For the reaction 4 NH3 (g) + 5 O2 (g) ⇌ 4 NO (g) + 6 H2O (g) ΔH∘ = - 904 kJ which of the following changes will shift the equilibrium to the right, toward the formation of more products? For the reaction which of the following changes will shift the equilibrium to the right, toward the formation of more products? Increasing the volume of the reaction vessel Adding more water vapor Removing O2 (g) Adding 1 atm of Ne (g) to the reaction vessel Increasing the temperature SubmitRequest Answer

Increasing the volume of the reaction vessel.

Explanation:

Let us consider the following reaction:

4 NH₃ (g) + 5 O₂ (g) ⇌ 4 NO (g) + 6 H₂O (g) ΔH° = - 904 kJ

To assess the effect of different changes on the equilibrium we need to consider Le Chatelier Principle: if a system at equilibrium undergoes a change (in pressure, volume, etc.), it will react to counteract such change.

Increasing the volume of the reaction vessel

If volume increases, pressure decreases according to Boyle's Law. Then, the system will try to increase pressure by shifting the equilibrium towards where there are more moles of gases, that is, to the right (10 moles of gases).

Adding more water vapor

If water vapor is added, the system will try to reduce its concentration by shifting the equilibrium towards the left.

Removing O₂ (g)

If oxygen is removed, the system will try to increase its concentration by shifting the equilibrium towards the left.

Adding 1 atm of Ne (g) to the reaction vessel

Adding inert gases has no effect on the equilibrium.

Increasing the temperature

This is an exothermic reaction (ΔH° < 0). If temperature is raised, the system will try to decrease it, shifting the equilibrium towards the left to absorb heat.