What is meant by chemical equilibrium? Why is it described as a dynamic state? Consider the following reversible chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92.4 kJ (b) Use Le Chatelier's principle to predict the levels (high or low) of temperature and pressure needed to maximise the yield of ammonia when equilibrium is established - Leaving Cert Chemistry - Question 9 - 2004

For maximizing the yield of ammonia:

• Predicted Temperature Level: Low temperature.
• Predicted Pressure Level: High pressure.

Reasoning: (i) The forward reaction, which produces ammonia, is exothermic (it releases heat). Therefore, lowering the temperature shifts the equilibrium to the right, favoring the production of ammonia.

(ii) The forward reaction results in fewer moles of gas (from 4 moles of reactants to 2 moles of products). According to Le Chatelier's principle, increasing pressure shifts the equilibrium towards the side with fewer moles, hence high pressure favors ammonia production.

In industry, lower temperatures are not used to maximize ammonia yield despite predictions from Le Chatelier's principle. This is because the rate of reaction decreases significantly at lower temperatures, making the process impractically slow. Therefore, a compromise temperature is selected that balances a reasonable yield of ammonia with an acceptable reaction rate.

A catalyst lowers the activation energy for both the forward and reverse reactions, allowing them to occur more easily and quickly. However, it does not affect the position of equilibrium or the yield of products; it merely allows equilibrium to be reached more rapidly.

The equilibrium constant expression (Kc) for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is:

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

To calculate Kc with the provided data:

• Moles of nitrogen (N₂): 6.0 moles
• Moles of hydrogen (H₂): 18.0 moles
• Moles of ammonia (NH₃): 6.0 moles
• Volume of the vessel: 5.0 L

Calculating the concentrations:

• Concentration of N₂: ( \frac{6.0}{5.0} = 1.2 , M )
• Concentration of H₂: ( \frac{18.0}{5.0} = 3.6 , M )
• Concentration of NH₃: ( \frac{6.0}{5.0} = 1.2 , M )

Substituting these values into the Kc expression:

$K_c = \frac{(1.2)^2}{(1.2)(3.6)^3} = \frac{1.44}{(1.2)(46.656)} \approx 0.4$

Thus, the value of the equilibrium constant Kc at this temperature is approximately 0.4.