The equilibrium of the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` will be shifted to the right when:

Options

A. `K_(p) lt Q`
B. `Q lt K_(p)`
C. `Q=K_(p)`
D. `Q lt 2K_(p)`

Answer

To determine when the equilibrium of the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) will shift to the right, we can analyze the factors that affect the equilibrium position according to Le Chatelier's principle.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction involves nitrogen gas (\( N_2 \)) and hydrogen gas (\( H_2 \)) reacting to form ammonia gas (\( NH_3 \)). The balanced equation shows that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

2. **Equilibrium Constant (K)**:
The equilibrium constant \( K \) for the reaction can be expressed in terms of the concentrations of the products and reactants:
\[ K = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Here, \( [NH_3] \) is the concentration of ammonia, \( [N_2] \) is the concentration of nitrogen, and \( [H_2] \) is the concentration of hydrogen.

3. **Reaction Quotient (Q)**:
The reaction quotient \( Q \) is calculated in the same way as \( K \), but it can be evaluated at any point in time, not just at equilibrium:
\[ Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

4. **Le Chatelier's Principle**:
According to Le Chatelier's principle, if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium.

5. **Shifting the Equilibrium to the Right**:
The equilibrium will shift to the right (toward the formation of products) under the following conditions:
- **Increase in Reactants**: Adding more \( N_2 \) or \( H_2 \) will shift the equilibrium to the right to produce more \( NH_3 \).
- **Decrease in Products**: Removing \( NH_3 \) will also shift the equilibrium to the right to produce more \( NH_3 \).
- **Increase in Pressure**: Since there are fewer moles of gas on the product side (2 moles of \( NH_3 \)) compared to the reactant side (4 moles total: 1 mole of \( N_2 \) and 3 moles of \( H_2 \)), increasing the pressure will favor the formation of \( NH_3 \).
- **Increase in Temperature**: If the reaction is exothermic, increasing the temperature will shift the equilibrium to the left. If it is endothermic, increasing the temperature will shift it to the right.

6. **Conclusion**:
The equilibrium of the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) will shift to the right when:
- The concentration of reactants \( N_2 \) or \( H_2 \) is increased.
- The concentration of products \( NH_3 \) is decreased.
- The pressure is increased.
- The temperature is adjusted depending on the nature of the reaction (exothermic or endothermic).