Scientific Understanding

**How do you convert between Kc and Kp?
**

Another problem you might come across is how to convert between Kp and Kc. That is to say how to convert between the equilibrium constant from pressure units and the equilibrium constant from concentration units. The formula is below.

Kp = Kc (RT)^{Δ}^{ n}

Above the symbols represent:

Kp is the equilibrium constant for pressure

Kc is the equilibrium constant for molar concentration

R is our old friend the gas constant from the Ideal gas law section . R = 0.0821 when using atm as pressure units.

T is temperature in Kelvin.

Δ n is the change in moles. This the exponent of R and T multiplied together. If you need to know how to calculate Δ n then go back to the section on Calculating Δ n (delta n).

**VIDEO Converting Between Kc and Kp Demonstrated Example 1**: If the Kc for the chemical equation below is 0.5 at a temperature of 300K, then what is the Kp?

2 OF_{2(g)} + 2 NH_{3(g)} <—> N_{2}F_{4(g)} + O_{2(g)} + 3 H_{2(g)}

What information does the problem give you?

Answer:

Kc = 0.5

T = 300K

R = 0.0821 (you should always have this constant)

What is the Δ n for this chemical equation?

Answer: 5 – 4 = 1

Set up the math equation.

Kp = | Kc * (R* T)^{Δ}^{ n} |

1 |

Put in all the values (numbers).

Kp = | 0.5 * (0.0821 * 300K)^{1} |

1 |

Multiply what is in the parenthesis.

Kp = | 0.5 * (24.63)^{1} |

1 |

Distribute the exponent inside the parenthesis.

Kp = | 0.5 * 24.63 |

1 |

Multiply the rest together.

Kp = | 12.3 |

1 |

COMPLETE ANSWER: Kp = 12.3

**VIDEO Converting Between Kc and Kp Demonstrated Example 2**: If the Kp for the chemical equation below is 8.7 * 10^{-3} at a temperature of 280K, then what is the Kc?

2 C_{4}H_{10}_{(l)} + 13 O_{2}_{(g)} <—-> 8 CO_{2}_{(g)} + 10 H_{2}O_{(g)}

What information does the problem give you?

Answer:

Kp = 8.7 * 10^{-3}

T = 280K

R = 0.0821 (you should always have this constant)

What is the Δ n for this chemical equation?

Answer: 18 – 13 = 5 don’t count the liquid

Set up the math equation.

Kp = | Kc * (R* T)^{Δ}^{ n} |

1 |

Put in all the values (numbers).

8.7 * 10^{-3} = |
Kc * (0.0821 * 280K)^{5} |

1 |

Multiply what is in the parenthesis.

8.7 * 10^{-3} = |
Kc * (23)^{5} |

1 |

Distribute the exponent inside the parenthesis.

8.7 * 10^{-3} = |
Kc * (6.44 * 10^{6}) |

1 |

Divide both sides by 6.44 * 10^{6}

8.7 * 10^{-3} = |
Kc * (6.44 * 10^{6}) |

6.44 * 10^{6} |
6.44 * 10^{6} |

Cancel out the right side.

8.7 * 10^{-3} = |
Kc * (6.44 * 10^{6}) |

6.44 * 10^{6} |
6.44 * 10^{6} |

Simplify

8.7 * 10^{-3} = |
Kc |

6.44 * 10^{6} |

Divide

1.35 * 10^{-9} = |
Kc |

1 |

COMPLETE ANSWER: Kc = 1.35 * 10^{-9}

**PRACTICE PROBLEMS**: Solve the question below involving Kp and Kc.

If the Kc for the chemical equation below is 25 at a temperature of 400K, then what is the Kp?

N_{2(g)} + 3 H_{2(g)} <—-> 2 NH_{3(g)}

Answer: Kp = 9.05 * 10^{-4}

If the Kc for the chemical equation below is 6.9 * 10^{-4} at a temperature of 250K, then what is the Kp?

2 N_{2}O_{3(g)} <——> 2 N_{2(g)} + 3 O_{2(g)}

Answer: Kp = 8.6 * 10^{3}

If the Kp for the chemical equation below is 5.7 * 10^{-2} at a temperature of 273K, then what is the Kc?

Ba(OH)_{2}_{(aq)} + 2 HCl_{(aq)} <—-> 2 H_{2}O_{(l)} + BaCl_{2}_{(aq)}

Answer: Kc =5.7 * 10^{-2}

If the Kp for the chemical equation below is 3.2 * 10^{4} at a temperature of 120C, then what is the Kc?

6 C_{4}_{(s)} + 12 O_{2}_{(g)} + 24 H_{2}_{(g)} <—-> 4 C_{6}H_{12}O_{6}_{(s)}

Answer: Kc = 6.6 * 10^{58}