K_{c}, the molar concentration equilibrium constant

K_{c} is the value of K_{eq} when all reactant and product concentrations are given in concentration units, such as mol/l.

When reactant and product concentrations are given in these units, K_{c} will be the calculated value of equilibrium.

K_{p}, the partial pressure equilibrium constant

K_{p} is the value of K_{eq} when all reactant and product concentrations are given in pressure units, such as atm.

When reactant and product concentrations are given in these units, K_{p} will be the calculated value of equilibrium.

How can the value of K_{p} be calculated, given the value of K_{c}?

The equation relating K_{p} to K_{c} is:

K_{p} = K_{c} (RT)^{Δn}

where:

R = the ideal gas constant in l-atm/K-mol

T = the absolute temperature in K

Δn = the change in moles of gas (moles product gas - moles reactant gas)

At STP, if the number of moles of gas decreases significantly in a reaction, what will be the relationship between K_{p} and K_{c}?

K_{c} will be greater than K_{p}.

The equation relating K_{p} to K_{c} is:

K_{p} = K_{c} (RT)^{Δn}

Since at STP, R*T is greater than 1, and since Δn is negative if number of moles of gas decreases, this gives :

K_{p} = K_{c} (1 / [number greater than 1]), hence K_{p} is a fraction of the size of K_{c}.

At STP, if the number of moles of gas remains the same in a reaction, what will be the relationship between K_{p} and K_{c}?

K_{c} will be equal to K_{p}.

The equation relating Kp to Kc is:

K_{p} = K_{c} (RT)^{Δn}

If the number of moles of gas is unchanged, Δn must equal zero, this gives:

K_{p} = K_{c} ([RT]^{0})=K_{c} * 1,

hence K_{p} = K_{c}.