partial pressures for \(K_{p2}\)
- The partial pressure of \(N_2O_4\) is:
\[ P_{N_2O_4} = \frac{y}{2} = \frac{2/3}{2} = \frac{1}{3} \text{ bar} \]
- The partial pressure of \(NO_2\) is:
\[ P_{NO_2} = x - y = 1.5 - \frac{2}{3} = \frac{4.5 - 2}{3} = \frac{2.5}{3} \text{ bar} \]

### Step 6: Write the expression for \(K_{p2}\)
Using the expression for \(K_{p2}\):
\[ K_{p2} = \frac{P_{N_2O_4}}{(P_{NO_2})^2} \]
Substituting the values:
\[ K_{p2} = \frac{\frac{1}{3}}{\left(\frac{2.5}{3}\right)^2} \]
Calculating:
\[ K_{p2} = \frac{\frac{1}{3}}{\frac{6.25}{9}} = \frac{1}{3} \cdot \frac{9}{6.25} = \frac{3}{6.25} = \frac{3 \cdot 100}{625} = \frac{300}{625} = 0.48 \text{ bar}^{-1} \]

### Final Answer
The value of \(K_{p2}\) is \(0.48 \text{ bar}^{-1}\).

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