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Wize University Chemistry Textbook > Equilibrium

Kc vs Kp

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Solving for Kc vs Kp

The equilibrium constants Kc and Kp are related by the following equation:

Kp=Kc(RT)Δn\boxed{K_p=K_c(RT)^{\Delta n}}

Kp is the equilibrium constant we can calculate when given partial pressures of products and reactants
Kc is the equilibrium constant we can calculate when given concentrations of products and reactants
R is the ideal gas constant (0.0821L atm/mol K)
T is temperature in K
Δn =moles of gas on products side - moles of gas on reactants side (use coefficients in equation)


Example:
In the synthesis of ammonia from nitrogen and hydrogen, N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g), Kc = 9.60 at 300oC. Calculate Kp for this reaction at this temperature. R=0.0821 Latm mol-1K-1



Δ\Deltan = 2 - (1 + 3) = 2 - 4 = -2
Kp = Kc (RT)Δn
= (9.60) ((0.0821L atm mol−1 K−1)(573 K))−2\left((0.0821L\ atm\ mol^{-1}\ K^{-1})(573\ K)\right)^{-2}
Kp = 4.34 ×\times 10-3

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Example:
And now what if we are told that the pressure at equilibrium for H2(g) is 3x10-1 atm and N2(g) is 2.5x10-2atm. What is the pressure of NH3 at equilibrium?


Kp=(PNH3 )2(PN2)(PH2)3\frac{\left(PNH_3\ \right)^2}{\left(PN_2\right)\left(PH_2^{ }\right)^3}

4.23x10−3=(PNH3)2(2.5x10−2)(3x10−1)34.23x10^{-3}=\frac{\left(PNH_3\right)^2}{\left(2.5x10^{-2}\right)\left(3x10^{-1}\right)^3}

4.34x10-3 (6.75x10-4)=[NH3]2

[NH3]=1.7x10-3 atm
Extra Practice
Calculating Kc from Kp Practice Problem
At 25oC, the Kp for the reaction N2O4(g) ⇌ 2 NO2(g) is 0.31, what is the value of Kc?
Consider the reaction of nitrogen and oxygen gases to produce nitric oxide:
N2(g)+O2(g)⇌2NO(g)                   Kc=4.0×10−4  at  1750 oCN_2(g)+O_2(g) \rightleftharpoons 2NO(g)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_c=4.0\times 10^{-4}\ \ at\ \ 1750 \ ^oC
What is the value of Kp for this reaction at 1750 °C?
Equilibrium constants (Kc vs Kp)
At 25oC, the Kp for the reaction N2O4 (g) ⇌ 2 NO2 (g) is 0.31. What is the value of Kc?
For which of the following reactions are Kp and Kc equal?
At 25oC, the Kp for the reaction N2O4 (g) ⇌ 2 NO2 (g) is 0.31. What is the value of Kc?
Calculating Kc from Kp Practice Problem
At 25oC, the Kp for the reaction N2O4(g) ⇌ 2 NO2(g) is 0.31, what is the value of Kc?
The reaction for the reforming of methane from steam is given by:
CH4(g)+H2O(g)⇌CO(g)+3H2(g)CH_4(g)+H_2O(g)⇌CO(g)+3H_2(g)
A closed reaction vessel starts with initial concentrations of 1.00 M for every gas. After reaching equilibrium at a particular temperature, the final CO concentration was measured to be 0.80 M. Determine KC for this reaction at this temperature.