Q.On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = – 92.38 kJ mol^-1
What will be the effect of addition of argon to the above reaction mixture at constant volume?

The key idea here is Le Chatelier's Principle, which states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

1. Effect of Temperature: The reaction $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ has $\Delta H = -92.38 \text{ kJ mol}^{-1}$, indicating it is an exothermic reaction. To increase the yield of ammonia, the equilibrium must shift to the right (product side). According to Le Chatelier's principle, for an exothermic reaction, decreasing the temperature favors the forward reaction, thus increasing the ammonia yield.

2. Effect of Pressure: The number of moles of gaseous reactants is $1 (\text{N}_2) + 3 (\text{H}_2) = 4$ moles, while the number of moles of gaseous products is $2 (\text{NH}_3) = 2$ moles. To increase the yield of ammonia, the equilibrium must shift to the side with fewer moles of gas. According to Le Chatelier's principle, increasing the pressure favors the reaction that leads to a decrease in the total number of gaseous moles, which in this case is the forward reaction, thus increasing the ammonia yield. …

Le Chatelier's principle is a fundamental concept in chemical equilibrium that helps predict how a system at equilibrium responds to changes in conditions. It states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that counteracts the change, thereby re-establishing a new equilibrium. We will use this principle to determine the optimal conditions for ammonia production.

The given reaction is:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

$\Delta H = – 92.38 \text{ kJ mol}^{-1}$

Let's analyze the reaction characteristics first:

1. Nature of the reaction (based on $\Delta H$): The negative value of $\Delta H$ ($-92.38 \text{ kJ mol}^{-1}$) indicates that the forward reaction (formation of ammonia) is exothermic. This means heat is released when ammonia is formed. Conversely, the reverse reaction (decomposition of ammonia) is endothermic.
2. Change in the number of moles of gas:
   • Moles of gaseous reactants = $1 \text{ mol } N_2 + 3 \text{ mol } H_2 = 4 \text{ moles of gas}$
   • Moles of gaseous products = $2 \text{ mol } NH_3 = 2 \text{ moles of gas}$
   • The forward reaction leads to a decrease in the total number of moles of gas (from 4 moles to 2 moles).

Now, let's apply Le Chatelier's principle to understand the effect of temperature and pressure.

Effect of Temperature on Ammonia Yield

1. Reasoning: Since the forward reaction ($N_2 + 3H_2 \rightarrow 2NH_3$) is exothermic ($\Delta H < 0$), it releases heat. To increase the yield of ammonia, we want to shift the equilibrium to the right (towards products).
2. Applying Le Chatelier's Principle: If the temperature of an exothermic reaction at equilibrium is decreased, the system will try to counteract this change by shifting in the direction that produces heat. This means the equilibrium will shift towards the products (ammonia).
3. Conclusion: To increase the yield of ammonia, the temperature should be decreased.

Effect of Pressure on Ammonia Yield

1. Reasoning: The forward reaction involves a decrease in the number of moles of gas (from 4 moles of reactants to 2 moles of products). To increase the yield of ammonia, we want to shift the equilibrium to the right (towards products).
2. Applying Le Chatelier's Principle: If the pressure of a gaseous system at equilibrium is increased, the system will try to counteract this change by shifting in the direction that reduces the total number of moles of gas. This means the equilibrium will shift towards the side with fewer moles of gas.
3. Conclusion: Since the product side ($2NH_3$) has fewer moles of gas (2 moles) than the reactant side ($N_2 + 3H_2$, 4 moles), increasing the pressure will shift the equilibrium to the right, thereby increasing the yield of ammonia.