The value of `K_P` for the reactions `N_2(g) +3H_2(g) hArr 2NH_3(g), " is " 4.28 xx 10 ^(-5) " at " 450 ^(@) C`. A reaction mixture contains `N_2,H_2 and NH_3` at partial pressures of 0.6 atm ,2.5 atm and 0.50 atm respectively . In which direction the reaction will proceed?

Answer

To determine the direction in which the reaction will proceed, we need to compare the reaction quotient \( Q_p \) with the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The balanced equation for the reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

2. **Identify the Given Values:**
- \( K_p = 4.28 \times 10^{-5} \) at \( 450^\circ C \)
- Partial pressures:
- \( P_{N_2} = 0.6 \, \text{atm} \)
- \( P_{H_2} = 2.5 \, \text{atm} \)
- \( P_{NH_3} = 0.5 \, \text{atm} \)

3. **Write the Expression for \( K_p \):**
The expression for \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

4. **Calculate the Reaction Quotient \( Q_p \):**
The reaction quotient \( Q_p \) is calculated using the current partial pressures:
\[ Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(0.5)^2}{(0.6)(2.5)^3} \]

- Calculate \( (0.5)^2 = 0.25 \)
- Calculate \( (2.5)^3 = 15.625 \)
- Now substitute these values into the equation:
\[ Q_p = \frac{0.25}{0.6 \times 15.625} \]
- Calculate \( 0.6 \times 15.625 = 9.375 \)
- Thus,
\[ Q_p = \frac{0.25}{9.375} \approx 0.02667 \, \text{(or } 2.67 \times 10^{-2}) \]

5. **Compare \( Q_p \) with \( K_p \):**
- We have \( Q_p \approx 2.67 \times 10^{-2} \)
- And \( K_p = 4.28 \times 10^{-5} \)

Since \( Q_p > K_p \), it indicates that the reaction mixture has more products than at equilibrium.

6. **Determine the Direction of the Reaction:**
According to Le Chatelier's principle, if \( Q_p > K_p \), the reaction will shift to the left (towards the reactants) to reach equilibrium.

### Conclusion:
The reaction will proceed in the backward direction (towards the reactants).