7.7 - Calculating Equilibrium Concentrations

Overview

If you know the balanced equation, the starting amounts, and K, you can predict exactly what the mixture will look like at equilibrium. The work has two parts: compare Q with K to decide which way the reaction has to go, then use an ICE table to find how far it goes. This skill shows up on both sections of the AP exam, and it's the foundation for the acid-base calculations in Unit 8.

Which way will the reaction go?

Start by comparing the reaction quotient Q (Topic 7.3) with K. Q uses the concentrations or partial pressures right now; K is the value Q will have once the system reaches equilibrium. Q always moves toward K.

Comparison	What happens	Why
$Q < K$	The net reaction runs forward: reactants are used up and products form until Q rises to K.	There are too few products compared with equilibrium, so the forward rate is faster than the reverse rate.
$Q > K$	The net reaction runs in reverse: products are used up and reactants form until Q falls to K.	There are too many products, so the reverse rate is faster.
$Q = K$	No net change. The system is at equilibrium.	The forward and reverse reactions still happen, at equal rates, so the proportions of reactants and products stay constant.

7.7 - Calculating Equilibrium Concentrations

A quick way to remember it: products are on top of the Q expression. If Q is too small, the system makes more products to raise it. If Q is too big, it makes more reactants to lower it.

Write Q exactly the way you'd write K: products over reactants, each raised to its coefficient, leaving out solids and pure liquids. If no product is present,

Q = 0

and the reaction must run forward. If a reactant is missing, Q can't be calculated (you'd divide by zero), but the reaction clearly must run in reverse.

Example 1: Predicting the direction

At 1000 K,

K_c = 2.8 \times 10^{2}

for

\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}

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

Predict the direction for each mixture at 1000 K.

(a)

[\mathrm{SO_2}] = 0.10

[\mathrm{O_2}] = 0.050

[\mathrm{SO_3}] = 0.50

Q_c = \frac{(0.50)^2}{(0.10)^2(0.050)} = 5.0 \times 10^{2}

Q > K

(500 > 280), so the net reaction runs in reverse: some

\mathrm{SO_3}

breaks down into

\mathrm{SO_2}

and

\mathrm{O_2}

(b) 0.40 mol

\mathrm{SO_2}

, 0.20 mol

\mathrm{O_2}

, and 0.60 mol

\mathrm{SO_3}

in a 4.00 L flask. Convert to molarity first:

[\mathrm{SO_2}] = 0.40\ \mathrm{mol} / 4.00\ \mathrm{L} = 0.10

[\mathrm{O_2}] = 0.050

M, and

[\mathrm{SO_3}] = 0.15

Q_c = \frac{(0.15)^2}{(0.10)^2(0.050)} = 45

Q < K

, so the net reaction runs forward, making more

\mathrm{SO_3}

. (Plugging in moles instead of molarity gives 11, which is the wrong value. It happens to point the same way here, but it won't always.)

(c)

[\mathrm{SO_2}] = 0.050

[\mathrm{O_2}] = 0.0070

[\mathrm{SO_3}] = 0.070

Q_c = \frac{(0.070)^2}{(0.050)^2(0.0070)} = 2.8 \times 10^{2}

Q = K

, so this mixture is already at equilibrium. Both reactions keep running at equal rates, and the concentrations stay constant.

Finding equilibrium concentrations with an ICE table

In Topic 7.4 you used an ICE table to find K from measured data. Here it runs the other way: K is known and the equilibrium concentrations are the unknowns. ICE stands for Initial, Change, Equilibrium.

Write the balanced equation and the K expression.

Fill in the Initial row with concentrations in M (for $K_c$ ) or partial pressures in atm (for $K_p$ ). Convert moles to molarity first.

Decide the direction by comparing Q with K. If one side starts at zero, the reaction can only go one way.

Fill in the Change row with one unknown, $x$ . Each species changes by its coefficient times $x$ : minus on the side being used up, plus on the side being made.

Write the Equilibrium row as Initial + Change.

Substitute into the K expression and solve for $x$ . Then calculate every equilibrium concentration.

Check by plugging your answers back into the K expression. You should get K back, within rounding.

Coefficients show up twice, and it's easy to forget one: once as the multiplier in the Change row (

2x

for a coefficient of 2), and again as the exponent in the K expression (

(2x)^2

Example 2: A reaction that runs in reverse

At 430 °C,

K_c = 54.3

for

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

A flask at 430 °C starts with

[\mathrm{H_2}] = [\mathrm{I_2}] = 0.0200

M and

[\mathrm{HI}] = 0.400

M. Find the equilibrium concentrations.

Step 1: Direction.

Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \frac{(0.400)^2}{(0.0200)(0.0200)} = 400

Q > K

(400 > 54.3), so the reaction runs in reverse: HI breaks down, and

\mathrm{H_2}

and

\mathrm{I_2}

form.

Step 2: ICE table. Let

x

be the concentration of

\mathrm{H_2}

(and of

\mathrm{I_2}

) that forms. Each

\mathrm{H_2}

that forms uses up 2 HI, so HI drops by

2x

	$[\mathrm{H_2}]$ (M)	$[\mathrm{I_2}]$ (M)	$[\mathrm{HI}]$ (M)
Initial	0.0200	0.0200	0.400
Change	$+x$	$+x$	$-2x$
Equilibrium	$0.0200 + x$	$0.0200 + x$	$0.400 - 2x$

Step 3: Solve. Both the top and bottom of the expression are perfect squares, so take the square root of both sides instead of using the quadratic formula:

54.3 = \frac{(0.400 - 2x)^2}{(0.0200 + x)^2} \quad \Rightarrow \quad \sqrt{54.3} = 7.369 = \frac{0.400 - 2x}{0.0200 + x}

7.369(0.0200 + x) = 0.400 - 2x \quad \Rightarrow \quad 0.1474 + 7.369x = 0.400 - 2x

9.369x = 0.2526 \quad \Rightarrow \quad x = 0.02696\ \mathrm{M}

Step 4: Equilibrium concentrations.

[\mathrm{H_2}] = [\mathrm{I_2}] = 0.0200 + 0.02696 = 0.0470\ \mathrm{M} \qquad [\mathrm{HI}] = 0.400 - 2(0.02696) = 0.346\ \mathrm{M}

Step 5: Check.

\frac{(0.346)^2}{(0.0470)^2} = 54.2 \approx 54.3

The tiny difference comes from rounding. Because

Q > K

, the reactants went up and the product went down, the opposite of what you might expect from the words "reactant" and "product." The graph shows the concentrations moving to their equilibrium values and leveling off.

When K is small: the small-x approximation

Many equilibrium problems lead to a quadratic (or worse) equation. When K is small, there's a shortcut. A small K means the reaction barely proceeds (Topic 7.5), so

x

is tiny compared with the starting concentration, and you can drop it where it's subtracted from (or added to) a much larger number:

[\mathrm{A}]_0 - x \approx [\mathrm{A}]_0

The approximation has to be justified, so always check it afterward with the 5% rule:

\frac{x}{[\mathrm{A}]_0} \times 100\% < 5\%

x

is less than 5% of the starting value, the approximation is fine: the error it adds is smaller than the uncertainty in a typical K value. If not, go back and solve the full equation. A calculator is allowed on the whole AP exam, so solving a quadratic is always an option.

Only drop

x

where it's added to or subtracted from a larger number. Never drop an

x

that stands alone, like the

x

for a product that started at zero; that

x

is the answer you're looking for.

Example 3: The approximation works

Phosgene breaks down into carbon monoxide and chlorine:

\mathrm{COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)}

At a certain temperature,

K_c = 2.2 \times 10^{-10}

. (This is the reverse of the phosgene reaction in Topic 7.5, Example 2, so its K is

1/(4.6 \times 10^{9})

.) A 0.600 L flask starts with 0.300 mol of

\mathrm{COCl_2}

and nothing else. Find the equilibrium concentrations.

Step 1: Initial concentration and direction.

[\mathrm{COCl_2}]_0 = 0.300\ \mathrm{mol} / 0.600\ \mathrm{L} = 0.500

M. No products are present, so

Q = 0 < K

and the reaction runs forward.

Step 2: ICE table.

	$[\mathrm{COCl_2}]$ (M)	$[\mathrm{CO}]$ (M)	$[\mathrm{Cl_2}]$ (M)
Initial	0.500	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.500 - x$	$x$	$x$

2.2 \times 10^{-10} = \frac{[\mathrm{CO}][\mathrm{Cl_2}]}{[\mathrm{COCl_2}]} = \frac{x^2}{0.500 - x}

Step 3: Approximate. K is tiny, so assume

0.500 - x \approx 0.500

\frac{x^2}{0.500} = 2.2 \times 10^{-10} \quad \Rightarrow \quad x^2 = 1.1 \times 10^{-10} \quad \Rightarrow \quad x = 1.0 \times 10^{-5}\ \mathrm{M}

Step 4: Check the approximation.

\frac{1.0 \times 10^{-5}}{0.500} \times 100\% = 0.002\%

That's far below 5%, so the approximation holds.

Answer:

[\mathrm{CO}] = [\mathrm{Cl_2}] = 1.0 \times 10^{-5}

M and

[\mathrm{COCl_2}] = 0.500

M (0.500 minus 0.000010 still rounds to 0.500). As Topic 7.5 predicts for a K this small, the reaction barely proceeds: only about 0.002% of the phosgene breaks down.

Example 4: When the approximation fails (AP-style)

At a certain temperature,

K_p = 0.150

for

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

A rigid container is filled with

\mathrm{N_2O_4}

at a pressure of 1.000 atm and no

\mathrm{NO_2}

, then held at that temperature. (This

K_p

uses pressures and is for a different temperature from the

K_c = 0.360

used elsewhere in this unit, so the two numbers don't need to match.)

(a) Which way will the reaction go? No

\mathrm{NO_2}

is present, so

Q_p = 0 < K_p

. The reaction runs forward.

(b) Set up the ICE table. This is

K_p

, so work in partial pressures (atm). Each

\mathrm{N_2O_4}

that reacts makes 2

\mathrm{NO_2}

	$P_{\mathrm{N_2O_4}}$ (atm)	$P_{\mathrm{NO_2}}$ (atm)
Initial	1.000	0
Change	$-x$	$+2x$
Equilibrium	$1.000 - x$	$2x$

K_p = \frac{(P_{\mathrm{NO_2}})^2}{P_{\mathrm{N_2O_4}}} = \frac{(2x)^2}{1.000 - x} = 0.150

(c) Try the approximation. Assume

1.000 - x \approx 1.000

4x^2 = 0.150 \quad \Rightarrow \quad x = 0.194 \qquad \frac{0.194}{1.000} \times 100\% = 19.4\%

That's well over 5%, so the approximation fails. This makes sense:

K_p = 0.150

isn't far below 1, so a lot of

\mathrm{N_2O_4}

reacts.

(d) Solve the quadratic.

4x^2 = 0.150(1.000 - x) \quad \Rightarrow \quad 4x^2 + 0.150x - 0.150 = 0

x = \frac{-0.150 + \sqrt{(0.150)^2 - 4(4)(-0.150)}}{2(4)} = \frac{-0.150 + 1.556}{8} = 0.176\ \mathrm{atm}

Take the positive root. The other root is negative, which would mean a negative pressure.

P_{\mathrm{N_2O_4}} = 1.000 - 0.176 = 0.824\ \mathrm{atm} \qquad P_{\mathrm{NO_2}} = 2(0.176) = 0.352\ \mathrm{atm}

(e) Check.

\frac{(0.352)^2}{0.824} = 0.150

(f) Explain why the total pressure changed. At equilibrium,

P_{\text{total}} = 0.824 + 0.352 = 1.176

atm, higher than the starting 1.000 atm. Each

\mathrm{N_2O_4}

molecule that reacts becomes two

\mathrm{NO_2}

molecules, so the number of gas particles in the fixed volume goes up. At constant volume and temperature, pressure is proportional to the number of gas particles.

(g) Graph it. The partial pressures change until Q reaches K, then level off. The

\mathrm{NO_2}

curve rises twice as much as the

\mathrm{N_2O_4}

curve falls, because of the 1 : 2 ratio in the equation.

Common mistakes

Plugging moles into the K expression. Use molarity (mol/L) for $K_c$ and partial pressures for $K_p$ .

Mixing up the direction. $Q < K$ means the reaction makes more products (forward). $Q > K$ means it makes more reactants (reverse).

Skipping the Q vs. K comparison when the mixture starts with both reactants and products, and then using the wrong signs in the Change row.

Forgetting coefficients: a coefficient of 2 means $2x$ in the Change row and a squared term in the K expression.

Using the small-x approximation without the 5% check, or dropping an $x$ that stands alone.

Reporting $x$ as the answer. $x$ is a change; the question usually wants the equilibrium concentrations.

Thinking that the reactions stop when $Q = K$ . They keep running, at equal rates.

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