In a reaction mixture containing `H_(2),N_(2)` and `NH_(3)` at partial pressure of 2 atm, 1 atm and 3 atm respectively, the value of `K_(p)` at `725K` is `4.28xx10^(-5)atm^(-2)` . In which direction the net reaction will go ?
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`

Options

A. Forward
B. Backward
C. No net reaction
D. Direction cannot be predicted

Explanation

To determine the direction of the net reaction for the given system, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Identify the partial pressures
From the question, we have the following partial pressures:
- \( P_{H_2} = 2 \, \text{atm} \)
- \( P_{N_2} = 1 \, \text{atm} \)
- \( P_{NH_3} = 3 \, \text{atm} \)

### Step 3: Calculate the reaction quotient \( Q_p \)
The reaction quotient \( Q_p \) is calculated using the formula:
\[ Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
Substituting the values:
\[ Q_p = \frac{(3)^2}{(1)(2)^3} = \frac{9}{1 \times 8} = \frac{9}{8} = 1.125 \]

### Step 4: Compare \( Q_p \) with \( K_p \)
The equilibrium constant \( K_p \) is given as:
\[ K_p = 4.28 \times 10^{-5} \, \text{atm}^{-2} \]

Now we compare \( Q_p \) and \( K_p \):
- \( Q_p = 1.125 \)
- \( K_p = 4.28 \times 10^{-5} \)

### Step 5: Determine the direction of the reaction
Since \( Q_p > K_p \), the reaction will shift to the left (towards the reactants) to reach equilibrium.

### Conclusion
The net reaction will proceed in the backward direction.

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