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

Before we start, earn every symbol you will meet on every line — nothing is used blind:

Definition The quantities in play

$K$ (plain, no subscript) — the generic equilibrium constant: whenever a statement is true for both the concentration form and the pressure form, we write just $K$ as a stand-in for "either $K_{c}$ or $K_{p}$ , whichever applies." When the distinction matters we use the subscript.
$K_{c}$ — the equilibrium constant written with concentrations (moles per litre), products over reactants, each raised to its coefficient.
$[i]$ — shorthand for the molar concentration of species $i$ , i.e. its moles $n_{i}$ divided by the container volume $V$ : $[i] = n_{i} / V$ . So $[A]$ means "concentration of A."
$K_{p}$ — the same ratio but written with partial pressures of the gases. Linked by $K_{p} = K_{c} (R T)^{Δ n_{g}}$ .
$Q$ — the reaction quotient: the exact same ratio as $K$ , but evaluated at any moment, not only at equilibrium. If $Q = K$ we are at equilibrium; otherwise the reaction is still moving. (More in Reaction Quotient Q vs K.)
$Δ n_{g}$ — gaseous moles of products minus gaseous moles of reactants (solids and liquids never counted).

Three pictures to anchor every trap

Before the questions, look at three plots. The traps below are just words describing what these curves show — keep glancing back at them.

Picture 1 — equilibrium is dynamic, not dead. The forward rate falls as reactants are used up; the backward rate rises as products build up. Where the two curves cross, the rates are equal (so no net change), yet both are still large and non-zero — molecules keep converting both ways forever. That crossing point is equilibrium.

Picture 2 — $Q$ tells you which way to move; $K$ is the finish line. $Q$ is the same ratio as $K$ but computed at any instant. If $Q < K$ the reaction runs forward (toward products); if $Q > K$ it runs backward; when $Q = K$ it stops moving net. The horizontal line is $K$ — fixed at a given temperature — and the reaction always slides toward it.

Picture 3 — the $(R T)^{Δ n_{g}}$ factor. $K_{p} = K_{c} (R T)^{Δ n_{g}}$ . Because $R T > 1$ in ordinary units, the sign of $Δ n_{g}$ decides everything: positive $Δ n_{g}$ pushes $K_{p}$ above $K_{c}$ , negative pushes it below, and $Δ n_{g} = 0$ makes them equal. This single curve replaces a dozen "is $K_{p}$ bigger or smaller?" traps.

Formula Why the

$(R T)^{Δ n_{g}}$ factor appears — full expansion Write the general reaction as $\sum_{i} ν_{i} X_{i}$ where $ν_{i}$ is the coefficient of species $X_{i}$ (products count as $+ ν_{i}$ , reactants as $- ν_{i}$ ). Then $K_{p}$ is the product over all gaseous species: $K_{p} = \prod_{i} (p_{i})^{ν_{i}}$ Step 1 — replace each partial pressure using $p_{i} = [i] R T$ (why: ideal gas law, $p_{i} = (n_{i} / V) R T$ ): $K_{p} = \prod_{i} ([i] R T)^{ν_{i}}$ Step 2 — split each factor, since $(x y)^{ν} = x^{ν} y^{ν}$ (why: separate the concentration part from the $R T$ part): $K_{p} = \prod_{i} [i]^{ν_{i}} \cdot \prod_{i} (R T)^{ν_{i}}$ Step 3 — the first product is $K_{c}$ ; in the second, multiplying powers of the same base adds exponents, $\prod_{i} (R T)^{ν_{i}} = (R T)^{\sum_{i} ν_{i}}$ : $K_{p} = K_{c} \cdot (R T)^{\sum_{i} ν_{i}}$ Step 4 — and $\sum_{i} ν_{i}$ (products minus reactants) is exactly $Δ n_{g}$ : $K_{p} = K_{c} (R T)^{Δ n_{g}}$

True or false — justify

K

dimensionless in general?

False. Only when

Δ n_{g} = 0

do the concentration units cancel. In general

K_{c}

carries units

(mol L^{- 1})^{Δ n_{g}}

; the "dimensionless

K

" you use in

Δ G^{\circ} = - R T ln K

is really

K

with each term divided by its standard state, a subtler object.

A large

K

means the reaction is fast.

False.

K = k_{f} / k_{b}

is a ratio of rate constants and tells you only how far the reaction goes (products vs reactants at equilibrium), not how quickly it gets there. Speed is set by

k_{f}

itself — see Rate Constants and Arrhenius Equation.

Adding a catalyst increases

K

False. A catalyst multiplies

k_{f}

and

k_{b}

by the same factor, so their ratio

k_{f} / k_{b} = K

is untouched. It only makes equilibrium arrive sooner.

K_{p}

is always greater than

K_{c}

False.

K_{p} = K_{c} (R T)^{Δ n_{g}}

. If

Δ n_{g} < 0

and

R T > 1

, then

(R T)^{Δ n_{g}} < 1

K_{p} < K_{c}

(see Picture 3). Sign of

Δ n_{g}

decides direction.

At equilibrium the forward and backward reactions have stopped.

False. They run at equal rates, not zero rate — equilibrium is dynamic, not static (see Picture 1). Net change is zero but molecules keep converting both ways (Chemical Equilibrium – Dynamic Nature).

Doubling the volume of a gas mixture changes

K_{c}

False. Volume changes concentrations, which changes

Q

and shifts the position of equilibrium, but

K_{c}

itself depends only on temperature. The system re-adjusts so the ratio returns to the same

K_{c}

For

H_{2} + I_{2} ⇌ 2 H I

K_{p} = K_{c}

exactly.

True. Here

Δ n_{g} = 2 - 2 = 0

, so

(R T)^{0} = 1

and

K_{p} = K_{c}

. This is one of the few reactions where the numbers coincide.

Increasing temperature always increases

K

False. It increases

K

for endothermic reactions and decreases

K

for exothermic ones, because heating shifts equilibrium toward the heat-absorbing side (Le Chatelier's Principle).

K

for the reverse reaction equals

- K

of the forward.

False. Reversing swaps numerator and denominator, giving

K_{rev} = 1/ K

, never a negative number.

K

is a ratio of positive quantities, so it is always positive.

Spot the error

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

has

K_{c} = \frac{[ C a O ] [ C O _{2} ]}{[ C a C O _{3} ]}

Error: solids

C a O

and

C a C O_{3}

have fixed (constant) activity and must be dropped. Correct:

K_{c} = [C O_{2}]

and

K_{p} = p_{C O_{2}}

"For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

Δ n_{g} = 2 - 4 = - 2

, so

K_{p} = K_{c} (R T)^{2}

Error: the exponent is

Δ n_{g} = - 2

, not

+ 2

. Correct:

K_{p} = K_{c} (R T)^{- 2}

. Sign flip is the classic slip.

"Rate of a reaction

\propto [A]^{a} [B]^{b}

where

a, b

are the coefficients, for any reaction."

Error: that power law holds only for an elementary reaction. For a multi-step mechanism the rate law's exponents are experimental orders, unrelated to stoichiometric coefficients — yet the equilibrium expression

K_{c}

still uses the coefficients.

Q

and

K

are different formulas."

Error: they are the identical algebraic expression. The only difference is when you plug in the concentrations —

Q

at any instant,

K

at equilibrium (see Picture 2).

"Since a catalyst is written as a species, include it in

K

Error: a catalyst is neither consumed nor produced net, so it never appears in the balanced equation as reactant/product and is absent from

K

Δ n_{g}

for

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

1 - 1 = 0

because there are two solids and one gas... equal counts."

Error: only gaseous moles count. Products have 1 gas (

C O_{2}

), reactants have 0 gas, so

Δ n_{g} = 1 - 0 = 1

"Multiplying every coefficient by 2 doubles

K

Error: it raises

K

to the power 2, giving

K^{2}

, because every concentration term gets its exponent doubled. Multiplication (

2 K

) is wrong.

Why questions

Why are the exponents in

K_{c}

the stoichiometric coefficients?

For an elementary reaction, a forward event needs those many molecules to meet at once, so equating

k_{f} [A]^{a} [B]^{b} = k_{b} [C]^{c} [D]^{d}

carries those powers into

K_{c}

— but note this collision picture is an assumption valid only for elementary steps. The general proof that the coefficients appear in

K_{c}

for any reaction comes from thermodynamics, not collisions.

Why does

K

depend only on temperature and nothing else?

K = k_{f} / k_{b}

, and each rate constant depends on

T

through Arrhenius. Pressure, concentration and catalysts change

k_{f}

and

k_{b}

equally (or not at all), so the ratio is invariant; only

T

shifts them by different amounts.

Why do we omit pure solids and liquids from

K

Their "concentration" (density ÷ molar mass) is a fixed number set by the substance, not by how much is present. Being constant, it is absorbed into

K

; their activity is defined as 1.

Why can we replace concentration with partial pressure for gases?

The ideal gas law gives

p_{i} = (n_{i} / V) R T = [i] R T

, so partial pressure is directly proportional to concentration

[i]

at fixed

T

. Substituting into

\prod_{i} (p_{i})^{ν_{i}}

and adding the

R T

exponents (see the boxed expansion above) produces

K_{p} = K_{c} (R T)^{Δ n_{g}}

Why is the sign of

Δ G^{\circ}

tied to whether

K > 1

Because

Δ G^{\circ} = - R T ln K

. If

K > 1

then

ln K > 0

Δ G^{\circ} < 0

(products favoured);

K < 1

gives

Δ G^{\circ} > 0

(Gibbs Free Energy and K).

Why does adding an inert gas at constant volume leave equilibrium unchanged?

It raises total pressure but not the partial pressures or concentrations of the reacting species, so

Q

is unchanged and no shift occurs.

K_{p}

and

K_{c}

stay the same regardless.

Why must

K

always be a positive number?

It is a ratio of concentrations (or pressures), all of which are physically non-negative and non-zero at equilibrium, so no combination can produce a zero or negative value.

Edge cases

If a reaction has no gaseous species at all, does

K_{p}

exist?

Not meaningfully —

K_{p}

is defined through partial pressures of gases. For an all-aqueous or all-solid/liquid system you use

K_{c}

(with pure solids/liquids omitted); there are no pressures to write.

What is

Δ G^{\circ}

when

K = 1

Exactly zero, since

Δ G^{\circ} = - R T ln 1 = 0

. Neither direction is thermodynamically favoured — reactants and products are equally stable at standard states.

What happens to

Q

the instant you mix pure reactants with no product?

The numerator (products) is zero, so

Q = 0 < K

: the reaction must run forward. This is a legal degenerate value of

Q

, whereas

K

itself is never zero at equilibrium.

Δ n_{g} = 0

, what is

(R T)^{Δ n_{g}}

and what does it imply?

(R T)^{0} = 1

, so

K_{p} = K_{c}

and both are dimensionless. The units of concentration and pressure cancel completely across products and reactants.

Can

K

be infinite?

Not for a genuinely reversible reaction; a finite backward rate keeps some reactant present, so the denominator stays non-zero. A "reaction that goes to completion" is the limit where

K

is astronomically large but still finite in principle.

For a very small

K

(say

1 0^{- 30}

), is the forward reaction impossible?

No — it still proceeds to a tiny but non-zero extent. Small

K

means equilibrium lies far toward reactants, not that no product ever forms.

If temperature is not specified, is a numeric

K

meaningful?

No. Since

K

changes with

T

, a bare number like "

K = 16

" is incomplete without stating the temperature at which it was measured.

Recall One-line self-test before moving on

Cover this and recite: What changes $K$ , what doesn't, and why? Answer ::: Only temperature changes $K$ (it shifts $k_{f}$ and $k_{b}$ unequally). Concentration, volume, pressure, inert gas and catalysts do not change $K$ — they only move the system toward the same fixed ratio.

Connections

2.6.02 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