7.10 - Reaction Quotient and Le Châtelier's Principle

Overview

Le Châtelier's principle tells you which way a system shifts. The reaction quotient, Q, tells you why. Every stress either changes Q while K stays the same, or changes K while Q stays the same for a moment. Either way, Q and K no longer match, and the system shifts until they do. This topic lets you back up any Le Châtelier prediction with a calculation.

Q, K, and the direction of a reaction

Recall from Topic 7.7: Q has the same form as K, but it uses the concentrations (or partial pressures) right now. If

Q < K

, there are too few products, so the net reaction runs forward (the system shifts right). If

Q > K

, there are too many, so it runs in reverse (shifts left). If

Q = K

, the system is at equilibrium.

A stress pushes Q away from K

At equilibrium, Q = K. A stress makes Q differ from K, which takes the system out of equilibrium. The system responds the only way it can: the forward or reverse reaction runs faster for a while, changing the concentrations until Q matches K again. That's the new equilibrium state.

Stresses fall into two groups:

Stresses that change Q only: adding or removing a species, changing the volume of a gas mixture, and diluting a solution. K stays the same.

A stress that changes K: temperature. At the instant you heat or cool the system, the concentrations haven't changed yet, so Q is the same. But K is now different.

In both cases, the concentrations or partial pressures redistribute until Q and K are equal again.

Stresses that change Q

Adding or removing a species

Adding a reactant makes the denominator of Q bigger, so Q drops below K and the system shifts right. Adding a product makes the numerator bigger, so Q rises above K and the system shifts left. Removing a species does the opposite.

Example 1: Adding hydrogen

At 430 °C,

K_c = 54.3

for

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

An equilibrium mixture has

[\mathrm{H_2}]

= 0.0200 M,

[\mathrm{I_2}]

= 0.00500 M, and

[\mathrm{HI}]

= 0.0737 M. Hydrogen is injected until

[\mathrm{H_2}]

= 0.0400 M. (a) Calculate Q right after the addition and predict the shift. (b) The new equilibrium has

[\mathrm{H_2}]

= 0.0379 M,

[\mathrm{I_2}]

= 0.00294 M, and

[\mathrm{HI}]

= 0.0778 M. Show that this mixture is at equilibrium.

(a) Right after the addition, only

[\mathrm{H_2}]

has changed:

Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \frac{(0.0737)^2}{(0.0400)(0.00500)} = 27.2

Q_c = 27.2 < K_c = 54.3

, so the system shifts right and makes more

\mathrm{HI}

. Doubling

[\mathrm{H_2}]

cut Q exactly in half, because

[\mathrm{H_2}]

appears to the first power in the denominator.

(b) Substitute the new concentrations:

Q_c = \frac{(0.0778)^2}{(0.0379)(0.00294)} = 54.3 = K_c

Q equals K again, so this is the new equilibrium. The changes fit the 1 : 1 : 2 ratio:

[\mathrm{H_2}]

dropped by 0.0021 M,

[\mathrm{I_2}]

dropped by 0.00206 M, and

[\mathrm{HI}]

rose by 0.0041 M, twice as much (within rounding).

Changing the volume of a gas mixture

Halving the volume doubles every gas concentration and every partial pressure at once. Whether that changes Q depends on the moles of gas on each side. For

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

, right after the volume is halved:

Q_c = \frac{(2[\mathrm{NO_2}])^2}{2[\mathrm{N_2O_4}]} = 2 \times \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = 2K_c

The numerator has the higher power (2 mol of gas versus 1), so doubling everything doubles Q. Now Q > K, and the system shifts left, toward fewer moles of gas. That's the Topic 7.9 rule, now with a reason. For

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

, the powers match (2 on top, 1 + 1 on the bottom), the factors of 2 cancel, Q doesn't change, and there's no shift.

Example 2: Compressing $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$

At a certain temperature,

K_c = 0.360

for

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

. An equilibrium mixture has

[\mathrm{N_2O_4}]

= 0.0400 M and

[\mathrm{NO_2}]

= 0.120 M. The volume is suddenly halved. (a) Calculate Q right after and predict the shift. (b) The new equilibrium has

[\mathrm{N_2O_4}]

= 0.103 M and

[\mathrm{NO_2}]

= 0.193 M. Show that Q = K, and compare

[\mathrm{NO_2}]

with its value before the volume change.

(a) Each concentration doubles, to 0.0800 M and 0.240 M:

Q_c = \frac{(0.240)^2}{0.0800} = 0.720

Q_c = 0.720 > K_c = 0.360

, so the system shifts left, toward

\mathrm{N_2O_4}

(b)

Q_c = \frac{(0.193)^2}{0.103} = 0.362 \approx K_c

That's equal to

K_c = 0.360

within rounding (the concentrations are rounded to 3 significant figures).

[\mathrm{NO_2}]

went from 0.120 M (before) to 0.240 M (right after the squeeze) to 0.193 M (new equilibrium). It ends up between the two, which is why the gas in Topic 7.9's syringe darkens, lightens a little, and stays darker than it started.

Diluting a solution

Diluting a solution to twice its volume halves every dissolved concentration. For

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

, right after the dilution:

Q_c = \frac{\tfrac{1}{2}[\mathrm{FeSCN^{2+}}]}{\left(\tfrac{1}{2}[\mathrm{Fe^{3+}}]\right)\left(\tfrac{1}{2}[\mathrm{SCN^{-}}]\right)} = 2K_c

Q > K, so the system shifts left, toward the side with more dissolved particles. Whenever the two sides have different numbers of dissolved particles, dilution changes Q.

Temperature changes K

Temperature is the one stress that changes K. Which way K moves depends on the sign of

\Delta H^\circ

Forward reaction	Raise the temperature	Lower the temperature
Endothermic ( $\Delta H^\circ > 0$ )	K increases	K decreases
Exothermic ( $\Delta H^\circ < 0$ )	K decreases	K increases

Right after a temperature change at constant volume, the concentrations haven't had time to change, so Q is the same as before. But it no longer equals the new K:

Endothermic reaction, heated: K goes up, so Q < K and the system shifts right.

Exothermic reaction, heated: K goes down, so Q > K and the system shifts left.

That matches what Le Châtelier's principle says when you treat heat as a reactant or product. It also works backward: if K values at two temperatures show K growing as the temperature rises, the forward reaction is endothermic. Unit 9 explains why K depends on temperature, through

\Delta G^\circ

Stress	What changes at that instant	Result
Add a reactant	Q decreases (K is the same)	Q < K, shifts right
Add a product	Q increases (K is the same)	Q > K, shifts left
Halve the volume of a gas mixture	Q changes unless the moles of gas are equal on both sides	Shifts toward fewer moles of gas
Dilute a solution	Q changes unless the dissolved particles are equal on both sides	Shifts toward more dissolved particles
Heat an endothermic reaction	K increases (Q is the same)	Q < K, shifts right
Heat an exothermic reaction	K decreases (Q is the same)	Q > K, shifts left
Add a catalyst	Neither Q nor K	No shift

Example 3: Reading a concentration-time graph (AP-style)

The graph shows

[\mathrm{N_2O_4}]

and

[\mathrm{NO_2}]

in a sealed flask. The mixture starts at equilibrium with

K_c = 0.360

. At times

t_1

t_2

, and

t_3

, one of three stresses is applied:

\mathrm{NO_2}

is added, the volume is halved, or the temperature is raised.

7.10 - Reaction Quotient and Le Châtelier's Principle

(a) Match each time to its stress, and justify each choice with the graph. (b) Calculate Q right after

t_1

, and use it to explain the changes after

t_1

. (c) After

t_3

, the mixture settles at

[\mathrm{N_2O_4}]

= 0.112 M and

[\mathrm{NO_2}]

= 0.335 M. Calculate K at the new temperature and decide whether the forward reaction is endothermic or exothermic.

(a)

$t_1$ , $\mathrm{NO_2}$ added: only $[\mathrm{NO_2}]$ jumps. $[\mathrm{N_2O_4}]$ doesn't change at that instant.

$t_2$ , volume halved: both concentrations jump at the same instant, and each one doubles (0.0641 M to 0.128 M, and 0.152 M to 0.304 M).

$t_3$ , temperature raised: neither concentration jumps. Both start curving smoothly at $t_3$ , because only K changed.

(b) Right after

t_1

[\mathrm{N_2O_4}]

= 0.0400 M and

[\mathrm{NO_2}]

= 0.200 M:

Q_c = \frac{(0.200)^2}{0.0400} = 1.00

Q_c = 1.00 > K_c = 0.360

, so the system shifts left:

[\mathrm{NO_2}]

falls and

[\mathrm{N_2O_4}]

rises until Q is back to 0.360. That happens at 0.0641 M and 0.152 M, and

(0.152)^2/0.0641 = 0.360

(c)

K_c = \frac{(0.335)^2}{0.112} = 1.00

K went up from 0.360 to 1.00 when the temperature was raised. K increases with temperature only for an endothermic reaction, so the forward reaction is endothermic. The graph agrees: heating shifted the system right, toward

\mathrm{NO_2}

, the endothermic direction.

Common mistakes

Saying that adding a reactant changes K. It changes Q. Only temperature changes K.

Using the old concentrations in Q. Use the values right after the stress, before the system shifts.

Changing only one concentration when the volume changes. A volume change scales every gas concentration by the same factor.

Drawing a jump at a temperature change. At constant volume, concentrations change gradually after heating. Only K jumps.

Reading Q > K as "shift right." Q > K means there's too much product, so the reverse reaction wins and the system shifts left.

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