1.1 What is Chemistry?

http://www.ck12.org Chapter 19. Chemical Equilibrium

P= (RT)(Vn)

What is the significance of this? Take a look at the fraction on the right. We are dividing the amount of gas (in
moles) by the volume (in liters). The fraction n/V is simply the concentration of the gas. At a given temperature,
the partial pressure of any gas is directly proportional to its concentration. Because of this, we can write an alternate
equilibrium constant expression in which partial pressures are used in place of concentrations. The pressure-based
equilibrium constant, indicated by the abbreviation Kp, can be written as follows for a generic reaction in which all
components are present in the gas state:

aA(g)+bB(g)cC(g)+dD(g)

Kp=

(PC)c(PD)d

(PA)a(PB)b

For reactions that include solids or liquids in addition to gases, those components are left out of the equilibrium
constant expression, just as they are for a concentration-based equilibrium constant. For reactions that include both
gases and aqueous components, the Kpform of the equilibrium constant is generally not used.

Example Problem 19.3

Write the pressure-based equilibrium constant expression (Kp) for the following reaction:

2O 3 (g)3O 2 (g)

Answer:

The process here is exactly the same as for a normal equilibrium constant expression, except that we are using
partial pressures instead of concentrations. Reactants go in the denominator, and products go in the numerator. Each
component is raised to an exponent that is equal to its coefficient in the balanced equation.

Kp=

(PO 2 )^3

(PO 3 )^2

Converting Between Kcand Kp

For a gaseous substance, the conversion factor between concentration and partial pressure can be derived from the
ideal gas law:

PAV=nART

PA= (

nA

V

)(RT)

PA= A

where [A] is the molar concentration of A. Let’s look again at the generic expression for Kp:

aA(g)+bB(g)cC(g)+dD(g)

Kp=

(PC)c(PD)d

(PA)a(PB)b

If we substitute in concentrations for each of the partial pressures, we would get the following: