Reaction quotient Q vs K — direction of shift: Question bank

Definition Symbols used on this page (all earned before use)

$R$ = the gas constant, $R = 8.314 J mol^{- 1} K^{- 1}$ — a fixed number of nature that converts "temperature" into "energy per mole."
$T$ = the absolute temperature measured in kelvin (K), always a positive number, so $R T > 0$ always.
$Δ G$ = the Gibbs free-energy change for the forward reaction right now at the current concentrations — the "downhill push." Negative $Δ G$ means the forward reaction is spontaneous; positive $Δ G$ means the reverse reaction is spontaneous.
$Δ G^{\circ}$ = the standard Gibbs free-energy change, i.e. $Δ G$ measured under standard conditions (all species at $1 M$ or $1 atm$ ). It depends only on $T$ , not on the live mixture, and links to $K$ by $Δ G^{\circ} = - R T ln K$ . (See Gibbs free energy and spontaneity.)
$Δ n_{g a s}$ = the change in moles of gas = (moles of gaseous product) − (moles of gaseous reactant) from the balanced equation. It controls how volume/pressure changes affect $Q$ (see Relation between Kp and Kc).

This page assumes only the parent note: $Q$ has the same algebraic form as $K$ (products over reactants, each raised to its coefficient), but $Q$ uses whatever concentrations you have right now, while $K$ is the fixed target ratio at a given temperature. The master rule is $Δ G = R T ln (Q / K)$ : $Q < K$ forward, $Q > K$ backward, $Q = K$ rest.

The figure below is the single mental picture behind every item on this page: a number line of $Q$ with $K$ as the target, colour-coded by which way the reaction rolls.

Figure — Reaction quotient Q vs K — direction of shift

True or false — justify

True or false: If

Q > K

, the reaction has stopped.

False — a shift backward is still net motion; the reaction only truly stops when

Q

falls back to equal

K

, where

Δ G = 0

True or false: A large value of

Q

always means lots of product will form.

False — a large

Q

means product is already over-represented; if

Q > K

the reaction actually consumes product going backward.

True or false:

K

can change if you add more reactant.

False —

K

depends only on temperature; adding reactant lowers

Q

(more denominator), so the system shifts, but

K

itself is untouched.

True or false:

Q

and

Δ G

always have the same sign.

False —

Δ G = R T ln (Q / K)

depends on

Q

relative to

K

, not on

Q

alone;

Δ G

is negative whenever

Q < K

even if

Q

is a large number.

True or false: At the instant of mixing pure reactants,

Q = 0

and the reaction must go forward.

True — with no product the numerator is

0

, so

Q = 0 < K

, giving

Δ G = R T ln (0/ K) \to - \infty

: strongly forward. (See Gibbs free energy and spontaneity.)

True or false: If

Q = K

you can never make the reaction shift again.

False — you're at equilibrium now, but changing concentration, pressure, or temperature moves

Q

(or

K

) and restarts a shift, exactly as Le Chatelier's principle predicts.

True or false: For

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

, comparing a pressure-based

Q_{p}

to a concentration-based

K_{c}

tells you the direction.

False — you must compare like with like:

Q_{p}

against

K_{p}

Q_{c}

against

K_{c}

, since their numerical values differ (see Relation between Kp and Kc).

True or false: A reaction with

Q = K

has

Δ G^{\circ} = 0

False —

Q = K

makes

Δ G = 0

, but the standard change

Δ G^{\circ} = - R T ln K

is generally non-zero; the two are different quantities.

True or false: Adding a catalyst changes

Q

and lets the system reach a new equilibrium ratio.

False — a catalyst speeds the forward and reverse reactions equally, so it changes neither

Q

nor

K

; it only reaches the same equilibrium faster (see Position of equilibrium vs rate of reaction).

Spot the error

"For

C (s) + C O_{2} (g) ⇌ 2 C O (g)

Q = \frac{[ C O ] ^{2}}{[ C ] [ C O _{2} ]}

." Find the mistake.

The solid carbon

C (s)

must be omitted (its activity is

1

); correct form is

Q = \frac{[ C O ] ^{2}}{[ C O _{2} ]}

(see Activity and why pure solids-liquids are omitted).

"For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

Q = \frac{[ N H _{3} ]}{[ N _{2} ] [ H _{2} ]}

." Find the mistake.

The stoichiometric coefficients were dropped; each species is raised to its coefficient:

Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

Q < K

Δ G > 0

and the reaction runs backward." Find the mistake.

The sign is flipped —

Q < K

gives

ln (Q / K) < 0

, so

Δ G < 0

and the reaction runs forward.

"Since

Q = K

at equilibrium, both the forward and reverse reactions have stopped." Find the mistake.

Only the net change stops; forward and reverse reactions continue at equal rates (dynamic equilibrium) —

Q

is a position, not a rate (see Position of equilibrium vs rate of reaction).

"I added a catalyst, so

Q

jumped closer to

K

instantly." Find the mistake.

A catalyst leaves

Q

and

K

untouched; it only lowers the activation barrier so both directions go faster, reaching the same equilibrium sooner — it does not move the position.

"I diluted the mixture, but

Q

stays the same because I diluted everything equally." Find the mistake.

Dilution does not generally preserve

Q

unless the moles of gas are equal on both sides; a non-zero

Δ n_{g a s}

makes

Q

shift, so the equilibrium moves.

Δ G = R T ln (Q / K)

, and since

Q

can be negative when there's no product,

ln

is undefined." Find the mistake.

Concentrations are never negative, so

Q \geq 0

; the true boundary case is

Q = 0

(only reactants), giving

ln 0 \to - \infty

, a strong forward push — not an undefined value.

Why questions

Why does the ratio

Q / K

, not

Q

itself, decide direction?

Because

Δ G = R T ln (Q / K)

; only when the current ratio

Q

differs from the target ratio

K

is there free energy to release, and the sign of that difference sets the direction.

Why can

Q

range from

0

\infty

while

K

is a single number?

Q

reads off whatever concentrations happen to exist (all reactant

\to 0

, all product

\to \infty

), whereas

K

is the one fixed ratio the system settles into at a given temperature.

Why does adding an inert (non-reacting) gas at constant volume not shift

Q_{p}

An inert gas at fixed volume doesn't change any reacting species' partial pressure, so every term in

Q_{p}

is unchanged and

Q_{p} = K_{p}

still holds.

Why is

Δ G^{\circ}

constant while

Δ G

changes as the reaction proceeds?

Δ G^{\circ} = - R T ln K

fixes the standard reference (depends only on

T

), but

Δ G = Δ G^{\circ} + R T ln Q

carries the live

Q

term, which drifts toward zero as

Q \to K

Why does comparing

Q

K

give the same prediction as Le Chatelier for adding a product?

Adding product raises

Q

above

K

, so

Δ G > 0

and the system shifts backward — exactly what Le Chatelier's principle says about relieving an added stress.

Edge cases

If the reaction has equal moles of gas on both sides (e.g.

H_{2} + I_{2} ⇌ 2 H I

), what happens to

Q

when you halve the volume?

Nothing — every concentration doubles but the exponents cancel (

Δ n_{g a s} = 0

), so

Q

is unchanged and no shift occurs.

You start with only products and no reactants. What is

Q

and which way does it go?

Q \to \infty

(zero denominator), so

Q > K

Δ G

for the forward reaction is positive, which means the reverse reaction is spontaneous — the system runs backward to make reactants.

Adding inert gas at constant volume vs constant pressure — does it matter?

At constant volume nothing shifts (partial pressures unchanged); at constant pressure the total volume must grow, diluting every reacting gas, so if

Δ n_{g a s} \neq = 0

then

Q_{p}

moves and the equilibrium shifts toward the side with more moles of gas.

Two flasks have the same

Q

but different temperatures. Can they shift in opposite directions?

Yes —

K

is temperature-dependent, so the same

Q

may sit below

K

in one flask (forward) and above

K

in the other (backward).

At exactly

Q = K

, what is the numerical value of

ln (Q / K)

and of

Δ G

ln (1) = 0

, so

Δ G = R T \cdot 0 = 0

regardless of temperature — the defining condition of equilibrium.

K

is enormous (say

1 0^{20}

), does that guarantee the forward reaction is fast?

No — a huge

K

only means the equilibrium position lies far toward products; the rate of getting there is a separate matter, best fixed by a catalyst (see Position of equilibrium vs rate of reaction).

For a reaction where all species are pure solids and liquids, what is

Q

Q = 1

(every activity is

1

), so the comparison to

K

is fixed and no meaningful concentration-driven shift is defined.

Connections

Reaction quotient Q vs K — direction of shift (index 2.6.4)
Equilibrium constant K (Kc and Kp)
Le Chatelier's principle
Gibbs free energy and spontaneity
Relation between Kp and Kc
Activity and why pure solids-liquids are omitted
Position of equilibrium vs rate of reaction