Consider 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g). Delta H=-904.4 kJ. How does each of the following changes affect the yield of NO at equilibrium? Increase, decrease or no changea) increase [NH3]b) increase [H2O]c) decrese [O2]d) decrease the volume of the container in which the reaction occurse) add a catalystf) increase temperature

a) increase; b) decrease; c) decrease; d) decrease; e) same; b) decrease

Explanation:

4NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O (g); ΔH = - 904.4 kJ

The enthalpy of reaction is negative, so the reaction is exothermic, and we can consider the heat as a "product".

We can rewrite the equation as

4NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O (g) + 904.4 kJ

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

a) Increase [NH₃]

If we increase the concentration of ammonia, the position of equilibrium (POE) will move to the right to get rid of the added ammonia. The yield of NO will increase.

b) Increase [H₂O]

If we add increase the concentration of water vapour, the POE will move to the left to get rid of the added water. The yield of NO will decrease.

c) Decrease [O₂]

If we add decrease the concentration of oxygen, the POE will move to the left to get to replace the removed oxygen. The yield of NO will decrease.

d) Decrease the volume of the container

We have nine moles of reactants and 10 mol of products.

If we decrease volume of the container, the POE will move to the left (the direction of fewer moles) to reduce the increased pressure. The yield of NO will decrease.

e) Add a catalyst

Adding a catalyst will speed up the approach to equilibrium, but it has no effect on the POE. The POE will remain unchanged. The yield of NO will stay the same.

f) Increase temperature

If we increase the temperature, the POE will move to the left to get rid of the added heat. The yield of NO will decrease.