Relationship Kp = Kc(RT)^Δn: Question bank — Chemistry deep dive

This bank hunts the ideas behind the parent topic, not the arithmetic. Every prompt below is a place students slip: a sign, a phase, a unit, a limiting case. Read the question, answer out loud, then reveal.

Definition The three symbols this whole page rests on

Before any question, pin down the vocabulary so nothing below is a mystery:

General reaction. Whenever we speak of a "general gas reaction" we mean $a A + b B ⇌ c C + d D$ , where $A, B$ are reactants, $C, D$ are products, and the small letters $a, b, c, d$ are the stoichiometric coefficients — the counting numbers in front of each species that say how many moles react.
$Δ n$ (read "delta n") = (moles of gaseous products) − (moles of gaseous reactants) $= (c + d) - (a + b)$ , counting only gases. It is the exponent on $R T$ in the bridge formula.
$R$ = the ideal gas constant, the fixed number linking pressure, volume, amount and temperature in $p V = n R T$ . Its numerical value depends on the units you use — e.g. $0.0821 L atm K^{- 1} mol^{- 1}$ for atm and mol L⁻¹.
$T$ = temperature, and it must always be in kelvin (absolute temperature), because the gas law is only valid on that scale.

True or false — justify

K_{p} = K_{c}

can only happen when the reaction has no gases at all.

False — it happens whenever ==

Δ n = 0

==, i.e. equal moles of gas on both sides (e.g.

H_{2} + I_{2} ⇌ 2 H I

), regardless of how many gases are present.

For any reaction, raising temperature makes

(R T)^{Δ n}

larger, so

K_{p}

always grows relative to

K_{c}

False — only if

Δ n > 0

. If

Δ n < 0

, larger

R T

makes

(R T)^{Δ n}

smaller, widening the gap the other way; if

Δ n = 0

the factor stays

1

K_{c}

is dimensionless then

K_{p}

must also be dimensionless.

False in the "with units" convention — when

Δ n \neq = 0

the factor

(R T)^{Δ n}

carries pressure/volume units, so

K_{p}

and

K_{c}

generally differ in units. (In strict thermodynamics both are dimensionless via activities.)

The formula

K_{p} = K_{c} (R T)^{Δ n}

works for a reaction in aqueous solution with no gases.

False — it is derived from

p_{X} = [X] R T

, which is the ideal gas law. With no gases there are no partial pressures, so

K_{p}

is not even defined.

Doubling the container volume changes

K_{p}

because pressures drop.

False —

K_{p}

(like

K_{c}

) depends only on temperature. Volume changes shift the position of equilibrium (Le Chatelier), not the constant. See Le Chatelier's Principle.

Δ n

in this formula is the same

Δ n

used everywhere else in the reaction.

False — here

Δ n

counts only gaseous species; a mole balance for stoichiometry or enthalpy may count solids/liquids, but this bridge formula does not.

Because

R

appears in the formula, using a different value of

R

changes the true value of

K_{p}

False — the number changes only because you changed the units of

p

and

[X]

. Match

R

to those units and the physical equilibrium is identical.

Spot the error

Student writes: "For

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

Δ n = 3 - 2 = + 1

Error: they put reactants − products backwards and miscounted. Products

= 2

, reactants

= 2 + 1 = 3

, so

Δ n = 2 - 3 = - 1

(products − reactants).

Student writes: "

K_{p} = K_{c} (R T)^{Δ n}

with

T = 25

for room temperature."

Error:

T

must be in kelvin. Room temperature is

25 + 273 = 298 K

; the gas law only holds for absolute temperature.

Student writes: "For

C (s) + O_{2} (g) ⇌ C O_{2} (g)

Δ n = 1 - 2 = - 1

Error: solid carbon is not counted. Gaseous products

= 1

(

C O_{2}

), gaseous reactants

= 1

(

O_{2}

), so

Δ n = 1 - 1 = 0

and

K_{p} = K_{c}

Student writes: "

p_{X} = [X] / R T

, so pressure is small when

R T

is large."

Error: the gas law gives

p_{X} = [X] R T

(multiply, not divide). More thermal push

R T

means higher pressure for the same crowding. See Ideal Gas Law PV=nRT.

Student writes: "

K_{p} = K_{c} \cdot R T \cdot Δ n

Error:

Δ n

is an exponent on

R T

, not a multiplier. Correct form is

K_{p} = K_{c} (R T)^{Δ n}

; the whole

R T

is raised to the power

Δ n

Student writes: "

Δ n = - 2

, and since

R, T > 0

this makes

(R T)^{- 2}

negative, so

K_{p} < 0

Error: a negative exponent gives a reciprocal, not a negative number.

(R T)^{- 2} = 1/ (R T)^{2} > 0

;

K_{p}

stays positive (it is a ratio of positive pressures).

Why questions

Why does the ideal gas law, of all tools, bridge

K_{p}

and

K_{c}

Because it is the one relation linking pressure to amount-per-volume:

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

. Partial pressure is concentration scaled by

R T

, so it converts one constant's variables into the other's. See Partial Pressure and Dalton's Law.

Why does the whole conversion collapse to depend only on

Δ n

and nothing else about the reaction?

Every partial pressure contributes one factor of

R T

per mole, so the total power of

R T

is (product mole powers) − (reactant mole powers)

= Δ n

; all the concentration parts regroup exactly into

K_{c}

Why is temperature the only thing that changes

K_{p}

K_{c}

, but pressure or volume are not?

K

is fixed by the reaction's energetics at a given

T

(via

Δ G^{\circ} = - R T ln K

, see Gibbs Free Energy and K). Changing

V

or total

p

moves concentrations/partial pressures around until the ratio returns to that same

K

Why must

R

"match" the units of

K_{p}

and

K_{c}

Because

p_{X} = [X] R T

must be dimensionally consistent: if

p

is in atm and

[X]

in mol L⁻¹, then

R

must carry L atm K⁻¹ mol⁻¹ so the units cancel correctly.

Why is

Δ n

defined as products minus reactants and not the reverse?

To mirror the constant itself, which is "products over reactants." With the general reaction

a A + b B ⇌ c C + d D

, the products

C, D

sit on top, so their coefficients get the positive sign:

Δ n = (c + d) - (a + b)

Why can two reactions with the same

K_{c}

have very different

K_{p}

Because they can have different

Δ n

. The

(R T)^{Δ n}

factor depends on the gas-mole change, so a

Δ n = + 2

reaction and a

Δ n = - 2

reaction diverge sharply even from identical

K_{c}

Edge cases

What happens to the formula when

Δ n = 0

(R T)^{0} = 1

, so

K_{p} = K_{c}

exactly, for every temperature and every value of

R

— the pressure and concentration pictures coincide.

For

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

, what is

Δ n

and why?

Δ n = 1 - 0 = + 1

. Only

C O_{2}

is gaseous; both solids are excluded, so gaseous products

= 1

, gaseous reactants

= 0

What is

K_{p}

for a reaction with no gases (e.g. all-aqueous)?

Undefined — there are no partial pressures to form

K_{p}

. The bridge formula simply does not apply; use $K_c$ alone.

At the limiting case

R T = 1

(in whatever units), what happens for any

Δ n

(R T)^{Δ n} = 1^{Δ n} = 1

, so

K_{p} = K_{c}

numerically regardless of

Δ n

— a unit-dependent coincidence, not a physical rule.

If a reaction has

Δ n < 0

and

R T > 1

, order

K_{p}

and

K_{c}

(R T)^{Δ n}

is a positive number less than

1

, so

K_{p} < K_{c}

— fewer gas moles on the product side pulls the pressure constant below the concentration constant.

What if some reactant is a gas but a product is a pure liquid at that temperature?

Only the gas is counted; the pure liquid contributes

0

to the gas-mole tally (and drops out of

K

entirely), so it lowers the product count in

Δ n

Can

Δ n

be a fraction?

Yes, if the balanced equation uses fractional coefficients (e.g.

\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3}

gives

Δ n = 1 - 2 = - 1

Δ n

tracks whatever coefficients you wrote — keep the whole equation consistent.

Recall One-line self-check

If you can state, without the note, (1) which species count in $Δ n$ , (2) the sign convention, (3) why the bridge is the ideal gas law, and (4) what happens when $Δ n = 0$ — you own this topic.

Connections

Parent — Hinglish version
Ideal Gas Law PV=nRT — the reason the bridge exists.
Equilibrium Constant Kp — what $K_{p}$ actually measures.
Law of Chemical Equilibrium — home of $K_{c}$ .
Partial Pressure and Dalton's Law — meaning of $p_{X}$ .
Le Chatelier's Principle — why $V$ / $p$ shift position but not $K$ .
Gibbs Free Energy and K — $Δ G^{\circ} = - R T ln K$ .