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Calculating equilibrium constant Kp using partial pressures

Definition of equilibrium constant Kp for gas phase reactions, and how to calculate Kp from Kc. 

Key points

  • The equilibrium constant, Kp, describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
  • For a gas-phase reaction, aA(g)+bB(g)cC(g)+dD(g), the expression for Kp is
  • Kp is related to the equilibrium constant in terms of molar concentration, Kc, by the equation below:
where Δn is
Δn=mol of product gasmol of reactant gas

Introduction: a short review of equilibrium and Kc

When a reaction is at equilibrium, the forward reaction and reverse reaction have the same rate. The concentrations of the reaction components stay constant at equilibrium, even though the forward and backward reactions are still occurring.
Why penguins, you ask? Keep reading!! Photo credit: Wikimedia Commons, CC BY-SA 3.0.
Equilibrium constants are used to define the ratio of concentrations at equilibrium for a reaction at a certain temperature. In general, we use the symbol K or Kc to represent equilibrium constants. When we use the symbol Kc, the subscript c means that all concentrations are being expressed in terms of molar concentration, or mol soluteL of solution.

Kp vs. Kc: using partial pressure instead of concentration

When a reaction component is a gas, we can also express the amount of that chemical at equilibrium in terms of its partial pressure. When the equilibrium constant is written with the gases in terms of partial pressure, the equilibrium constant is written as the symbol Kp. The subscript p stands for penguins.
For example, let's say we have the generic balanced gas-phase reaction below:
In this equation, a moles of reactant A react with b moles of reactant B to make c moles of product C and d moles of product D.
If we know the partial pressures for each component at equilibrium, where the partial pressure of A(g) is abbreviated as PA, then the expression for Kp for this reaction is
Remember the following important points when calculating Kp:
  • Make sure the reaction is balanced! Otherwise, the stoichiometric coefficients and the exponents in the equilibrium constant will be incorrect.
  • Pure liquids or solids have a concentration of 1 in the equilibrium expression. This is the same as when calculating Kc.
  • Kp is often written without units. Since the value of Kp depends on the units used for the partial pressure, you will need to check the pressure units used in your textbook when solving a Kp problem.
  • All the partial pressures used for calculating Kp should have the same units.
  • We can write Kp for reactions that include solids and pure liquids since they do not appear in the equilibrium expression.

Converting between gas concentration and partial pressure

Close up of someone pouring light brown soda from a can into a glass. The bubbles coming out of the soda create a thick layer of froth at the top of the glass.
Soda is pressurized with carbon dioxide, which is slightly soluble in the soda liquid. When the can is opened, the gas partial pressure above the liquid surface decreases, which causes the dissolved carbon dioxide to go from the aqueous to the gas phase. Therefore, bubbles! Photo credit: Marnav Sharma, CC BY 2.0
We can convert between gas concentration—in units of M or molL—and partial pressure using the ideal gas equation. Since molar concentration is the number of moles of gas per volume, or nV, we can rearrange the ideal gas equation to get the relationship between P and nV as follows:
PV=nRT         Divide both sides by V.P=(nV)RT
We can use this relationship to derive an equation to convert directly between Kc and Kp at temperature T, where R is the gas constant:
The symbol Δn is the number of moles of gas on the product side minus the number of moles of gas on the reactant side in the balanced reaction:
Δn=mol of product gasmol of reactant gas
Let's practice using these equations in some examples!

Example 1: finding Kp from partial pressures

Let's try finding Kp for the following gas-phase reaction:
We know the partial pressures for each component at equilibrium for some temperature T:
At temperature T, what is Kp for this reaction?
First we can write the Kp expression for our balanced equation:
We can now solve for Kp by plugging in the equilibrium partial pressures in the equilibrium expression:

Example 2: finding Kp from Kc

Now let's look at a different reversible reaction:
If Kc for this reaction is 4.5×104 at 400K, what is the equilibrium constant, Kp, at the same temperature?
Use the gas constant that will give Kp for partial pressure units of bar.
To solve this problem, we can use the relationship between the two equilibrium constants:
To find Δn, we compare the moles of gas from the product side of the reaction with the moles of gas on the reactant side:
Δn=mol of product gasmol of reactant gas=2mol NH3(1mol N2+3mol H2)=2mol gas
We can now substitute in our values for Kc, T, and Δn to find Kp. We will want to keep track of the units of the gas constant R in our equation since that will determine if we are calculating Kp for partial pressures of bar or atm. Since we want to calculate Kp for when partial pressure has units of bar, we will use R=0.08314LbarKmol.
Note that if we had used a gas constant defined in terms of atm, we would have gotten a different value for Kp.

Example 3: find Kp from total pressure

Finally, let's consider the equilibrium reaction for the decomposition of water:
Assume that initially there is no hydrogen or oxygen gas present. As the reaction proceeds to equilibrium, however, the total pressure increases by 2.10atm.
Based on this information, what is Kp for the reaction?
To do this problem, it might be helpful to visualize our partial pressures using an ICE table.
Note that we don't include pure liquids in our calculations for Kp; the table only includes partial pressure information for the two gaseous products. Since initially there are no products in our system, we can fill in the first row of our table with zeros.
Next, we look at the balanced equation to describe how the partial pressures change when the reaction reaches equilibrium. Based on the stoichiometric coefficients, we know that if the value for PO2 increases by x, the change for PH2 will be twice that much, 2x. The third row in the table sums up the expressions in the first two rows to describe the partial pressures at equilibrium.
At this point, Dalton's Law can help us solve for x. We know from Dalton's Law that the total pressure of a system, Ptotal, is equal to the sum of the partial pressures for each of the components in the system:
Using our equilibrium values, we can express the total pressure for our reaction as follows:
Using our observed total pressure of 2.10atm, we can solve for x:
By substituting in 0.70atm for x in the last row of our ICE table, we can now find the equilibrium partial pressures for the two gases:
Now we can set up an equilibrium expression for the reaction and solve for Kp:


  • The equilibrium constant Kp describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
  • For a gas-phase reaction aA(g)+bB(g)cC(g)+dD(g), the expression for Kp is
  • Kp is related to the equilibrium constant in terms of molar concentration, Kc, by the equation below
where Δn is
Δn=mol of product gasmol of reactant gas

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