An unfamiliar reversible reaction between two gases, D and E, reaches dynamic equilibrium in a closed container at constant temperature: 2D(g) + E(g) 2F(g) (a) A small additional amount of gas E is injected into the container, while the temperature and the volume of the containe…

Answer: ((a)) The position of equilibrium shifts to the right, towards F; the system opposes the increase in E concentration by using up some of the added E (together with some D) to form more F. Because some D is used up in this shift, the concentration of D decreases, falling below the valu… ((b)) Continuously removing F keeps its concentration below the value it would otherwise reach at equilibrium, so the equilibrium keeps shifting right to replace it, converting more D and E into F. Because this shift happens repeatedly rather than stopping once at a single equilibrium… ((c)) There are 2 + 1 = 3 moles of gas on the reactant side and 2 moles of gas on the product side, so there are fewer gas molecules on the product side. Doubling the volume decreases the concentration (and pressure) of all the gases; the equilibrium shifts towards the side with the g…. Part (a): Increasing the concentration of E By Le Chatelier's principle, if a system at equilibrium is disturbed, it responds in a way that opposes the change. Injecting more gas E increases its concentration. The equilibrium responds by s…

IGCSE Chemistry 0620: Effect of Concentration and Volume on the Position of Equilibrium

An unfamiliar reversible reaction between two gases, $\text{D}$ and $\text{E}$ , reaches dynamic equilibrium in a closed container at constant temperature:

$2\text{D}(g) + \text{E}(g) \rightleftharpoons 2\text{F}(g)$

(a) A small additional amount of gas $\text{E}$ is injected into the container, while the temperature and the volume of the container are kept constant. Predict, using Le Chatelier's principle, the effect this has on the position of equilibrium, and state what happens to the concentration of $\text{D}$ as a new equilibrium is reached. [3]

(b) $\text{F}(g)$ is then removed continuously from the container as it forms, by a separate process that does not affect $\text{D}$ or $\text{E}$ , while the temperature is kept constant. Predict and explain the effect of continuously removing $\text{F}(g)$ on the overall amount of $\text{F}$ produced over time, compared with simply allowing the mixture to settle at a single, static equilibrium. [2]

(c) Starting again from the original equilibrium mixture, the volume of the container is instead doubled at constant temperature, without adding or removing any gas. Using the number of moles of gas on each side of the equation, predict and explain the effect this has on the position of equilibrium. [2]

Part (a): Increasing the concentration of E

By Le Chatelier’s principle, if a system at equilibrium is disturbed, it responds in a way that opposes the change.

Injecting more gas $\text{E}$ increases its concentration. The equilibrium responds by shifting in the direction that uses up some of this extra $\text{E}$ . That is, towards the right, converting more $\text{D}(g)$ and $\text{E}(g)$ into $\text{F}(g)$ :

$2\text{D}(g) + \text{E}(g) \rightleftharpoons 2\text{F}(g)$

Because this rightward shift also consumes some $\text{D}(g)$ , the concentration of $\text{D}$ decreases, ending up lower than it was in the original equilibrium mixture, even though no $\text{D}$ was added directly.

Part (b): Continuously removing the product

Normally, once equilibrium is reached, the concentrations of $\text{D}$ , $\text{E}$ and $\text{F}$ stop changing. But if $\text{F}(g)$ is constantly being removed as it forms, its concentration is kept below the value it would otherwise settle at.

By Le Chatelier’s principle, the equilibrium continually shifts to the right to try to replace the $\text{F}$ that has been removed, converting more $\text{D}$ and $\text{E}$ into product. Because this removal-and-replacement happens repeatedly, over time a greater overall amount of $\text{F}$ is produced than if the mixture were simply left to reach a single, static equilibrium.

Part (c): Doubling the volume of the container

Counting moles of gas in the equation:

Reactant side: $2\text{D}(g) + \text{E}(g)$ = $2 + 1 = 3$ moles of gas
Product side: $2\text{F}(g)$ = $2$ moles of gas

The product side has fewer moles of gas than the reactant side.

Doubling the volume of the container, at constant temperature and with no gas added or removed, decreases the concentration (and pressure) of every gas present. By Le Chatelier’s principle, the equilibrium shifts towards the side with the greater number of gas molecules, since that partly opposes the decrease in concentration/pressure by increasing the total number of gas particles present. Here, that is the reactant side, so the equilibrium shifts to the left: the amount of $\text{F}$ decreases, and the amounts of $\text{D}$ and $\text{E}$ increase.

Final answers

(a) Equilibrium shifts right; concentration of $\text{D}$ decreases.
(b) Continual removal of $\text{F}$ keeps shifting equilibrium right, giving a greater overall amount of $\text{F}$ produced.
(c) Fewer gas moles on the product side (2 vs 3), so doubling the volume shifts equilibrium left, decreasing the amount of $\text{F}$ .