AP Chemistry 7.3-7.4: Reaction Quotient and Calculating K

Reaction quotient $Q_c$

For a reaction $mA+nB\rightleftharpoons xC+yD$, $Q_c=\frac{[C]^x[D]^y}{[A]^m[B]^n}$. The exponents are the stoichiometric coefficients from the balanced equation, and the concentrations describe the system at any moment.

How is a partial-pressure reaction quotient written for a gaseous reaction?

For $mA(g)+nB(g)\rightleftharpoons xC(g)+yD(g)$, $Q_p=\frac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n}$, where each $P$ is the species' partial pressure.

How are the numerator and denominator of a reaction quotient determined from a balanced chemical equation?

Species on the product side appear in the numerator, and species on the reactant side appear in the denominator. Each species is raised to the power of its coefficient in the balanced equation.

Write $Q_c$ for $2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)$.

$Q_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$.

Write $Q_c$ for $N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$.

$Q_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$.

Write $Q_c$ for $4NH_3(g)+7O_2(g)\rightleftharpoons4NO_2(g)+6H_2O(g)$.

$Q_c=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}$.

What is the relationship between the reaction quotient and the equilibrium constant?

The equilibrium constant is the reaction quotient evaluated at equilibrium: $K=Q_{\text{eq}}$. At a fixed temperature, every equilibrium state for the same reaction has the same value of $K$.

What happens to $Q_c$ when a reaction starts with reactants only?

If products are initially absent, the numerator of $Q_c$ is zero, so $Q_c=0$. The reaction initially proceeds forward, increasing $Q_c$ until $Q_c=K_c$.

What happens to $Q_c$ when a reaction starts with products only?

If reactants are initially absent, the denominator of $Q_c$ is zero, so the simplified quotient is undefined or approaches infinity. The reaction initially proceeds in the reverse direction until $Q_c=K_c$.

How can $Q$ and $K$ predict the direction of a reaction?

If $Q<K$, the system proceeds in the forward direction to form more products. If $Q>K$, it proceeds in the reverse direction to form more reactants; if $Q=K$, the system is at equilibrium and has no net reaction.

A reaction has $K_c=0.640$ and a mixture has $Q_c=0.040$. In which direction will the net reaction proceed?

Because $Q_c<K_c$, the net reaction proceeds forward, toward products.

A reaction has $K_c=0.640$ and a mixture has $Q_c=140$. In which direction will the net reaction proceed?

Because $Q_c>K_c$, the net reaction proceeds in reverse, toward reactants.

What does a large equilibrium constant indicate about the equilibrium composition?

A large $K$ indicates that products are favored at equilibrium and that a relatively large fraction of reactants is converted to products. It does not mean the reaction occurs quickly.

What does a small equilibrium constant indicate about the equilibrium composition?

A small $K$ indicates that reactants are favored at equilibrium and that relatively little reactant is converted to product.

Why does the magnitude of $K$ not indicate reaction rate?

The equilibrium constant describes the composition at equilibrium, not how rapidly equilibrium is reached. Reaction speed depends on kinetics, whereas $K$ depends on thermodynamic factors such as temperature.

Dynamic equilibrium

A state in which the forward and reverse reactions continue to occur at equal rates. Because the rates are equal, macroscopic concentrations and pressures remain constant even though molecular reactions continue.

Does the equilibrium constant depend on the initial concentrations of reactants and products?

At fixed conditions, especially fixed temperature, $K$ is independent of the initial composition. Different starting mixtures approach the same $K$, although they may have different equilibrium compositions.

Homogeneous equilibrium

An equilibrium in which all reactants and products are in the same phase, commonly an aqueous solution or a gas mixture.

For $HF(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+F^-(aq)$, what is the equilibrium-constant expression?

$K_c=\frac{[H_3O^+][F^-]}{[HF]}$. Liquid water is omitted because it is a pure liquid with essentially constant activity.

Which species are included in an introductory $K_c$ or $Q_c$ expression?

Include aqueous solutes and gases. Omit pure solids and pure liquids because their activities are treated as constant and incorporated into the equilibrium constant.

Heterogeneous equilibrium

An equilibrium involving reactants or products in two or more phases, such as a solid in equilibrium with an aqueous ion or gas.

Write the equilibrium-constant expression for $PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$.

$K_c=[Pb^{2+}][Cl^-]^2$. The solid $PbCl_2$ is omitted.

Write the equilibrium-constant expression for $CaO(s)+CO_2(g)\rightleftharpoons CaCO_3(s)$.

$K_c=\frac{1}{[CO_2]}$ and $K_p=\frac{1}{P_{CO_2}}$. Both pure solids are omitted.

Write the equilibrium-constant expression for $Br_2(l)\rightleftharpoons Br_2(g)$.

$K_c=[Br_2(g)]$; the pure liquid $Br_2(l)$ is omitted.

How are $K_c$ and $K_p$ related for a gas-phase reaction?

$K_p=K_c(RT)^{\Delta n}$, where $\Delta n=$ moles of gaseous products minus moles of gaseous reactants, using stoichiometric coefficients.

Why does $K_p=K_c$ when $\Delta n=0$?

Using the ideal-gas relationship $P=[\text{gas}]RT$, the factors of $RT$ cancel when the total gaseous moles of products and reactants are equal. Thus $K_p=K_c(RT)^0=K_c$.

Find the relationship between $K_p$ and $K_c$ for $N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$.

$\Delta n=2-(1+3)=-2$, so $K_p=K_c(RT)^{-2}=\frac{K_c}{(RT)^2}$.

Find the relationship between $K_p$ and $K_c$ for $C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$.

$\Delta n=2-1=1$, so $K_p=K_c(RT)$.

What is the effect on an equilibrium constant of reversing a chemical equation?

Reversing the reaction exchanges products and reactants, so the new constant is the reciprocal: $K_{\text{reverse}}=\frac{1}{K_{\text{forward}}}$.

What is the effect on $K$ of multiplying every coefficient in a reaction by a factor $x$?

The new equilibrium constant is raised to that factor: $K_{\text{new}}=(K_{\text{original}})^x$. This follows because every exponent in the equilibrium expression is multiplied by $x$.

What happens to equilibrium constants when chemical equations are added?

The equilibrium constant for the net reaction equals the product of the constants for the added reactions: $K_{\text{net}}=K_1K_2\cdots$. Any species that appears identically on both sides cancels from the net equation.

How can the equilibrium constant for $H_2(g)+CO_2(g)\rightleftharpoons CO(g)+H_2O(g)$ be found from these reactions: $CoO+CO\rightleftharpoons Co+CO_2$ with $K_1=490$, and $CoO+H_2\rightleftharpoons Co+H_2O$ with $K_2=67$?

Reverse the first reaction, giving $1/K_1$, and add it to the second reaction. Therefore, $K=K_2/K_1=67/490\approx0.14$.

How do you calculate a reaction quotient from measured concentrations?

First write the balanced reaction and its $Q_c$ expression. Then substitute the current molar concentrations, raise each to its stoichiometric power, and evaluate the products-over-reactants ratio. These concentrations need not be equilibrium values.

For $2NO_2(g)\rightleftharpoons N_2O_4(g)$, calculate $Q_c$ when $[NO_2]=0.100\,M$ and $[N_2O_4]=0$ before the reaction begins.

$Q_c=\frac{[N_2O_4]}{[NO_2]^2}=\frac{0}{(0.100)^2}=0$.

How do you calculate $K_c$ from equilibrium concentrations?

Write the equilibrium expression from the balanced equation and substitute the concentrations measured at equilibrium. For $2NO_2(g)\rightleftharpoons N_2O_4(g)$ with $[NO_2]=0.016\,M$ and $[N_2O_4]=0.042\,M$, $K_c=\frac{0.042}{(0.016)^2}\approx1.6\times10^2$.

Calculate $K_c$ for $2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)$ when $[SO_2]=0.90\,M$, $[O_2]=0.35\,M$, and $[SO_3]=1.1\,M$ at equilibrium.

$K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}=\frac{(1.1)^2}{(0.90)^2(0.35)}\approx4.3$.