Equilibrium constants play a pivotal role in understanding the behavior of chemical reactions.
The symbol Kp represents the equilibrium constant (K) of a gaseous reaction in terms of partial pressures (P). It is directly related to the equilibrium constant in terms of concentrations (Kc) by the formula –
⇒ Kp = Kc (RT)∆n.
In this article, you will learn how to find Kp from Kc via plenty of examples, so let’s start. Happy learning!
The general chemical equation for a reversible gas-phase reaction, in which a moles of A react with b molesof B to produce c moles of C and d moles of D, is:
a A(g) + b B(g)⇌ c C(g) + d D(g)
An equilibrium state is achieved when the rate of forward reaction is equal to the rate of backward reaction.
The equilibrium constant (Kp) is defined as the ratio of the product of the partial pressures of Products to the product of the partial pressures of Reactants raised to the number of moles of each in the balanced chemical equation.
Mathematically represented as:
Where;
PA = Partial pressure of A (in Pa or atm)
PB = Partial pressure of B (in Pa or atm)
PC = Partial pressure of C (in Pa or atm)
PD = Partial pressure of D (in Pa or atm)
Partial pressure is defined as the pressure exerted (force/area) by one of the gases in a gaseous mixture.
The total pressure of the mixture of gases is equal to the sum of the partial pressures of each gas.
Pmixture = PA + PB + PC + PD
Therefore, you must keep in mind that equation (i) is only applicable to find the equilibrium constant if the components of a reaction mixture are gaseous in nature.
What is Kc?
For a general chemical reaction a A (g) + b B(g)⇌c C(g) + d D(g), the equilibrium constant (Kc) is defined as the ratio of the product of the molar concentrations of Products to the product of the molar concentrations of Reactants.
Mathematically expressed as:
Where,
[A]= Concentration of A (in mol/L, mol/dm3 or M)
[B] = Concentration of B (in mol/L, mol/dm3 or M)
[C] = Concentration of C (in mol/L, mol/dm3 or M)
[D] = Concentration of D (in mol/L, mol/dm3 or M)
a, b, c, and d are the number of moles of A, B, C, and D in the balanced chemical equation.
How to find Kp from Kc? – (Kc to Kp)
Kc can be used to find Kp as per the formula given below:
Where,
Kp = equilibrium constant in terms of partial pressure
Kc = equilibrium constant in terms of molar concentrations
R= ideal gas constant
T = absolute temperature (measured in Kelvins)
∆n = change in the number of moles of gas
∆n is calculated as follows:
∆n = total number of moles of gas on the product side (nP) – total number of moles of gas on the reactant side (nR)
R = 8.314 J/K.mol if the pressure is measured in Pascals (Pa) while R= 0.08206 L.atm/mol.K if the pressure is given in atm.
There are no fixed units for Kp and Kcas these are calculated based on the components involved on the reactant and product sides, respectively.
Kp and Kc can also be unitless entities in case all the units in the ratio are canceled out completely.
Thus, we can use the balanced chemical equation of a reaction and the values of Kc and T given in the question statement to find unknown Kp.
Let’s explore this concept further through the examples given in the next section.
Solved Examples for finding Kp from Kc?
Example # 1:Find the value of Kp for the reversible reaction shown below, which is taking place at 300 K, given that its Kc value is equal to 668 mol-2dm6.
N2 (g) + 3 H2 (g) ⇌2 NH3 (g)
Solution:
As per the balanced chemical equation,
∆n = 2 – 4 = -2
As per the question statement,
Kc = 668
T = 300 K
Kp =?
We already know that R= 0.08206 L.atm/mol.K
Applying the formula:
⇒ Kp = Kc (RT) ∆n
Substituting all the known values to find unknown Kp:
Kp = (668)[(0.08206)(300)]-2
Kp = 668(24.618)-2
∴ Kp = 1.10
In this example, the units of Kp are:
Kp = (atm)2/(atm)4
Kp = atm-2
Result: The equilibrium constant w.r.t partial pressures (Kp) is 1.10 atm-2.
Example # 2:Kc for the reaction given below is 3.5 x 10-3 mol2.dm-6. Find its Kp at 400 K.
CH4 (g) + H2O (g) ⇌ CO(g) + 3 H2 (g)
Solution:
As per the balanced chemical equation,
∆n = 4 – 2 = 2
As per the question statement,
Kc = 3.5 x 10-3 mol2.dm-6
T = 400 K
Kp = ?
We already know that R= 0.08206 L.atm/mol.K
Applying the formula:
⇒ Kp = Kc (RT) ∆n
Substituting all the known values to find unknown Kp:
Kp = (3.5 x 10-3)[(0.08206)(400)]2
Kp = (3.5 x 10-3)(32.824)2
∴ Kp = 3.77
In this example, the units of Kp are:
Kp = (atm)4/(atm)2
Kp = atm2
Result: The Kp value for the given reaction at 400 K is 3.77 atm2.
Example # 3: Calculate Kp for the following reaction taking place at 333K, given that its Kc value is 6.96 x 10-5 mol.dm-3.
PH3BCl3(s) ⇌ PH3(g) + BCl3 (g)
Solution:
In this reaction, PH3BCl3 is present in the solid state; thus, there is no gaseous reactant. Therefore,
∆n = 2 – 0 = 2
As per the question statement,
Kc = 6.96 x 10-5 mol.dm-3
T = 333 K
Kp =?
We already know that R= 0.08206 L.atm/mol.K
Applying the formula:
⇒ Kp = Kc (RT) ∆n
Substituting all the known values to find unknown Kp:
Kp = (6.96 x 10-5)[(0.08206)(333)]2
Kp = (6.96 x 10-5)(27.32598)2
∴ Kp = 0.052
In this example, the units of Kp are:
Kp = atm2
Result: The Kp value for the given reaction at 400 K is 0.052 atm2.
FAQ
How to find Kp from Kc?
We can find Kp from Kc using the formula given below:
Kp = Kc (RT)∆n
Where,
R = Ideal gas constant
T= Absolute temperature
∆n = Change in the number of moles of the gas
Which Universal constant (R) should we use when converting from Kc to Kp (0.082 or 8.314)?
We should use R = 0.082 L.atm/mol.K if the partial pressures of gaseous reactants and products are measured in atm while the molar concentrations are given in mol/dm3or mol/L.
Contrarily, R = 8.314 J/K.mol in case the pressures are measured in Pa while the concentrations are in mol/m3.
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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