Introduction to the Equilibrium Constant K

Different starts, the same ratio

At 25 °C, different amounts of N₂O₄ are sealed in a 1.00 L flask and left to reach equilibrium. Here are the equilibrium concentrations:

Initial [N₂O₄] (M)	Equilibrium [N₂O₄]	Equilibrium [NO₂]	[NO₂]/[N₂O₄]	[NO₂]²/[N₂O₄]
0.020	0.01514	0.009726	0.64	6.25 × 10⁻³
0.040	0.03284	0.01433	0.44	6.25 × 10⁻³
0.100	0.08826	0.02349	0.27	6.25 × 10⁻³

The equilibrium concentrations differ, and so does [NO₂]/[N₂O₄]. But [NO₂]²/[N₂O₄] is always the same number: the equilibrium constant.

The equilibrium constant

For a reaction at equilibrium at a given temperature, the products’ concentrations raised to their coefficients, divided by the reactants’ concentrations raised to their coefficients, is a constant. This is the equilibrium constant化学平衡常数 K. For aA + bB ⇌ cC + dD:

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

All the concentrations are equilibrium concentrations. When you write K:

Leave out pure solids and pure liquids (including water as the solvent), whose concentrations are effectively constant. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K = [CO₂].
K depends only on temperature, not on concentrations or pressure.
K depends on how the equation is written: for the reverse reaction, K is the reciprocal.

What the size of K tells you

A large K means the products are favored at equilibrium; a small K means the reactants are favored. Roughly, K > 10⁵ means the reaction goes essentially to completion, and K < 10⁻⁵ means very little product forms.

For example, even at 2000 °C, N₂(g) + O₂(g) ⇌ 2NO(g) has K = 4.1 × 10⁻⁴. Air in a car engine makes only a little NO, but that is enough to cause smog.

The reaction quotient Q

Put the concentrations at any moment, not necessarily at equilibrium, into the K expression and you get the reaction quotient浓度商 Q. Comparing Q with K tells you which way the reaction will go:

Q < K: too few products, so the reaction goes forward.
Q = K: the system is at equilibrium.
Q > K: too many products, so the reaction goes in reverse.

Pick a reaction and change the concentrations to compare Q with K. Then find the equilibrium concentrations, set out as an ICE table:

Try it: reaction quotient Q and equilibrium constant K

How Chinese textbooks differ

Chinese textbooks write concentrations as c(A) instead of [A], with the power on the c, as in K = c²(NO₂)/c(N₂O₄). They sometimes give K units; AP treats K as unitless.
Chinese students set out equilibrium problems in a “three-line” table (三段式) of initial, change, and equilibrium amounts, which is the same as an ICE table.
Chinese problems often ask for a reactant’s conversion (转化率): the amount that reacted ÷ the starting amount × 100%. It looks at a reactant, while percent yield (2.6.7) looks at a product.

Example 1: writing K expressions

Write the equilibrium-constant expressions for (a) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), (b) C(s) + H₂O(g) ⇌ CO(g) + H₂(g), and (c) CaCO₃(s) ⇌ CaO(s) + CO₂(g).

Step-by-step solution

The rule. K = products over reactants, each concentration raised to the power of its coefficient. Pure solids and liquids are left out.
(a) SO₃ has a coefficient of 2, SO₂ has 2, and O₂ has 1:
$K = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$
(b) C is a solid, so leave it out. The water here is a gas, so it stays in:
$K = \frac{[\ce{CO}][\ce{H2}]}{[\ce{H2O}]}$
(c) CaCO₃ and CaO are both solids: K = [CO₂]. At a given temperature, [CO₂] at equilibrium has a fixed value.
A common mistake. Writing a coefficient as a multiplier, as in 2[SO₃], or including solids. Also note that liquid water as the solvent is left out, but water vapor is included.

Example 2: comparing Q and K

At 400 °C, H₂(g) + I₂(g) ⇌ 2HI(g) has K = 50. At one moment [H₂] = [I₂] = 0.10 M and [HI] = 0.50 M. (a) Which way will the reaction go? (b) What if [HI] = 1.0 M instead?

Step-by-step solution

Q. Substitute the concentrations at this moment into the K expression:
$\begin{aligned} Q &= \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} \\ &= \frac{0.50^2}{0.10 \times 0.10} \\ &= 25 \end{aligned}$
Compare. Q = 25 < K = 50: there isn’t enough product yet, so the reaction goes forward and more HI forms.
ICE table for the equilibrium [HI]. Let x M of H₂ react; I₂ also drops by x and HI rises by 2x. H₂ and I₂ have the same numbers, so they share a column:

(M) H₂ or I₂ HI
I 0.10 0.50
C −x +2x
E 0.10 − x 0.50 + 2x
Substitute into K.
$\frac{(0.50 + 2x)^2}{(0.10 - x)^2} = 50$
Take the square root of both sides. Both the top and the bottom are positive:
$\begin{aligned} \frac{0.50 + 2x}{0.10 - x} &= \sqrt{50} \\ &= 7.07 \end{aligned}$
Solve for x.
$\begin{aligned} 0.50 + 2x &= 7.07(0.10 - x) \\ 0.50 + 2x &= 0.707 - 7.07x \\ 9.07x &= 0.207 \\ x &= 0.0228\ \text{M} \end{aligned}$
Equilibrium concentrations.
$\begin{aligned} [\ce{HI}] &= 0.50 + 2 \times 0.0228 \\ &= 0.546 \approx \boxed{0.55\ \text{M}} \\ [\ce{H2}] &= 0.10 - 0.0228 \\ &= 0.0772\ \text{M} \end{aligned}$
[I₂] is the same as [H₂].
Check. 0.546² ÷ 0.0772² ≈ 50, which is K.
(b) If [HI] = 1.0 M. Q = 1.0² ÷ (0.10 × 0.10) = 100 > 50: too much product, so the reaction goes in reverse and HI breaks down until Q = K. The same kind of ICE table (HI drops by 2x, H₂ and I₂ each rise by x) gives [HI] ≈ 0.94 M at equilibrium.
A common mistake. Writing the change in HI as +x. Changes are in the ratio of the coefficients, and HI has a coefficient of 2, so it is +2x.

(M)	H₂ or I₂	HI
I	0.10	0.50
C	−x	+2x
E	0.10 − x	0.50 + 2x

Practice: compare Q with K. Which way does the reaction go?

Q = 0.5, K = 2
Q = 4, K = 2
Q = 50, K = 50
Only reactants are in the container
Only products are in the container
H₂ + I₂ ⇌ 2HI with K = 50, and all three at 0.10 M
N₂O₄ ⇌ 2NO₂ with K = 6.3 × 10⁻³, and [N₂O₄] = [NO₂] = 0.010 M
CaCO₃(s) ⇌ CaO(s) + CO₂(g) with K = 0.010 at some temperature, and [CO₂] = 0.020 M

Common mistakes

Mistake 1: “Every substance goes into the K expression, including solids and water.” Pure solids and pure liquids, including water as the solvent, are left out.

Mistake 2: “Adding reactant increases K.” K depends only on temperature. Adding reactant lowers Q, and the system shifts, but K stays the same.

Mistake 3: “A large K means a fast reaction.” K tells you how far a reaction goes, not how fast. Hydrogen and oxygen forming water has an enormous K, yet a mixture of the two gases at room temperature shows no visible reaction.

Quick check

What is the equilibrium-constant expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

For a reaction at some temperature, K = 1.0 and the current Q = 2.5. The reaction will:

Summary

K = (products’ equilibrium concentrations raised to their coefficients) ÷ (reactants’ likewise). Pure solids and liquids are left out, and K depends only on temperature.
Large K favors products and small K favors reactants; K > 10⁵ is essentially complete, and K < 10⁻⁵ means very little product.
If Q < K the reaction goes forward, if Q = K it is at equilibrium, and if Q > K it goes in reverse.

On to AP (for later)

AP topic 7.3 compares Q and K to predict direction, 7.4 calculates K from equilibrium concentrations or partial pressures, and 7.5 interprets the size of K. Topic 7.6 covers how K changes when you rewrite an equation (reverse it and K becomes 1/K; multiply it by n and K is raised to the n). Topic 7.7 uses ICE tables to find equilibrium concentrations. As here, pure solids and liquids are left out of K.

Data sources: K values from OpenStax Chemistry 2e, chapter 13; the N₂O₄ concentrations in the table are calculated from the equilibrium constant at 25 °C.

Introduction to the Equilibrium Constant K平衡常数 K 入门

Learning goals

Before you start

Different starts, the same ratio

The equilibrium constant

What the size of K tells you

The reaction quotient Q

How Chinese textbooks differ

Example 1: writing K expressions

Example 2: comparing Q and K

Common mistakes

Quick check

Summary

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