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Calculating K for a Reaction
Using Known K's for Other Reactions

K for a Reversed Reaction

The equilibrium expression written for a reaction written in the reverse direction is the reciprocal of the one for the forward reaction.

K' = 1/K

K' is the constant for the reverse reaction and K is that of the forward reaction.

Example:  What is the value of the equilibrium constant for the reaction 2 NO2(g) double arrows N2O4(g) at 100o C?

N2O4(g) double arrows 2 NO2(g)   Kc = 0.212 @ 100oC?

The desired reaction is the reverse of the reaction for which the Kc is known.  The equilibrium expression is the reciprocal of that given.

K'c = 1/Kc = 1/0.212 = 4.72

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Reaction Coefficients Multiplied by a Number

If the coefficients in a balanced equation are multiplied by a factor, n, the equilibrium expression is raised to the nth power.

K' = (K)n

K' is the constant for the reaction multiplied by n and K is the constant of the original reaction.

Example:  What is the value of the equilibrium constant for the reaction, 2 I2(g) + 2 Cl2(g) double arrows  4 ICl(g)?

I2(g) + Cl2 (g) double arrows  ICl(g)    Kc = 4.54 x 102 @ 25oC?

The desired reaction has been multiplied by 4.  The value of the equilibrium constant will be the 4th power of the given Kc.

K'c = Kc4 = (4.54 x 102)4 = 4.25 x 1010

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Adding Two or More Equations

If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added. 

K' = K1 x K2  . . .

K1 , K2, etc. represent the equilibrium constants for reactions being added together, and K' represents the equilibrium constant for the desired reaction.

Example: Calculate the value of Kc for the reaction:  2 NO(g) + Br2(g) double arrows 2 NOBr (g) using the following information.

2 NO(g) double arrows N2(g) + O2(g)                            Kc1 = 1 x 1030
N2(g) + Br2(g) + O2(g) double arrows 2 NOBr(g)          Kc2 = 2 x 10-27

The two equations can be added to yield the desired equation. The value of Kc for the reaction will be the product of the other two.

K'c = Kc1 x Kc2 = (1 x 1030)(1 x 10-27) = 2 x 103

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Calculations Incorporating Two or More of These Algebraic Manipulations

It is possible to combine more than one of these manipulations.

Example:   Calculate the value of Kc for the reaction:  2 N2O(g) + 3 O2(g) double arrows 2 N2O4(g), using the following information.
 

Equation Equilibrium Constant
2 N2 (g) + O2(g) double arrows 2 N2O(g) Kc = 1.2 x 10-35
N2O4(g) double arrows 2 NO2(g) Kc = 4.6 x 10-3
N2(g) + O2(g) double arrows NO2(g) Kc = 4.1 x 10-9
These three equations can be combined to get the desired reaction.

Checking addition of equations.
2 N2O(g) double arrows 2 N2(g) + O2(g)
Kc = 1/(1.2 x 10-35) =
8.3 x 1034
4 NO2(g) double arrows 2 N2O4(g)
Kc = 1/(4.6 x 10-3)2
4.7 x 104
2 N2(g) + 4 O2(g) double arrows 4 NO2(g)
Kc = (4.1 x 10-9)4
2.8 x 10-34

Kc = (8.3 x 1034)(4.7 x 104)(2.8 x 10-34) = 1.1 x 106

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