## Chem – Molar Concentration and the Equilirium Constant Kc

How do you use concentration values in equilibrium equations?  What is the equilibrium constant Kc?

Now that we are done writing equilibrium equations we can start using them with the molar concentration numbers and determine what numbers or values we have. If a book or teacher mentions an equilibrium problem where you have molar concentrations (Molarity) then they will sometimes call the equilibrium constant Kc (or Keq). C is refering to the molar concentration.  The main point of this section is to take the Molarity or K numbers you get from a question and put them into the equilibrium equation which you learned how to form in the previous sections……… Then you solve for whichever unknown you are left with.

Example: Given the balanced chemical equation below if the concentration of H2SO4 is 4M, H2 is 1M, and O2 is 2M at equilibrium, then what is the equilibrium constant K?

H2SO4(aq) <—–> H2(aq) + S(s) + 2 O2(aq)

 K = [ H2 ] [ O2 ]2 = [ 1 ] [ 2 ]2 = K = 1 [ H2SO4 ] 4

VIDEO Solving Equilibrium Equations (Kc) Demonstrated Example 1: Given the balanced chemical equation below. If the concentration at equilibrium of MgBr2 is 3M, NaI is 4M, MgI2 is 6M and NaBr is 2M at equilibrium. What is the the equilibrium constant K?

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

What is the equilibrium equation look like with chemicals?

 K = [ MgI2 ] [ NaBr ]2 [ MgBr2 ] [ NaI ]2

Replace the chemicals with their corresponding concentrations (numbers).

 K = [ 6M ] [ 2M ]2 [ 3M ] [ 4M ]2

Calculate in steps. First apply the exponents from above.

 K = [ 6M ] [ 4M2] [ 3M ] [ 16M2]

Then multiply top and bottom

 K = [ 24M3 ] [ 48M3 ]

Then divide for the complete answer.

 K = 0.5 1

Some questions don’t ask to solve for K but ask you instead to solve for one of the concentrations. Look at the next example to show you how to deal with those.

VIDEO Solving Equilibrium Equations (Kc) Demonstrated Example 2: Given the balanced chemical equation below. If the concentration at equilibrium of Fe3+ is 3M and Fe2(CO3)3 is 2M. What is the concentration of CO32- if the equilibrium constant is 0.5 M-5?

2 Fe3+(aq) + 3 CO32-(aq) <—-> Fe2(CO3)3(s)

What is the equilibrium equation look like with chemicals? (Remember to leave out the solid)

 K = 1 [ Fe3+ ]2 [ CO32- ]3

Replace the chemicals and equilibrium constant with their corresponding concentrations (numbers).

 0.5 M-5 = 1 [ 3M ]2 [ CO32- ]3

Calculate in steps. First isolate the variable. In this case the [ CO32- ]3

Multiply both sides by [ CO32- ]3

 0.5 M-5 * [ CO32- ]3 = 1 [ CO32- ]3 [ 3M ]2 [ CO32- ]3

Cross out like terms.

 0.5 M-5 * [ CO32- ]3 = 1 [ 3 CO32- ]3 [ 3M ]2 [ CO32- ]3

Then we simplify.

 0.5 M-5 * [ CO32- ]3 = 1 [ 3M ]2

Divide both sides by 0.5 M-5

 0.5 M-5 * [ CO32- ]3 = 1 0.5 M-5 [ 3M ]2 * 0.5 M-5

Cross out like terms.

 0.5 M-5 * [ CO32- ]3 = 1 0.5 M-5 [ 3M ]2 * 0.5 M-5

Simplify

 [ CO32- ]3 = 1 [ 3M ]2 * 0.5 M-5

Calculate starting with applying the exponents from above.

 [ CO32- ]3 = 1 [ 9M2] * 0.5 M-5

Multiply the bottom right.

 [ CO32- ]3 = 1 4.5 M-3

Divide 1 by 4.5 = 0.222

 [ CO32- ]3 = 0.222 M-3

Rotate the units of Molarity to the top by making the exponent positive on top.

 [ CO32- ]3 = 0.222 M3 1

Take the cube (3) root of both sides.

 [ CO32- ]= 0.6 M 1

COMPLETE ANSWER: [ CO32- ]= 0.6 M

PRACTICE PROBLEMS: Give the missing equilibrium constant (K) or concentration below.

What is the equilibrium constant of the equation below at equilibrium when the concentration of C4 is 4M, O2 is 0.5, and CO2 is 2M at equilibrium?

C4(s) + 4 O2(g) <—-> 4 CO2(g)

What is the concentration of O2 from the equation below if at equilibrium when the concentration of OF2 is 0.5M, NH3 is 0.25M, N2F4 is 10M, and H2 is 8M at equilibrium when the equilibrium constant is 6 * 103 M?

2 OF2(g) + 2 NH3(g) <—> N2F4(g) + O2(g) + 3 H2(g)

Answer: K = 1.8 * 10-2 M

What is the equilibrium constant of the equation below if at equilibrium the concentration of Ag+ is 2M, Ca3(PO4)2 is 4M, Ca2+ is 0.03M, and Ag3PO4 is 0.05M at equilibrium?

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

Answer: K = 4.22 * 10-7 M-3

What is the concentration of H2O from the equation below if at equilibrium the H2 is 0.02M and the O2 is 0.008M and the equilibrium constant is 0.015M?

2 H2O(g) <—-> 2 H2(g) + O2(g)  