7.3 - Reaction Quotient and Equilibrium Constant

Overview

To describe where a reaction stands, chemists use one number built from the concentrations: the reaction quotient, Q. When the reaction is at equilibrium, Q has a special value called the equilibrium constant, K. This topic shows how to write Q and K for any reaction, the difference between

K_c

and

K_p

, and which substances to leave out. Every equilibrium calculation in Units 7 and 8 starts with writing one of these expressions correctly.

The law of mass action

For a general reaction

a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}

the law of mass action says the equilibrium concentrations always fit this expression:

K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Products go on top, reactants on the bottom.

Each concentration is raised to the power of its coefficient in the balanced equation.

The terms are multiplied, not added.

Square brackets mean molarity (mol/L), and for $K_c$ they are the equilibrium values.

For example, for

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

K_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}

In Topics 7.1 and 7.2, the flasks settled at

0.0400

\mathrm{N_2O_4}

and

0.120

\mathrm{NO_2}

, so

K_c = (0.120)^2 / 0.0400 = 0.360

at that temperature. K for a given reaction is constant at a given temperature. On the AP exam, K values are written without units.

The reaction quotient, Q

The reaction quotient has exactly the same form as K, but you can plug in the concentrations at any moment, not just at equilibrium:

Q_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Q describes the mixture right now. It changes as the reaction proceeds. K is one fixed number for the reaction at that temperature.

Q tends toward K. As a system approaches equilibrium, Q gets closer and closer to K. At equilibrium, $Q_c = K_c$ .

Two special cases. With only reactants present, $Q = 0$ . With only products present, Q has zero in the denominator, so it's undefined (think of it as huge).

The graph shows

Q_c

over time for the two flasks from Topic 7.2. Starting with only

\mathrm{N_2O_4}

Q_c

starts at 0 and rises. Starting with only

\mathrm{NO_2}

Q_c

starts off the top of the graph and falls. Both level off at

K_c = 0.360

7.3 - Reaction Quotient and Equilibrium Constant

Comparing Q with K tells you which way a reaction will go to reach equilibrium. You'll use this in Topics 7.7 and 7.10.

Example 1: Calculating Q as a reaction proceeds (AP-style)

For

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

K_c = 0.360

at a certain temperature. A flask that started with only

\mathrm{N_2O_4}

is checked at two moments: at time A,

[\mathrm{N_2O_4}] = 0.0700

M and

[\mathrm{NO_2}] = 0.0600

M; at time B,

[\mathrm{N_2O_4}] = 0.0500

M and

[\mathrm{NO_2}] = 0.100

M. (a) Calculate

Q_c

at each time. (b) Is the system at equilibrium at either time? (c) Describe how

Q_c

changes from time A onward. (d) A student calculates

Q_c = 0.857

at time A. What mistake did the student make?

(a)

Q_A = \frac{(0.0600)^2}{0.0700} = 0.0514 \qquad Q_B = \frac{(0.100)^2}{0.0500} = 0.200

(b) No. Neither value equals

K_c = 0.360

(c)

Q_c

is below

K_c

and rising: 0.0514, then 0.200. It keeps rising toward 0.360 and then stays there, like the solid curve in the graph above. A rising

Q_c

means

[\mathrm{NO_2}]

is going up and

[\mathrm{N_2O_4}]

is going down, so the net reaction is forward.

(d) The student used

0.0600 / 0.0700 = 0.857

and forgot to square

[\mathrm{NO_2}]

. The coefficient 2 in the balanced equation becomes an exponent.

Kc and Kp

For reactions involving gases, you can use partial pressures (in atm) instead of concentrations. This gives

K_p

(and

Q_p

for any moment):

K_p = \frac{(P_\mathrm{C})^c(P_\mathrm{D})^d}{(P_\mathrm{A})^a(P_\mathrm{B})^b}

Both

K_c

and

K_p

are on the AP reference sheet.

	$K_c$ (and $Q_c$ )	$K_p$ (and $Q_p$ )
What you plug in	Molar concentrations, [X], in M	Partial pressures, $P_\mathrm{X}$ , in atm
Which species appear	Gases and aqueous species	Gases only
Typical use	Reactions in solution, or gas reactions	Gas reactions

How they're related. For a gas, the ideal gas law

PV = nRT

rearranges to

P_\mathrm{X} = \frac{n_\mathrm{X}}{V}RT = [\mathrm{X}]RT

So at constant temperature, a gas's partial pressure is proportional to its concentration.

K_c

and

K_p

describe the same equilibrium, but their numbers are usually different, because every concentration term picks up a factor of

RT

. They are equal only when both sides of the equation have the same total moles of gas, as in

\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}

What the AP exam expects. You won't be asked to convert between

K_c

and

K_p

. You do need to notice which one a question gives you and use matching quantities: molarities with

K_c

, partial pressures with

K_p

, never a mix of the two.

The exam also won't ask for equilibrium calculations on a dissolved substance in equilibrium with its own gas, like

\mathrm{CO_2(g) \rightleftharpoons CO_2(aq)}

in a sealed soda.

Example 2: Writing Kc and Kp for a gas reaction

Write

K_c

and

K_p

for the reaction

\mathrm{4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)}

Everything is a gas (including the water), so every species appears, each raised to its coefficient:

K_c = \frac{[\mathrm{NO}]^4[\mathrm{H_2O}]^6}{[\mathrm{NH_3}]^4[\mathrm{O_2}]^5} \qquad K_p = \frac{(P_\mathrm{NO})^4(P_\mathrm{H_2O})^6}{(P_\mathrm{NH_3})^4(P_\mathrm{O_2})^5}

There are 9 moles of gas on the left and 10 on the right, so

K_c

and

K_p

have different values for this reaction.

Leave out pure solids and pure liquids

Q and K do not include substances whose concentration doesn't depend on how much of them there is. That means pure solids and pure liquids.

Why: the "concentration" of a pure solid is the moles of solid divided by the volume of the solid itself. That equals its density divided by its molar mass, which is fixed. Doubling the amount of solid doubles both the moles and the volume, so the ratio doesn't change. The same goes for a pure liquid. Since these values never change, they're built into K instead of written in the expression.

Calcium carbonate decomposing in a sealed container shows this:

\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)} \qquad K_p = P_\mathrm{CO_2} \qquad K_c = [\mathrm{CO_2}]

At a given temperature, the pressure of

\mathrm{CO_2}

at equilibrium is the same whether there's a little solid or a lot, as long as both solids are present.

Water. When water is the solvent, as in most aqueous reactions,

\mathrm{H_2O(l)}

is left out: it's nearly pure, and its concentration barely changes. But

\mathrm{H_2O(g)}

is a gas, so it is included. Dissolved species, (aq), are always included. Check every phase label before you write an expression.

Reaction	Expression	Why it looks that way
$\mathrm{H_2O(l) \rightleftharpoons H_2O(g)}$	$K_p = P_\mathrm{H_2O}$	Liquid left out; K is the vapor pressure from Topic 7.1
$\mathrm{3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)}$	$K_p = \frac{(P_\mathrm{H_2})^4}{(P_\mathrm{H_2O})^4}$	Solids left out; water is a gas here
$\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$	$K_c = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}$	Water is the solvent, so it's left out
$\mathrm{Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s)}$	$K_c = \frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2}$	Redox: only the ions appear
$\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}$	$K_c = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2$	No denominator left (Topic 7.11 calls this $K_{sp}$ )

Example 3: Expressions with solids, liquids, and ions

Write the

K_c

expression for each reaction.

(a)

\mathrm{C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g)}

K_c = \frac{[\mathrm{CO}][\mathrm{H_2}]}{[\mathrm{H_2O}]}

Carbon is a solid, so it's left out. Water is a gas here, so it stays.

(b)

\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)}

K_c = \frac{[\mathrm{CH_3COO^-}][\mathrm{H_3O^+}]}{[\mathrm{CH_3COOH}]}

Liquid water is the solvent, so it's left out. (Unit 8 calls this

K_a

(c)

\mathrm{2Ag_2O(s) \rightleftharpoons 4Ag(s) + O_2(g)}

K_c = [\mathrm{O_2}] \qquad (\text{and } K_p = P_\mathrm{O_2})

(d)

\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}

K_c = \frac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}

All three species are dissolved ions, so all three appear. Topic 7.4 measures this K in the lab.

Common mistakes

Using coefficients as multipliers. $2[\mathrm{NO_2}]$ is wrong; $[\mathrm{NO_2}]^2$ is right.

Adding terms, or flipping the fraction. Multiply the terms, with products on top.

Including solids, pure liquids, or solvent water. Leave them out, but keep $\mathrm{H_2O(g)}$ and every (aq) species.

Mixing pressures and concentrations in one expression. Use all molarities for $K_c$ or all partial pressures for $K_p$ .

Plugging starting concentrations into K. Starting values give Q. K uses equilibrium values only.

Assuming $K_c = K_p$ . That's only true when the total moles of gas are the same on both sides.

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