Question Video: Determining How a Change in Pressure Affects the Equilibrium Position | Nagwa Question Video: Determining How a Change in Pressure Affects the Equilibrium Position | Nagwa

Question Video: Determining How a Change in Pressure Affects the Equilibrium Position Chemistry • Third Year of Secondary School

The following is part of the Ostwald process that produces nitric acid: 4 NH₃ (g) + 5 O₂ (g) ⇌ 4 NO (g) + 6 H₂O (g). Which of the following statements explains why the percentage of NO produced may decrease as the pressure increases? [A] The total volume of gas increases for the backward reaction. [B] Fewer moles of gas are on the reactant’s side. [C] The forward reaction is favored by an increase in pressure. [D] Increasing the pressure moves the equilibrium to the right.

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Video Transcript

The following is part of the Ostwald process that produces nitric acid: four NH3 gas plus five O2 gas in equilibrium with four NO gas plus six H2O gas. Which of the following statements explains why the percentage of NO produced may decrease as the pressure increases? (A) The total volume of gas increases for the backward reaction. (B) Fewer moles of gas are on the reactant’s side. (C) The forward reaction is favored by an increase in pressure. (D) Increasing the pressure moves the equilibrium to the right.

The Ostwald process is the most common method used in industry to synthesize nitric acid. The nitric acid that’s produced is mainly used in fertilizers. The part of the Ostwald process we’re looking at in this question is an equilibrium reaction. We’re being asked what would happen to this equilibrium system when the pressure changes.

If we want to understand how changing conditions will affect an equilibrium system, we should refer to Le Chatelier’s principle. Le Chatelier’s principle states that for a dynamic equilibrium, if the conditions change, the position of equilibrium will move to counteract the change. In a dynamic equilibrium, both the forward and backward reactions are occurring at the same rate. So, no change in concentration of the products and reactants is observed.

In this question, we want to know what will happen when the pressure increases. Le Chatelier’s principle tells us that if the pressure increases, the position of equilibrium will move to decrease the pressure. The pressure in a container of gas is due to gas particles colliding with the walls of the container. The more gas particles there are in the container, the higher the pressure will be. This means when the pressure increases, the position of equilibrium will shift so that fewer moles of gas are produced which will decrease the pressure.

There are nine moles of gas in total on the reactant side and 10 moles of gas in total on the product side. Since there are fewer moles of gas on the reactant side, we know that the backward reaction will be favored as the backward reaction produces fewer moles of gas. When the backward reaction becomes favored, we say that the position of equilibrium shifts to the left. This is because more reactants are formed and the reactants are on the left-hand side of the equation.

Now, we have enough information to explain why the percentage of NO produced may decrease as the pressure increases. When the pressure increases, equilibrium will shift towards the side of the reaction with fewer moles of gas, which will cause the pressure to decrease. In this case, that is the reactants side. NO is a product in this reaction, so the percentage of nitric oxide will decrease as it reacts to form the reactants. The answer that matches this is answer choice (B).

Answer choice (A) is wrong because the total volume of gas decreases for the backward reaction as there are fewer moles of gas produced. Answer choice (C) is wrong because the backward reaction is favored by an increase in pressure. Answer choice (D) is wrong because the increasing pressure will move the equilibrium to the left towards the reactants.

The statement that explains why the percentage of NO produced may decrease as the pressure increases is that there are fewer moles of gas on the reactant side, answer choice (B).

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