## Chem – College: The Relationship Between Different Equilibrium Constants

What is the relationship between different equilibrium constants?

Within the same chemical equation, questions can be asked about what happens to the equilibrium constant if we change something to that equation. There are two classic examples that pretty well illustrate the point of how it works.

The first is what happens when you multiply the chemical equation by a number like 2. Another way to say that is we double all of the coefficients in the chemical equation. If that happens then the equilibrium constant (K value) is taken to the second power (K2).

The second classic example is what happens when we flip the chemical equation over (reactants become products and products become reactants).  In that case the equilibrium constant (K value) becomes 1 divided by K…(1/K).

From there you can mix and match any combination of multiplying or flipping chemical equation over (reversing) to create problems of how K would be different.

NOTE:  The concept we just talked about above is different than changing the concentrations of a chemical equation. If you change the concentrations of a chemical equation by doubling all of them then you will double the equilibrium constant. If you change the concentrations of a chemical equation by dividing all of them by 2 then you will half the equilibrium constant.

Example 1: Solve for the new equilibrium constant given the changing conditions.

If the equilibrium constant for the chemical equation below is K = 3M2.

N2(g) + 3 H2(g) <—-> 2 NH3(g)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

2 N2(g) + 6 H2(g) <—-> 4 NH3(g)

Answer: K2 = (3M2)2 = 9M4

Example 2: Solve for the new equilibrium constant given the changing conditions.

If the equilibrium constant for the chemical equation below is K = 3M2

N2(g) + 3 H2(g) <—-> 2 NH3(g)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

2 NH3(g) <—-> N2(g) + 3 H2(g)

Answer: 1/K = 1/3M2 = 0.33M-2

PRACTICE PROBLEMS: Solve for how the equilibrium constant changes when the reaction changes.

If the equilibrium constant for the chemical equation below is K = 1.2 * 102M

2 H2O(g) <—-> 2 H2(g) + O2(g)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

2 H2(g) + O2(g) <—-> 2 H2O(g)

Answer: 1/K = 1 / (1.2 * 102M) = 8.3 * 10-3M-1

If the equilibrium constant for the chemical equation below is K = 0.6

MgBr2(aq) + 2 NaI(aq) <——> MgI2(aq) + 2 NaBr(aq)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

2 MgBr2(aq) + 4 NaI(aq) <——> 2 MgI2(aq) + 4 NaBr(aq)

Answer: K2 = (0.6)2 = 0.36

If the equilibrium constant for the chemical equation below is K = 2.5 * 10-3

C4(s) + 4 O2(g) <—-> 4 CO2(g)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

8 CO2(g) <—-> 2 C4(s) + 8 O2(g)

Answer: 1/K2 = 1 / (2.5 * 10-3)2 = 1.6 * 105

If the equilibrium constant for the chemical equation below is K = 9M3

6 Ag+(s) + Ca3(PO4)2(s) <—-> 3 Ca2+(aq) + 2 Ag3PO4(s)

Then what is the equilibrium constant for the next chemical equation at the same temperature?

18 Ag+(s) + 3 Ca3(PO4)2(s) lt;—-gt; 9 Ca2+(aq) + 6 Ag3PO4(s)

Answer: K3 = (9M3)3 = 729M9  