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.