In a vessel, two equilibrium are simultaneously established at the same temperature as follows
`N_2(g)+3H_2(g) hArr 2NH_3(g), K_(P_1)`
`N_2(g)+2H_2(g) hArr N_2H_4(g), K_(P_2)`
Initially , vessel contains `N_2 and H_2` in molar ratio 9:13. The equilibrium pressure is `7P_0` in which pressure due to `NH_3` is `P_0` and due to `H_2` to `2P_0`
Find the value of `K_(P_2)/K_(P_1)`

Answer

To solve the problem, we need to find the ratio \( \frac{K_{P_2}}{K_{P_1}} \) given the two equilibrium reactions and the initial conditions. Let's break it down step by step.

### Step 1: Understanding the Initial Conditions
We have two reactions:
1. \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) with equilibrium constant \( K_{P_1} \)
2. \( N_2(g) + 2H_2(g) \rightleftharpoons N_2H_4(g) \) with equilibrium constant \( K_{P_2} \)

Initially, the molar ratio of \( N_2 \) to \( H_2 \) is 9:13. We can express the initial moles of \( N_2 \) and \( H_2 \) in terms of a variable \( x \):
- Let the initial moles of \( N_2 = 9x \)
- Let the initial moles of \( H_2 = 13x \)

### Step 2: Establishing the Equilibrium Pressures
At equilibrium, we are given:
- Total pressure \( P_{total} = 7P_0 \)
- Pressure of \( NH_3 = P_0 \)
- Pressure of \( H_2 = 2P_0 \)

Let \( P_{N_2} \) be the pressure of \( N_2 \) and \( P_{N_2H_4} \) be the pressure of \( N_2H_4 \). We can express the total pressure as:
\[ P_{total} = P_{N_2} + P_{H_2} + P_{NH_3} + P_{N_2H_4} \]
Substituting the known pressures:
\[ 7P_0 = P_{N_2} + 2P_0 + P_0 + P_{N_2H_4} \]
This simplifies to:
\[ P_{N_2} + P_{N_2H_4} = 4P_0 \quad \text{(Equation 1)} \]

### Step 3: Expressing the Changes in Pressure
From the first reaction, we can denote the change in pressure due to the formation of \( NH_3 \) and \( N_2H_4 \):
- Let the change in pressure for \( NH_3 \) be \( +P_0 \)
- Let the change in pressure for \( N_2H_4 \) be \( +Z \)

From the stoichiometry of the reactions:
- For every 2 moles of \( NH_3 \) formed, 1 mole of \( N_2 \) is consumed, and 3 moles of \( H_2 \) are consumed.
- For every mole of \( N_2H_4 \) formed, 1 mole of \( N_2 \) and 2 moles of \( H_2 \) are consumed.

### Step 4: Setting Up the Equations for \( K_{P_1} \) and \( K_{P_2} \)
Using the equilibrium pressures, we can express \( K_{P_1} \) and \( K_{P_2} \):
1. For \( K_{P_1} \):
\[ K_{P_1} = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(P_0)^2}{(P_{N_2})(2P_0)^3} = \frac{P_0^2}{P_{N_2} \cdot 8P_0^3} = \frac{1}{8P_{N_2}P_0} \]

2. For \( K_{P_2} \):
\[ K_{P_2} = \frac{(P_{N_2H_4})}{(P_{N_2})(P_{H_2})^2} = \frac{(Z)}{(P_{N_2})(2P_0)^2} = \frac{Z}{4P_{N_2}} \]

### Step 5: Solving for the Ratio \( \frac{K_{P_2}}{K_{P_1}} \)
Now we can find the ratio:
\[ \frac{K_{P_2}}{K_{P_1}} = \frac{\frac{Z}{4P_{N_2}}}{\frac{1}{8P_{N_2}P_0}} = \frac{Z \cdot 8P_{N_2}P_0}{4P_{N_2}} = 2ZP_0 \]

### Step 6: Finding \( Z \)
From Equation 1, we have \( P_{N_2} + P_{N_2H_4} = 4P_0 \). If we let \( P_{N_2H_4} = Z \), we can express \( P_{N_2} \) as:
\[ P_{N_2} = 4P_0 - Z \]

### Conclusion
To find \( \frac{K_{P_2}}{K_{P_1}} \), we need the value of \( Z \). However, we can express the final answer in terms of \( Z \):
\[ \frac{K_{P_2}}{K_{P_1}} = 2ZP_0 \]