Scientific Understanding

**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 H_{2}SO_{4 }is 4M, H_{2} is 1M, and O_{2} is 2M at equilibrium, then what is the equilibrium constant K?

H_{2}SO_{4(aq)} <—–> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)}

K = | [ H_{2} ] [ O_{2} ]^{2} = |
[ 1 ] [ 2 ]^{2} = |
K = 1 |

[ H_{2}SO_{4} ] |
4 |

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

MgBr_{2(aq)} + 2 NaI_{(aq)} <——> MgI_{2(aq)} + 2 NaBr_{(aq)}

What is the equilibrium equation look like with chemicals?

Answer:

K = | [ MgI_{2} ] [ NaBr ]^{2} |

[ MgBr_{2} ] [ 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 ] [ 4M^{2}] |

[ 3M ] [ 16M^{2}] |

Then multiply top and bottom

K = | [ 24M^{3} ] |

[ 48M^{3} ] |

Then divide for the complete answer.

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 Fe^{3+} is 3M and Fe_{2}(CO_{3})_{3} is 2M. What is the concentration of CO_{3}^{2-} if the equilibrium constant is 0.5 M^{-5}?

2 Fe^{3+}_{(aq)} + 3 CO_{3}^{2-}_{(aq)} <—-> Fe_{2}(CO_{3})_{3(s)}

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

Answer:

K = | 1 |

[ Fe^{3+} ]^{2} [ CO_{3}^{2-} ]^{3} |

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

0.5 M^{-5} = |
1 |

[ 3M ]^{2} [ CO_{3}^{2-} ]^{3} |

Calculate in steps. First isolate the variable. In this case the [ CO_{3}^{2-} ]^{3}

Multiply both sides by [ CO_{3}^{2-} ]^{3}

0.5 M^{-5} * [ CO_{3}^{2-} ]^{3 }= |
1 [ CO_{3}^{2-} ]^{3} |

[ 3M ]^{2} [ CO_{3}^{2-} ]^{3} |

Cross out like terms.

0.5 M^{-5} * [ CO_{3}^{2-} ]^{3 }= |
1 [ 3 CO_{3}^{2-} ]^{3} |

[ 3M ]^{2} [ CO_{3}^{2-} ]^{3} |

Then we simplify.

0.5 M^{-5} * [ CO_{3}^{2-} ]^{3 }= |
1 |

[ 3M ]^{2} |

Divide both sides by 0.5 M^{-5}

0.5 M^{-5} * [ CO_{3}^{2-} ]^{3 }= |
1 |

0.5 M^{-5} |
[ 3M ]^{2} * 0.5 M^{-5} |

Cross out like terms.

0.5 M^{-5} * [ CO_{3}^{2-} ]^{3 }= |
1 |

0.5 M^{-5} |
[ 3M ]^{2} * 0.5 M^{-5} |

Simplify

[ CO_{3}^{2-} ]^{3 }= |
1 |

[ 3M ]^{2} * 0.5 M^{-5} |

Calculate starting with applying the exponents from above.

[ CO_{3}^{2-} ]^{3 }= |
1 |

[ 9M^{2}] * 0.5 M^{-5} |

Multiply the bottom right.

[ CO_{3}^{2-} ]^{3 }= |
1 |

4.5 M^{-3} |

Divide 1 by 4.5 = 0.222

[ CO_{3}^{2-} ]^{3 }= |
0.222 |

M^{-3} |

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

[ CO_{3}^{2-} ]^{3 }= |
0.222 M^{3} |

1 |

Take the cube (3) root of both sides.

[ CO_{3}^{2-} ]= |
0.6 M |

1 |

COMPLETE ANSWER: [ CO_{3}^{2-} ]= 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 C_{4} is 4M, O_{2} is 0.5, and CO_{2} is 2M at equilibrium?

C_{4(s)} + 4 O_{2(g)} <—-> 4 CO_{2(g)}

Answer: K = 256

What is the concentration of O_{2} from the equation below if at equilibrium when the concentration of OF_{2} is 0.5M, NH_{3} is 0.25M, N_{2}F_{4} is 10M, and H_{2} is 8M at equilibrium when the equilibrium constant is 6 * 10^{3} M?

2 OF_{2(g)} + 2 NH_{3(g)} <—> N_{2}F_{4(g)} + O_{2(g)} + 3 H_{2(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, Ca_{3}(PO_{4})_{2} is 4M, Ca^{2}^{+} is 0.03M, and Ag_{3}PO_{4} is 0.05M at equilibrium?

6 Ag^{+}_{(aq)} + Ca_{3}(PO_{4})_{2}_{(s)} <—-> 3 Ca^{2}^{+}_{(aq)} + 2 Ag_{3}PO_{4}_{(s)}

Answer: K = 4.22 * 10^{-7} M^{-3}

What is the concentration of H_{2}O from the equation below if at equilibrium the H_{2} is 0.02M and the O_{2} is 0.008M and the equilibrium constant is 0.015M?

2 H_{2}O_{(g)} <—-> 2 H_{2(g)} + O_{2(g)}

Answer: K = 0.0142 M