In a vessel, two equilibrium are simultanceously established at the same temperature as follows:
`N_(2)(g)+3H_(2)(g)hArr2 NH_(3)(g)`.... (1)
`N_(2)g)+2H_(2)(g)hARrN_(2)H_(4)(g)` ....(2)
Initially the vessel contains `N_(2) "and" H_(2)` in the molar ratio of `9:13`. The equilibrium pressure is `7P_(0)`, in wich pressure due to ammonia is `P_(o)` and due to hydrogen is `2P_(0)`. Find the values of equilibrium constats `(K_(p)'S)` for both the reactions

Answer

To solve the problem, we need to analyze the two simultaneous equilibria established in the vessel and calculate the equilibrium constants for both reactions.

### Step-by-Step Solution

1. **Identify Initial Conditions**:
- The initial molar ratio of \(N_2\) to \(H_2\) is \(9:13\).
- Let \(P_0\) be the pressure unit. Thus, we can express the initial pressures as:
\[ P_{N_2, \text{initial}} = 9P \quad \text{and} \quad P_{H_2, \text{initial}} = 13P \]

2. **Set Up the Equilibrium Expressions**:
- For the first reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
Let \(X\) be the change in pressure for \(N_2\) and \(H_2\) at equilibrium. The equilibrium pressures will be:
\[ P_{N_2} = 9P - X, \quad P_{H_2} = 13P - 3X, \quad P_{NH_3} = 2X \]

- For the second reaction:
\[ N_2 + 2H_2 \rightleftharpoons N_2H_4 \]
Let \(Y\) be the change in pressure for this reaction. The equilibrium pressures will be:
\[ P_{N_2} = 9P - X - Y, \quad P_{H_2} = 13P - 3X - 2Y, \quad P_{N_2H_4} = Y \]

3. **Use Given Equilibrium Pressures**:
- The problem states that the equilibrium pressure due to ammonia \(P_{NH_3} = P_0\) and due to hydrogen \(P_{H_2} = 2P_0\).
- From the first reaction, we have:
\[ 2X = P_0 \implies X = \frac{P_0}{2} \]

4. **Substitute \(X\) into the Hydrogen Pressure Equation**:
- From the first reaction:
\[ P_{H_2} = 13P - 3X = 2P_0 \]
Substituting \(X\):
\[ 13P - 3\left(\frac{P_0}{2}\right) = 2P_0 \implies 13P - \frac{3P_0}{2} = 2P_0 \]
Rearranging gives:
\[ 13P = 2P_0 + \frac{3P_0}{2} = \frac{4P_0 + 3P_0}{2} = \frac{7P_0}{2} \]
Thus:
\[ P = \frac{7P_0}{26} \]

5. **Calculate \(Y\)**:
- Substitute \(P\) into the equilibrium expression for \(H_2\) from the second reaction:
\[ P_{H_2} = 13P - 3X - 2Y = 2P_0 \]
Substituting \(P\) and \(X\):
\[ 13\left(\frac{7P_0}{26}\right) - 3\left(\frac{P_0}{2}\right) - 2Y = 2P_0 \]
Simplifying:
\[ \frac{91P_0}{26} - \frac{3P_0}{2} - 2Y = 2P_0 \]
Converting \(\frac{3P_0}{2}\) to a common denominator:
\[ \frac{91P_0}{26} - \frac{39P_0}{26} - 2Y = 2P_0 \]
Thus:
\[ \frac{52P_0}{26} - 2Y = 2P_0 \implies 2 - 2Y = 2 \implies Y = 0 \]

6. **Final Equilibrium Pressures**:
- Substitute \(X\) and \(Y\) back to find the final equilibrium pressures:
\[ P_{N_2} = 9P - X - Y = 9\left(\frac{7P_0}{26}\right) - \frac{P_0}{2} = \frac{63P_0}{26} - \frac{13P_0}{26} = \frac{50P_0}{26} = \frac{25P_0}{13} \]
\[ P_{H_2} = 13P - 3X - 2Y = 13\left(\frac{7P_0}{26}\right) - 3\left(\frac{P_0}{2}\right) = \frac{91P_0}{26} - \frac{39P_0}{26} = \frac{52P_0}{26} = 2P_0 \]

7. **Calculate Equilibrium Constants**:
- For the first reaction:
\[ K_{p1} = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(P_0)^2}{\left(\frac{25P_0}{13}\right)(2P_0)^3} = \frac{P_0^2}{\frac{25P_0}{13} \cdot 8P_0^3} = \frac{13}{200} = 0.065 \]

- For the second reaction:
\[ K_{p2} = \frac{P_{N_2H_4}}{(P_{N_2})(P_{H_2})^2} = \frac{0}{\left(\frac{25P_0}{13}\right)(2P_0)^2} = 0 \]

### Final Answer
- The equilibrium constant \(K_{p1}\) for the first reaction is \(0.065\).
- The equilibrium constant \(K_{p2}\) for the second reaction is \(0\).