# Question Video: Identifying a Reversible Reaction from Trends in Equilibrium Yield Chemistry • 10th Grade

The effects of temperature and pressure on the equilibrium yield of a reversible reaction are shown in the graph. In which of the following reactions would this behavior be observed? [A] CO + H₂O ⇌ CO₂ + H₂, Δ𝐻 ＜ 0 [B] N₂ + 3H₂ ⇌ 2NH₃, Δ𝐻 ＜ 0 [C] CH₄ + H₂O ⇌ CO + 3H₂, Δ𝐻 ＞ 0 [D] N₂ + 2O₂ ⇌ 2NO₂, Δ𝐻 ＞ 0 [E] 2NO₂ + O₂ ⇌ N₂ + 2O₃, Δ𝐻 ＞ 0

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### Video Transcript

The effects of temperature and pressure on the equilibrium yield of a reversible reaction are shown in the graph. In which of the following reactions would this behavior be observed? (A) CO plus H2O reacts reversibly to produce CO2 plus H2, where Δ𝐻 is less than zero. (B) N2 plus three H2 reacts reversibly to produce two NH3, where Δ𝐻 is less than zero. (C) CH4 plus H2O reacts reversibly to produce CO plus three H2, where Δ𝐻 is greater than zero. (D) N2 plus two O2 reacts reversibly to produce two NO2, where Δ𝐻 is greater than zero. Or (E) two NO2 plus O2 reacts reversibly to produce N2 plus two O3, where Δ𝐻 is greater than zero.

In this reaction, we observe two different changes to the equilibrium yield in response to two different stimuli, temperature and pressure. We note that at higher temperature, the yield is greater than at lower temperature. We also note that as pressure increases, yield decreases.

To figure out in which of the given reactions this behavior is observed, we first need to consider Le Chatelier’s principle. Le Chatelier’s principle states that for a dynamic equilibrium if the conditions change — concentration, temperature, or pressure — the position of equilibrium will move to counteract the change. So if temperature were to be increased, the position of the equilibrium would move to counteract the change. This means that the position of the equilibrium would shift in the endothermic direction, as in an endothermic reaction, more energy is taken in than given out. This would reduce the temperature.

If we look at the example reactions given to us, we can see that the first two reactions have a Δ𝐻 less than zero. This means that the forward reactions are exothermic; thus, the reverse reactions are endothermic. As previously stated, an increase in temperature would cause a shift in the endothermic direction. Thus, for options (A) and (B), the position of the equilibrium would shift to the left, towards the reactants. This would cause an increase in the concentration of reactants and a decrease in the yield of products. This is not consistent with the information given in the question, as an increase in temperature should give an increase in equilibrium yield. Therefore, neither option (A) or (B) can be correct.

For an increase in temperature to cause an increase in equilibrium yield, the forward reaction must be endothermic. So the answer will be (C), (D), or (E).

Next, let’s consider pressure. At higher pressures, the yield decreases. If the pressure is increased, the position of equilibrium will shift to counteract the change. So the equilibrium will shift to the side with fewer moles of gas. If we look at option (C), for example, there is a total of two moles of gas on the reactant side and four moles on the product side. So with an increase in pressure, the position of the equilibrium would shift to the left, causing an increase in the concentration of reactants or a decrease in the equilibrium yield. This is in line with what we know that an increase in pressure causes a decrease in equilibrium yield. Therefore, it seems as though the answer is option (C). But to confirm, let’s check the number of moles of gas in options (D) and (E).

Option (D) has fewer moles of gas on the product side, so it cannot be the answer to this question. An increase in pressure would cause an increase in equilibrium yield rather than a decrease. For option (E), there are three moles of gas on the reactant side and three moles of gas on the product side. So a change in pressure would have no impact on the position of the equilibrium. Therefore, option (E) cannot be the answer to this question.

Therefore, the reaction which would exhibit the behavior shown in the graph is (C). CH4 plus H2O reacts reversibly to produce CO plus three H2, where Δ𝐻 is greater than zero.