Law of mass action and Kc, Kp: Worked examples — Chemistry deep dive

Before anything: three symbols we lean on constantly.

Definition The three quantities, in plain words

$K_{c}$ — a plain number you get by putting product concentrations on top, reactant concentrations on the bottom, each raised to its coefficient. "Concentration" $[X]$ means moles of $X$ divided by the volume in litres.
$K_{p}$ — the same idea but with partial pressures $p_{X}$ instead of concentrations. Used when the species are gases, because a gas's pressure is easier to read off a gauge than its concentration.
$Q$ (the reaction quotient) — the exact same ratio as $K$ , but computed from whatever concentrations (or pressures) the mixture has right now, whether or not it is at equilibrium. $K$ is where the system is heading; $Q$ is where it stands. When $Q = K$ , the system is at equilibrium. (More in Reaction Quotient Q vs K.)
The bridge: $K_{p} = K_{c} (R T)^{Δ n_{g}}$ , where the key term == $Δ n_{g}$ == is (gas moles of products) − (gas moles of reactants). $R = 0.0821 L\cdotatm\cdotmol^{- 1} K^{- 1}$ when pressures are in atm.

The scenario matrix

Cell	Case class	What is special	Example
A	$Δ n_{g} = 0$	$K_{p} = K_{c}$ , units cancel	Ex 1
B	$Δ n_{g} < 0$ (fewer gas moles form)	$K_{p} < K_{c}$ (when $R T > 1$ )	Ex 2
C	$Δ n_{g} > 0$ (more gas moles form)	$K_{p} > K_{c}$	Ex 3
D	Degenerate — pure solids and liquids omitted	$K_{p}$ = a single pressure	Ex 4
E	Build $K$ from an ICE table (unknown extent $x$ )	you must find the concentrations first	Ex 5
F	Manipulating a known $K$ (reverse / scale / add)	$1/ K$ , $K^{n}$ , $K_{1} K_{2}$	Ex 6
G	Real-world word problem (gas–liquid)	translate words → equation → $K_{c}$	Ex 7
H	Exam twist — direction test with $Q$	compare $Q$ against $K$	Ex 8

Each example below is tagged with its cell letter. Together they touch every cell.

Cell A — $Δ n_{g} = 0$

Worked example Ex 1 (Cell A):

$H_{2} + I_{2} ⇌ 2 H I$ In a $2 L$ flask at equilibrium: $0.20 mol H_{2}$ , $0.20 mol I_{2}$ , $1.60 mol H I$ . Find $K_{c}$ and $K_{p}$ .

Forecast: Count the gas moles on each side — do you expect $K_{p}$ to differ from $K_{c}$ ?

Convert moles to concentrations: divide each by $V = 2$ L. $[H_{2}] = 0.10, [I_{2}] = 0.10, [H I] = 0.80$ . Why this step? $K_{c}$ is built from concentrations, not raw moles.
Apply the formula, products on top: $K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 0.80 ) ^{2}}{( 0.10 ) ( 0.10 )} = \frac{0.64}{0.01} = 64.$ Why this step? Coefficient of $H I$ is 2, so it is squared.
$Δ n_{g} = 2 - (1 + 1) = 0$ , so $K_{p} = K_{c} (R T)^{0} = K_{c} = 64$ . Why this step? Anything to the power zero is 1; the $R T$ factor vanishes.

Verify: $Δ n_{g} = 0$ means the $R T$ units cancel, so $K$ is a pure number — consistent with $64$ being dimensionless. Sanity: lots of product, little reactant → $K ≫ 1$ . ✓

Cell B — $Δ n_{g} < 0$ (gas moles shrink)

Worked example Ex 2 (Cell B): Haber

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ At $400 K$ , $K_{c} = 0.50 (mol/L)^{- 2}$ . Find $K_{p}$ .

Forecast: Four gas moles collapse into two. Should $K_{p}$ be bigger or smaller than $K_{c}$ ?

$Δ n_{g} = 2 - (1 + 3) = - 2$ . Why this step? $Δ n_{g}$ decides the power of $R T$ ; sign comes from products − reactants.
$K_{p} = K_{c} (R T)^{- 2}$ . Compute $R T = 0.0821 \times 400 = 32.84$ . Why this step? We must evaluate the bridge factor numerically.
$(R T)^{- 2} = 1/32.8 4^{2} = 1/1078.5 = 9.27 \times 1 0^{- 4}$ . So $K_{p} = 0.50 \times 9.27 \times 1 0^{- 4} = 4.64 \times 1 0^{- 4} atm^{- 2}$ . Why this step? Negative exponent means divide, shrinking $K_{p}$ ; the units are $atm^{Δ n_{g}} = atm^{- 2}$ .

Verify: Since $R T = 32.8 > 1$ and the exponent is negative, $K_{p} < K_{c}$ — exactly what "fewer gas moles form" predicts, because compressing 4 mol into 2 is disfavoured on a pressure scale. Units check: $atm^{- 2}$ matches $Δ n_{g} = - 2$ . ✓

Cell C — $Δ n_{g} > 0$ (gas moles grow)

Cell D — Degenerate: pure condensed phases omitted

Worked example Ex 4 (Cell D):

$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ A sealed flask holds only solid $N H_{4} H S$ at start. At equilibrium the total pressure is $0.60 atm$ . Find $K_{p}$ .

Forecast: The solid is omitted. Two gases form from nothing gaseous — in what ratio?

Write $K_{p}$ excluding the solid: $K_{p} = p_{N H_{3}} \cdot p_{H_{2} S} .$ Why this step? A pure solid has activity $= 1$ , so $N H_{4} H S$ never appears. The identical rule holds for pure liquids (activity $= 1$ ): e.g. in $F e (s) + H_{2} O (l)$ reactions or any equilibrium with a pure liquid solvent, that liquid is also left out of $K$ . Only gases and dissolved (aqueous) species carry a variable concentration.
Stoichiometry: one $N H_{3}$ and one $H_{2} S$ per decomposition, starting from pure solid, so $p_{N H_{3}} = p_{H_{2} S}$ . Why this step? Equal moles produced ⇒ equal partial pressures.
$p_{t o t a l} = p_{N H_{3}} + p_{H_{2} S} = 2 p$ . So $p = 0.30 atm$ each. $K_{p} = (0.30) (0.30) = 0.090 atm^{2} .$ Why this step? Split the measured total equally, then multiply.

Verify: Units are $atm^{2}$ because $Δ n_{g} = 2 - 0 = + 2$ (two gas moles from a solid) — consistent with a squared pressure. ✓

Cell E — Find $K$ from an ICE table

Worked example Ex 5 (Cell E):

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ Into a $1 L$ vessel we place $2.0 mol S O_{2}$ and $1.0 mol O_{2}$ . At equilibrium $1.6 mol S O_{3}$ has formed. Find $K_{c}$ .

Forecast: How many mol of $S O_{2}$ and $O_{2}$ are eaten to make $1.6$ mol $S O_{3}$ ?

Build the ICE (Initial–Change–Equilibrium) table. Let $x$ be the extent of reaction (see the definition above): one packet of the balanced equation makes $2$ mol $S O_{3}$ , so mol of $S O_{3}$ formed $= 2 x$ . Given $2 x = 1.6 \Rightarrow x = 0.8$ . Why this step? The extent $x$ ties every species to the coefficients through one number.
Changes dictated by coefficients: $S O_{2}$ (coeff 2) falls by $2 x = 1.6$ ; $O_{2}$ (coeff 1) falls by $x = 0.8$ ; $S O_{3}$ (coeff 2) rises by $2 x = 1.6$ . Equilibrium mol: $S O_{2} = 2.0 - 1.6 = 0.4$ ; $O_{2} = 1.0 - 0.8 = 0.2$ ; $S O_{3} = 1.6$ . Why this step? Coefficients set the ratio of amounts consumed/produced, all measured by $x$ .
Volume is $1$ L so mol = concentration. Apply the formula: $K_{c} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]} = \frac{( 1.6 ) ^{2}}{( 0.4 ) ^{2} ( 0.2 )} = \frac{2.56}{0.032} = 80 (mol/L)^{- 1} .$ Why this step? Coefficients 2, 2, 1 become the exponents; $Δ n_{g} = 2 - 3 = - 1$ gives the unit $(mol/L)^{- 1}$ .

Verify: All equilibrium amounts are positive (no negative concentrations) — physically valid. $K_{c} = 80 > 1$ : lots of product, matches a well-formed $S O_{3}$ mixture. ✓

Cell F — Manipulating a known $K$

Worked example Ex 6 (Cell F): three manipulations of one reaction

Given the concentration-based constant $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ has $K_{c} = 4.0 (mol/L)^{- 2}$ at some $T$ . Find the corresponding $K_{c}$ for: (a) reverse; (b) halve all coefficients; (c) add the reverse to itself.

Forecast: Reversing swaps top and bottom. Halving coefficients does what to the exponents?

(a) Reverse: $2 N H_{3} ⇌ N_{2} + 3 H_{2}$ . $K_{c, (a)} = \frac{1}{K _{c}} = \frac{1}{4.0} = 0.25 (mol/L)^{2} .$ Why this step? Reversing swaps numerator and denominator; the unit power flips sign too ( $Δ n_{g}$ goes $- 2 \to + 2$ ).
(b) Halve: $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3}$ . $K_{c, (b)} = K_{c}^{1/2} = 4.0 = 2.0 (mol/L)^{- 1} .$ Why this step? Scaling all coefficients by $n$ raises $K_{c}$ to the power $n$ ; here $n = \frac{1}{2}$ , and the unit power $- 2$ becomes $- 1$ .
(c) Add the reverse reaction (from (a)) to itself, i.e. multiply its $K_{c}$ twice: $K_{c, (c)} = (0.25) \times (0.25) = 0.0625 (mol/L)^{4} .$ Why this step? Adding reactions multiplies their $K_{c}$ s.

Verify: Consistency check — (a) then squaring (add to itself) gives $0.2 5^{2} = 0.0625$ ; and $K_{c, (a)} = 1/ K_{c} = 1/4$ is the reciprocal of the forward. All internally consistent. ✓

Cell G — Real-world word problem

Worked example Ex 7 (Cell G): a sealed bottle of soda

A sealed bottle holds carbonated water: $C O_{2} (a q) ⇌ C O_{2} (g)$ with $K_{c} = 1.2$ at room temperature. If $[C O_{2} (a q)] = 0.030 mol/L$ at equilibrium, find the effective concentration $[C O_{2} (g)] = n / V$ of the gas (recall from the definitions above that $n$ is moles and $V$ is volume in litres).

Forecast: Which side is "products" here — the gas above the liquid, right?

Identify products/reactants: the arrow points to gas, so $C O_{2} (g)$ is the product. Both species carry a variable concentration (dissolved gas and free gas), so both are included — neither is a pure solid or pure liquid, so nothing is omitted: $K_{c} = \frac{[ C O _{2} ( g )]}{[ C O _{2} ( a q )]} .$ Why this step? Here we deliberately use $K_{c}$ , not $K_{p}$ : even though one species is a gas, we are writing the equilibrium in terms of molar concentrations (mol/L) on both sides. A concentration $[C O_{2} (g)] = n / V$ is a perfectly legitimate way to describe a gas — $K_{c}$ just requires every varying species expressed the same way (as concentration). We could equally have used $K_{p}$ if we'd measured the gas by pressure; the two are linked by the usual $(R T)^{Δ n_{g}}$ bridge.
Rearrange for the unknown: $[C O_{2} (g)] = K_{c} [C O_{2} (a q)] = 1.2 \times 0.030 = 0.036 mol/L .$ Why this step? Solve the ratio for the top term.

Verify: $K_{c} > 1$ so more gas than dissolved — matches why an open bottle fizzes out ( $C O_{2}$ escapes as gas). Units cancel (mol/L over mol/L), so this particular $K_{c}$ is dimensionless ( $Δ n_{g}$ effectively $0$ for this concentration-based ratio). ✓

Cell H — Exam twist: which way will it go?

Worked example Ex 8 (Cell H): compare

$Q$ with $K$ For $H_{2} + I_{2} ⇌ 2 H I$ , $K_{c} = 64$ (from Ex 1). A fresh mix in $1$ L has $[H_{2}] = 0.20, [I_{2}] = 0.40, [H I] = 1.0$ . Is the system at equilibrium, and if not which way does it shift?

Forecast: Plug these current values into the same ratio — call it $Q$ (defined in the Definitions box above). If $Q < K$ , which direction restores balance?

Compute the reaction quotient $Q$ — same formula as $K_{c}$ but with the current, not equilibrium, concentrations: $Q = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 1.0 ) ^{2}}{( 0.20 ) ( 0.40 )} = \frac{1.0}{0.08} = 12.5.$ Why this step? $Q$ measures where we are; $K$ is where we want to be. (See Reaction Quotient Q vs K.)
Compare: $Q = 12.5 < K = 64$ . Why this step? The sign of $Q - K$ tells the direction.
Since $Q < K$ (too few products), the reaction shifts forward (left → right), making more $H I$ . Why this step? Nature drives $Q$ up toward $K$ by forming products.

Verify: $Q < K$ ⇒ forward, consistent with Le Chatelier's Principle "restore the ratio." If instead $Q > K$ , it would run backward. ✓

The direction rule at a glance

The figure below is a number line for the mixture in Ex 8. The amber vertical line marks $K = 64$ (the target). The cyan dot marks the current value $Q = 12.5$ , which sits to its left. The cyan arrow shows that a system with $Q < K$ is driven forward (left → right, making products) to climb toward $K$ ; the amber arrow on the right shows that any system with $Q > K$ is driven backward instead. To read any Cell-H problem: compute $Q$ , drop it on this line, and the nearest arrow tells you the direction.

The sign-of- $Δ n_{g}$ map

This last figure ties the three numeric examples together in one picture. The bars compare $K_{p}$ (amber) against a reference $K_{c}$ level (cyan, dashed line at "1") for the three cases A/B/C we just worked through:

Cell B (Ex 2, Haber, $Δ n_{g} < 0$ ): the amber bar is shorter than the cyan reference → $K_{p} < K_{c}$ , exactly the $4.64 \times 1 0^{- 4}$ vs $0.50$ shrink we computed.
Cell A (Ex 1, $H_{2} + I_{2}$ , $Δ n_{g} = 0$ ): the bars are equal → $K_{p} = K_{c} = 64$ .
Cell C (Ex 3, $P C l_{5}$ , $Δ n_{g} > 0$ ): the amber bar is taller → $K_{p} > K_{c}$ , the growth from $0.02$ up to $0.821$ .

So the sign of $Δ n_{g}$ alone decides whether $K_{p}$ lands below, on, or above $K_{c}$ — the picture is the visual summary of Ex 1–3.

Recall Quick self-test

Cell for $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ ? ::: D — degenerate, $K_{p} = p_{C O_{2}}$ If $Q > K$ , direction? ::: backward (toward reactants) Halving all coefficients does what to $K$ ? ::: takes its square root, $K^{1/2}$ Sign of $Δ n_{g}$ for $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ ? ::: negative, $2 - 3 = - 1$ When is $K_{p} = K_{c}$ ? ::: when $Δ n_{g} = 0$ Are pure liquids included in $K$ ? ::: no — like pure solids, their activity is 1, so they are omitted Units of $K_{p}$ when $Δ n_{g} = + 1$ ? ::: atm (i.e. $atm^{Δ n_{g}} = atm^{1}$ )

Connections

← Parent: Law of mass action and Kc, Kp
Reaction Quotient Q vs K
Le Chatelier's Principle
Chemical Equilibrium – Dynamic Nature
Gibbs Free Energy and K
Rate Constants and Arrhenius Equation
Ionic Equilibrium – Ka, Kb, Kw

Worked examples — Law of mass action and Kc, Kp