`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 equilibrium partiaal pressure of `N_(2)O_(3)(g)` is :

Options

A. `0.5`bar
B. `1.0` bar
C. `1.5` bar
D. `0.1` bar

Answer

To solve the problem step by step, we will analyze the decomposition of \( N_2O_3 \) and the subsequent reactions that occur at equilibrium.

### Step 1: Write the reactions and initial conditions
The decomposition of \( N_2O_3 \) can be represented as:
\[ N_2O_3(g) \rightleftharpoons NO_2(g) + NO(g) \]
The equilibrium constant \( K_{p1} \) for this reaction is given as \( 2.5 \) bar.

The second reaction is the dimerization of \( NO_2 \):
\[ 2NO_2(g) \rightleftharpoons N_2O_4(g) \]
Let \( K_{p2} \) be the equilibrium constant for this reaction (not provided in the question).

Initially, the flask is filled with pure \( N_2O_3 \) at a pressure of \( 2 \) bar. Therefore, the initial pressures are:
- \( P_{N_2O_3} = 2 \) bar
- \( P_{NO} = 0 \) bar
- \( P_{NO_2} = 0 \) bar

### Step 2: Define changes at equilibrium
Let \( x \) be the change in pressure of \( N_2O_3 \) that decomposes at equilibrium. Thus, at equilibrium:
- The pressure of \( N_2O_3 \) will be \( 2 - x \) bar.
- The pressure of \( NO \) will be \( x \) bar.
- The pressure of \( NO_2 \) will also be \( x \) bar.

### Step 3: Use given information about \( NO \)
We are given that the equilibrium partial pressure of \( NO \) is \( 1.5 \) bar. Therefore, we can set:
\[ x = 1.5 \text{ bar} \]

### Step 4: Calculate the equilibrium pressure of \( N_2O_3 \)
Now, we can find the equilibrium partial pressure of \( N_2O_3 \):
\[ P_{N_2O_3} = 2 - x = 2 - 1.5 = 0.5 \text{ bar} \]

### Final Answer
The equilibrium partial pressure of \( N_2O_3 \) is \( 0.5 \) bar.

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