AP Chemistry 7.9-7.10: Le Chatelier's Principle and Q

Dynamic equilibrium

A state in which the forward and reverse reactions continue simultaneously at equal rates. The concentrations remain constant, although reactions have not stopped.

Le Châtelier’s principle

When an equilibrium system is subjected to a stress, it shifts in the direction that partially counteracts the stress and establishes a new equilibrium.

What qualifies as a stress that can shift a chemical equilibrium?

A change in concentration, temperature, or—in a gas-phase equilibrium—a change in pressure or volume that changes the relevant partial pressures. A catalyst is not a stress because it does not change the equilibrium composition.

How does a concentration change affect an equilibrium system?

Adding a reactant generally shifts equilibrium toward products, while adding a product shifts it toward reactants. Removing a species causes a shift toward the side that produces that species.

For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}$, what happens when $\mathrm{H_2}$ is added?

The equilibrium shifts right, consuming some added $\mathrm{H_2}$ and some $\mathrm{I_2}$ while producing more $\mathrm{HI}$. At the new equilibrium, $[\mathrm{H_2}]$ is greater than before, $[\mathrm{I_2}]$ is lower, and $[\mathrm{HI}]$ is greater.

For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}$, what happens when $\mathrm{HI}$ is removed?

The equilibrium shifts right to produce more $\mathrm{HI}$. The concentrations of all three species may be lower than before if the amount of $\mathrm{HI}$ removed is sufficiently large, even though the direction of shift is toward products.

For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}$, what happens when reactant is removed or product is added?

The equilibrium shifts left, toward reactants, because the reverse reaction partially opposes the change.

Reaction quotient $Q_c$ for $\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}$

$Q_c=\dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$. At equilibrium, $Q_c=K_c$; pure solids and liquids are omitted from the expression.

How can comparing $Q$ with $K$ predict the direction of an equilibrium shift?

If $Q<K$, the system shifts toward products to increase $Q$. If $Q>K$, it shifts toward reactants to decrease $Q$. If $Q=K$, the system is at equilibrium.

What happens to $Q_c$ when reactant is added to an equilibrium mixture?

The denominator of $Q_c$ increases, so $Q_c<K_c$. The equilibrium shifts right until $Q_c$ again equals $K_c$.

What happens to $Q_c$ when product is removed from an equilibrium mixture?

The numerator of $Q_c$ decreases, so $Q_c<K_c$. The equilibrium shifts right to form more product.

Does changing concentration change the value of the equilibrium constant?

No. At constant temperature, changing concentrations changes $Q$ and causes a shift, but the system re-establishes equilibrium with the same $K$.

Why can a concentration change produce different final concentrations even when the shift direction is the same?

Adding different reactants or removing a product changes the starting composition in different ways. The equilibrium must then adjust stoichiometrically from each distinct composition, although $K$ remains the same at a fixed temperature.

How are gas partial pressure and molar concentration related for an ideal gas?

At constant temperature, $[\mathrm{gas}]=\dfrac{n}{V}=\dfrac{P}{RT}$, so partial pressure is proportional to molar concentration. Therefore, changing a gas's partial pressure has the same equilibrium effect as changing its concentration.

What is the effect of decreasing the volume of a container containing a gas-phase equilibrium?

All gas partial pressures increase, but the equilibrium shifts only if the change alters $Q_p$. If the side with fewer moles of gas is favored, decreasing volume shifts equilibrium toward that side.

How does increasing volume affect a gas-phase equilibrium with unequal total gas moles?

Increasing volume lowers all partial pressures, so equilibrium shifts toward the side with more moles of gas. This produces more gas particles and partially offsets the pressure decrease.

For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}$, what is the effect of changing volume?

There are two moles of gas on each side, so multiplying every partial pressure by the same factor leaves $Q_p$ unchanged. Changing volume causes no equilibrium shift.

For $\mathrm{2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)}$, what happens when volume decreases?

The equilibrium shifts left because the reactant side contains 2 moles of gas, whereas the product side contains 3 moles. The shift toward fewer gas moles partially counteracts the pressure increase.

What is the general rule for pressure changes caused by volume changes in gas equilibria?

Decreasing volume favors the side with fewer moles of gas, while increasing volume favors the side with more moles. If both sides have equal total gas moles, there is no shift.

Why does a volume change not always shift a gas-phase equilibrium?

A volume change scales every gas partial pressure by the same factor. If the total stoichiometric powers of gaseous reactants and products are equal, those factors cancel in $Q_p$, leaving $Q_p=K_p$.

What is the effect of adding an inert gas to a gas-phase equilibrium?

At constant volume, adding an inert gas increases total pressure but does not change the partial pressures of the reacting gases, so equilibrium does not shift. At constant pressure, the volume increases and the equilibrium may shift toward the side with more moles of gas.

How does temperature affect the equilibrium constant?

Temperature changes the forward and reverse rate constants, so it changes the value of $K$. This differs from concentration or pressure changes, which alter $Q$ but not $K$ at a fixed temperature.

How should heat be represented when applying Le Châtelier’s principle to an endothermic reaction?

Heat is treated as a reactant. Increasing temperature adds heat and shifts equilibrium toward products; decreasing temperature shifts it toward reactants.

How should heat be represented when applying Le Châtelier’s principle to an exothermic reaction?

Heat is treated as a product. Increasing temperature shifts equilibrium toward reactants, while decreasing temperature shifts it toward products.

For $\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}$ with $\Delta H=+57.20\ \mathrm{kJ}$, what happens when temperature increases?

The forward reaction is endothermic, so the equilibrium shifts right, producing more $\mathrm{NO_2}$. The equilibrium constant also increases.

For an exothermic equilibrium, what happens to $K$ when temperature increases?

The equilibrium shifts in the endothermic direction, toward reactants, so $K$ decreases when the reaction is written in the forward exothermic direction.

Catalyst

A substance that provides an alternative reaction pathway with lower activation energy. It speeds both the forward and reverse reactions, allowing equilibrium to be reached faster without changing $K$ or the equilibrium composition.

Why does a catalyst not shift an equilibrium?

A catalyst lowers the activation energies for both forward and reverse reactions, increasing both rate constants without changing their ratio in the way that determines $K$. Thus, it changes the time required to reach equilibrium, not the final equilibrium position.

What is the relationship between $K_c$ and rate constants for the elementary reaction $\mathrm{A\rightleftharpoons B}$?

At equilibrium, $k_f[A]=k_r[B]$, so $K_c=\dfrac{[B]}{[A]}=\dfrac{k_f}{k_r}$. Because rate constants depend on temperature, $K_c$ also depends on temperature.

How does opening a carbonated beverage affect the equilibrium $\mathrm{CO_2(g)\rightleftharpoons CO_2(aq)}$?

Opening the container lowers the partial pressure of gaseous $\mathrm{CO_2}$ above the liquid. The dissolution equilibrium shifts left, releasing dissolved $\mathrm{CO_2}$ and eventually making the beverage taste flat.

What sequence of equilibria explains carbonation and acidity in soft drinks?

Dissolved carbon dioxide can hydrate to form carbonic acid, which partially ionizes: $\mathrm{CO_2(g)\rightleftharpoons CO_2(aq)}$, $\mathrm{CO_2(aq)+H_2O(l)\rightleftharpoons H_2CO_3(aq)}$, and $\mathrm{H_2CO_3(aq)\rightleftharpoons HCO_3^-(aq)+H^+(aq)}$.

Why is carbon dioxide dissolved efficiently during soft-drink production?

A high $\mathrm{CO_2}$ pressure shifts $\mathrm{CO_2(g)\rightleftharpoons CO_2(aq)}$ to the right. The increased dissolved $\mathrm{CO_2}$ also promotes formation of carbonic acid and its ionization products.

Why does a sealed carbonated drink retain much of its dissolved carbon dioxide?

The container has a relatively small headspace, so only a limited amount of dissolved $\mathrm{CO_2}$ must leave before the gas pressure rises and the dissolution equilibrium is re-established.

Haber–Bosch equilibrium

The industrial ammonia synthesis is $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)}$, with $\Delta H=-92.2\ \mathrm{kJ}$. It is exothermic and produces fewer moles of gas, so high pressure and removal of ammonia favor product formation.

Why does the Haber–Bosch process use high pressure?

The reaction changes from 4 moles of gaseous reactants to 2 moles of gaseous product. Increasing pressure, or decreasing volume, shifts equilibrium right toward ammonia.

Why is ammonia continuously removed during the Haber–Bosch process?

Removing the product lowers its concentration or partial pressure, making $Q<K$. The equilibrium then shifts right, producing additional ammonia.

Why does the Haber–Bosch process use a moderate temperature rather than the lowest possible temperature?

Because ammonia formation is exothermic, lower temperature favors ammonia thermodynamically. However, the reaction becomes too slow at low temperature, so an industrial compromise uses moderate temperature along with a catalyst.

What are the three distinct roles of pressure, product removal, and a catalyst in ammonia synthesis?

High pressure shifts equilibrium toward $\mathrm{NH_3}$, continuous ammonia removal shifts it further toward products, and a catalyst increases the rate of reaching equilibrium. Only the first two change the equilibrium position; the catalyst does not.

How does a system's response to a stress relate to reaction rates?

A stress makes the forward and reverse rates temporarily unequal. The system undergoes a net reaction in the direction with the greater rate until the rates become equal again.