# 0.18 Reaction equilibrium in the gas phase  (Page 6/6)

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A look back at [link] and [link] shows that the equilibrium pressure of the product of the reaction increases with increasingthe initial quantity of reaction. This seems quite intuitive. Less intuitive is the variation of the equilibrium pressure of theproduct of this reaction with variation in the volume of the container, as shown in [link] . Note that the pressure ofNH 3 decreases by more than a factor of ten when the volume is increased by a factor of ten. This means that, at equilibrium, there arefewer moles of NH 3 produced when the reaction occurs in a larger volume.

To understand this effect, we rewrite the equilibrium constant in [link] to explicit show the volume of the container. This is done by applying Dalton's Law of Partial Pressures , so that each partial pressure is given by the Ideal Gas Law:

${K}_{p}=\frac{{n}_{\text{N}{\text{H}}_{3}}^{2}\left(\frac{RT}{V}\right)^{2}}{{n}_{{\text{N}}_{2}}\frac{RT}{V}{n}_{{\text{H}}_{2}}^{3}\left(\frac{RT}{V}\right)^{3}}=\frac{{n}_{\text{N}{\text{H}}_{3}}^{2}}{{n}_{{\text{N}}_{2}}{n}_{{\text{H}}_{2}}^{3}\left(\frac{RT}{V}\right)^{2}}$

Therefore,

${K}_{p}\left(\frac{RT}{V}\right)^{2}=\frac{{n}_{\text{N}{\text{H}}_{3}}^{2}}{{n}_{{\text{N}}_{2}}{n}_{{\text{H}}_{2}}^{3}}$

This form of the equation makes it clear that, when the volume increases, the left side of the equation decreases.This means that the right side of the equation must decrease also, and in turn, ${n}_{\text{N}{\text{H}}_{3}}$ must decrease while ${n}_{{\text{N}}_{2}}$ and ${n}_{{\text{H}}_{2}}$ must increase. The equilibrium is thus shifted from products toreactants when the volume increases for this reaction .

The effect of changing the volume must be considered for each specific reaction, because the effect dependson the stoichiometry of the reaction. One way to determine the consequence of a change in volume is to rewrite the equilibriumconstant as we have done in [link] .

Finally, we consider changes in temperature. We note that ${K}_{p}$ increases with $T$ for endothermic reactions and decreases with $T$ for exothermic reactions. As such, the products are increasinglyfavored with increasing temperature when the reaction is endothermic, and the reactants are increasingly favored withincreasing temperature when the reaction is exothermic. On reflection, we note that when the reaction is exothermic, thereverse reaction is endothermic. Putting these statements together, we can say that the reaction equilibrium always shifts in thedirection of the endothermic reaction when the temperature is increased.

All of these observations can be collected into a single unifying concept known as Le Châtelier's Principle .

Principle

## Le châtelier's principle

When a reaction at equilibrium is stressed by a change in conditions, the equilibrium will be reestablished insuch a way as to counter the stress.

This statement is best understood by reflection on the types of "stresses" we haveconsidered in this section. When a reactant is added to a system at equilibrium, the reaction responds by consuming some of that addedreactant as it establishes a new equilibrium. This offsets some of the stress of the increase in reactant. When the temperature israised for a reaction at equilibrium, this adds thermal energy. The system shifts the equilibrium in the endothermic direction, thusabsorbing some of the added thermal energy, countering the stress.

The most challenging of the three types of stress considered in this section is the change in volume. Byincreasing the volume containing a gas phase reaction at equilibrium, we reduce the partial pressures of all gases presentand thus reduce the total pressure. Recall that the response of this reaction to the volume increase was to create more of the reactants at theexpense of the products. One consequence of this shift is that more gas molecules are created, and this increases the total pressure inthe reaction flask. Thus, the reaction responds to the stress of the volume increase by partially offsetting the pressure decreasewith an increase in the number of moles of gas at equilibrium.

Le Châtelier's principle is a useful mnemonic for predicting how we might increase or decreasethe amount of product at equilibrium by changing the conditions of the reaction. From this principle, we can predict whether thereaction should occur at high temperature or low temperature, and whether it should occur at high pressure or low pressure.

## Review and discussion questions

1. In the data given for equilibrium of this reaction , there is no volume given. Show that changing the volume for the reactiondoes not change the number of moles of reactants and products present at equilibrium, i.e. changing the volume does not shift the equilibrium.
2. For this reaction the number of moles of NO 2 at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms ofdynamic equilibrium, give a reason why the rates of the forward and reverse reactions might be affected differently by changes in thevolume.
3. We could balance [link] by writing
$2{\text{N}}_{2}\left(g\right)+6{\text{H}}_{2}\left(g\right)\to 4\text{N}{\text{H}}_{3}\left(g\right)$
Write the form of the equilibrium constant for the reaction balanced as in [link] . What is the value of the equilibrium constant? (Refer to [link] .) Of course, the pressures at equilibrium do not depend on whether the reaction is balanced as in [link] or as in [link] . Explain why this is true, even though the equilibrium constant can be written differently and havea different value.
4. Show that the equilibrium constant ${K}_{p}$ in [link] for this reaction can be written in terms of the concentrations or particle densities, e.g. $\left[{\text{N}}_{2}\right]=\frac{{n}_{{\text{N}}_{2}}}{V}$ , instead of the partial pressures. In this form, we call theequilibrium constant ${K}_{c}$ . Find the relationship between ${K}_{p}$ and ${K}_{c}$ , and calculate the value of ${K}_{c}$ .
5. For each of these reactions, predict whether increases in temperature will shift the reaction equilibrium moretowards products or more towards reactants.

$2\text{C}\text{O}\left(g\right)+{\text{O}}_{2}\left(g\right)\to 2\text{C}{\text{O}}_{2}\left(g\right)$

${\text{O}}_{3}\left(g\right)+\text{N}\text{O}\left(g\right)\to \text{N}{\text{O}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)$

$2{\text{O}}_{3}\left(g\right)\to 3{\text{O}}_{2}\left(g\right)$

6. Plot the data in [link] on a graph showing ${K}_{p}$ on the y-axis and $T$ on the x-axis. The shape of this graph is reminiscent of the graph ofanother physical property as a function of increasing temperature. Identify that property, and suggest a reason why the shapes of thegraphs might be similar.
7. Using Le Châtelier's principle, predict whether the specified "stress" will produce anincrease or a decrease in the amount of product observed at equilibrium for the reaction:
$2{\text{H}}_{2}\left(g\right)+\text{C}\text{O}\left(g\right)\to \text{C}{\text{H}}_{3}\text{O}\text{H}\left(g\right)$

$\Delta ({H}^{°})=-91\frac{\mathrm{kJ}}{\mathrm{mol}}$

Volume of container is increased.

Temperature of container is raised.

CH 3 OH is extracted from container as it is formed.

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Maciej
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s.
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Tarell
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Damian
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s.
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Cied
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