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

Before we start, one reminder of the tools you already built in the parent note:

Level 1 — Recognition

Can you read the reaction and write the correct expression?

L1.1 Write $K_{c}$ for $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ .

Recall Solution L1.1

WHAT we do: products over reactants, each raised to its stoichiometric coefficient. WHY: that is exactly what the law of mass action prescribes. $K_{c} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]}$ The power $2$ on $S O_{3}$ and $S O_{2}$ comes from the coefficient $2$ ; $O_{2}$ has coefficient $1$ so power $1$ .

L1.2 Write $K_{p}$ for $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ .

Recall Solution L1.2

WHAT we do: drop every pure solid. WHY: the activity of a pure solid is fixed (its "concentration" is set by density, it cannot vary), so it is absorbed into the constant. Only the gas $C O_{2}$ survives: $K_{p} = p_{C O_{2}}$

L1.3 For which of these is $K_{p} = K_{c}$ ? (a) $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ (b) $H_{2} + I_{2} ⇌ 2 H I$ (c) $P C l_{5} ⇌ P C l_{3} + C l_{2}$

Recall Solution L1.3

WHAT we do: compute $Δ n_{g}$ ; $K_{p} = K_{c}$ only when $Δ n_{g} = 0$ .

(a) $Δ n_{g} = 2 - (1 + 3) = - 2$ → not equal.
(b) $Δ n_{g} = 2 - (1 + 1) = 0$ → equal. ✅
(c) $Δ n_{g} = (1 + 1) - 1 = + 1$ → not equal. Answer: (b).

Level 2 — Application

Plug numbers into a single formula.

L2.1 At equilibrium in a 2 L flask: $[N_{2}] = 0.10 M$ , $[H_{2}] = 0.30 M$ , $[N H_{3}] = 0.20 M$ for $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ . Find $K_{c}$ .

Recall Solution L2.1

WHAT: substitute into $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ . WHY the powers: coefficient $2$ on $N H_{3}$ , $1$ on $N_{2}$ , $3$ on $H_{2}$ . $K_{c} = \frac{( 0.20 ) ^{2}}{( 0.10 ) ( 0.30 ) ^{3}} = \frac{0.04}{( 0.10 ) ( 0.027 )} = \frac{0.04}{0.0027} \approx 14.8$ The flask volume is a distractor — concentrations are already given, so we never need $V$ .

L2.2 For $H_{2} + I_{2} ⇌ 2 H I$ , $K_{c} = 49$ at $700 K$ . What is $K_{p}$ ?

Recall Solution L2.2

WHAT: use $K_{p} = K_{c} (R T)^{Δ n_{g}}$ . WHY it's quick: $Δ n_{g} = 2 - 2 = 0$ , so $(R T)^{0} = 1$ . $K_{p} = 49 \times (R T)^{0} = 49$ When the number of gas molecules doesn't change, pressures and concentrations give the same constant.

L2.3 For $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , $K_{c} = 1.8 \times 1 0^{- 3}$ at $250 K$ . Find $K_{p}$ (use $R = 0.0821$ ).

Recall Solution L2.3

WHAT: $Δ n_{g} = (1 + 1) - 1 = + 1$ , so $K_{p} = K_{c} (R T)^{1}$ . WHY the sign is $+$ : more gas molecules are produced than consumed. $R T = 0.0821 \times 250 = 20.525$ $K_{p} = 1.8 \times 1 0^{- 3} \times 20.525 \approx 3.69 \times 1 0^{- 2}$

Level 3 — Analysis

Build the equilibrium concentrations yourself (ICE reasoning), then find $K$ .

L3.1 $2.0 mol$ $H I$ is placed in a $1 L$ flask and decomposes: $2 H I ⇌ H_{2} + I_{2}$ . At equilibrium $[H_{2}] = 0.20 M$ . Find $K_{c}$ .

Recall Solution L3.1

WHAT — build the ICE table (Initial, Change, Equilibrium). Look at the figure: two $H I$ break to make one $H_{2}$ and one $I_{2}$ , so the changes are in ratio $2 : 1 : 1$ .

	$2 H I$	$H_{2}$	$I_{2}$
I	$2.0$	$0$	$0$
C	$- 2 x$	$+ x$	$+ x$
E	$2.0 - 2 x$	$x$	$x$

WHY these signs: $H I$ is consumed (minus), products appear (plus); coefficient $2$ gives $- 2 x$ . Given $[H_{2}] = x = 0.20$ , so $[I_{2}] = 0.20$ and $[H I] = 2.0 - 2 (0.20) = 1.60$ . $K_{c} = \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}} = \frac{( 0.20 ) ( 0.20 )}{( 1.60 ) ^{2}} = \frac{0.04}{2.56} \approx 1.56 \times 1 0^{- 2}$

L3.2 In a $1 L$ flask, $1.0 mol$ $N_{2}$ and $3.0 mol$ $H_{2}$ react: $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ . At equilibrium $0.6 mol$ $N H_{3}$ has formed. Find $K_{c}$ .

Recall Solution L3.2

WHAT — ICE table with change ratio $1 : 3 : 2$ . $2 x = 0.6 \Rightarrow x = 0.3$ .

$[N_{2}] = 1.0 - 0.3 = 0.70$
$[H_{2}] = 3.0 - 3 (0.3) = 3.0 - 0.9 = 2.10$
$[N H_{3}] = 0.60$ WHY: each mole of $N_{2}$ consumed uses $3$ of $H_{2}$ and makes $2$ of $N H_{3}$ . $K_{c} = \frac{( 0.60 ) ^{2}}{( 0.70 ) ( 2.10 ) ^{3}} = \frac{0.36}{( 0.70 ) ( 9.261 )} = \frac{0.36}{6.4827} \approx 5.55 \times 1 0^{- 2}$

Level 4 — Synthesis

Combine several rules: manipulation, $K_{p} \leftrightarrow K_{c}$ , and mole fractions.

L4.1 Given $K_{c} = 16$ for $H_{2} + I_{2} ⇌ 2 H I$ . Find $K_{c}$ for (a) $2 H I ⇌ H_{2} + I_{2}$ and (b) $\frac{1}{2} H_{2} + \frac{1}{2} I_{2} ⇌ H I$ .

Recall Solution L4.1

WHAT: apply the two manipulation rules.

(a) Reversed reaction → $K^{'} = 1/ K$ : $K_{c}^{'} = \frac{1}{16} = 0.0625$ WHY: flipping swaps numerator and denominator.
(b) All coefficients multiplied by $\frac{1}{2}$ → $K^{''} = K^{1/2}$ : $K_{c}^{''} = 16 = 4$ WHY: raising every term to the power $\frac{1}{2}$ raises the whole ratio to $\frac{1}{2}$ .

L4.2 For $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ at $298 K$ , $K_{p} = 0.15 bar$ . Find $K_{c}$ (use $R = 0.0831$ ).

Recall Solution L4.2

WHAT: invert the link. $K_{p} = K_{c} (R T)^{Δ n_{g}} \Rightarrow K_{c} = K_{p} / (R T)^{Δ n_{g}}$ . $Δ n_{g} = 2 - 1 = + 1$ . $R T = 0.0831 \times 298 = 24.7638$ $K_{c} = \frac{0.15}{24.7638} \approx 6.06 \times 1 0^{- 3}$ WHY divide: since $K_{p} = K_{c} \cdot R T$ here, solving for $K_{c}$ divides by $R T$ .

L4.3 Two coupled equilibria at the same $T$ : $A ⇌ B, K_{1} = 4; B ⇌ C, K_{2} = 0.5$ Find $K$ for $A ⇌ C$ .

Recall Solution L4.3

WHAT: adding reactions multiplies their $K$ 's. $K = K_{1} \times K_{2} = 4 \times 0.5 = 2$ WHY: write $K_{1} = [B] / [A]$ and $K_{2} = [C] / [B]$ ; multiply and $[B]$ cancels, leaving $[C] / [A]$ .

Level 5 — Mastery

Everything at once: ICE + degree of dissociation + $K_{p}$ from mole fractions.

L5.1 $1 mol$ of $P C l_{5}$ is placed in a $2 L$ vessel at $500 K$ and $30%$ dissociates: $P C l_{5} ⇌ P C l_{3} + C l_{2}$ . Find $K_{c}$ and $K_{p}$ (use $R = 0.0821$ ).

Recall Solution L5.1

WHAT — degree of dissociation $α = 0.30$ . Moles at equilibrium:

$P C l_{5} : 1 - α = 0.70$
$P C l_{3} : α = 0.30$
$C l_{2} : α = 0.30$

Divide by $V = 2 L$ for concentrations:

$[P C l_{5}] = 0.35, [P C l_{3}] = 0.15, [C l_{2}] = 0.15$ $K_{c} = \frac{[ P C l _{3} ] [ C l _{2} ]}{[ P C l _{5} ]} = \frac{( 0.15 ) ( 0.15 )}{0.35} = \frac{0.0225}{0.35} \approx 6.43 \times 1 0^{- 2}$ WHY: coefficients are all $1$ , so no powers above $1$ .

Now $K_{p}$ : $Δ n_{g} = 2 - 1 = + 1$ . $R T = 0.0821 \times 500 = 41.05$ $K_{p} = K_{c} (R T)^{1} = 6.43 \times 1 0^{- 2} \times 41.05 \approx 2.64$

L5.2 $2 mol$ $S O_{2}$ and $1 mol$ $O_{2}$ are mixed at total pressure $P = 3 bar$ : $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ . At equilibrium $1 mol$ $S O_{3}$ forms. Find $K_{p}$ in terms of partial pressures.

Recall Solution L5.2

WHAT — ICE in moles, change ratio $2 : 1 : 2$ . If $S O_{3} = 2 x = 1$ then $x = 0.5$ .

$S O_{2} : 2 - 2 x = 2 - 1 = 1.0$
$O_{2} : 1 - x = 1 - 0.5 = 0.5$
$S O_{3} : 2 x = 1.0$
Total moles $= 1.0 + 0.5 + 1.0 = 2.5$

Mole fractions (why: partial pressure $=$ mole fraction $\times$ total $P$ ):

$y_{S O_{2}} = 1.0/2.5 = 0.40 \Rightarrow p_{S O_{2}} = 0.40 \times 3 = 1.20$
$y_{O_{2}} = 0.5/2.5 = 0.20 \Rightarrow p_{O_{2}} = 0.20 \times 3 = 0.60$
$y_{S O_{3}} = 1.0/2.5 = 0.40 \Rightarrow p_{S O_{3}} = 0.40 \times 3 = 1.20$ $K_{p} = \frac{p _{S O_{3}}^{2}}{p _{S O_{2}}^{2} p _{O_{2}}} = \frac{( 1.20 ) ^{2}}{( 1.20 ) ^{2} ( 0.60 )} = \frac{1}{0.60} \approx 1.67 bar^{- 1}$ WHY units: $Δ n_{g} = 2 - 3 = - 1$ , so $K_{p}$ carries $bar^{- 1}$ .

Recall One-line self-check summary

Reverse → invert $K$ ::: $K^{'} = 1/ K$ Scale coefficients by $n$ → ::: $K^{'} = K^{n}$ Add reactions → ::: multiply the $K$ 's $K_{p}$ from $K_{c}$ ::: $K_{p} = K_{c} (R T)^{Δ n_{g}}$ Mole fractions use which moles ::: the equilibrium total, not the initial total

Connections

← Back to parent: Law of Mass Action
Reaction Quotient Q vs K — the same ratio before equilibrium
Le Chatelier's Principle — how the ICE positions shift under stress
Gibbs Free Energy and K — where these $K$ values come from thermodynamically
Ionic Equilibrium – Ka, Kb, Kw — the same machinery for dissolved species