## Cracking the AP Chemistry Exam

**Part IV**

**Content Review for the AP Chemistry Exam**

**Chapter 7**

**Big Idea #5: Laws of Thermodynamics and Changes in Matter**

**HESS’S LAW**

Hess’s law states that if a reaction can be described as a series of steps, then ∆*H* for the overall reaction is simply the sum of the ∆*H* values for all the steps.

When manipulating equations for use in enthalpy calculations, there are three rules:

1. If you flip the equation, flip the sign on the enthalpy value

2. If you multiply or divide an equation by an integer, also multiply/divide the enthalpy value by that same integer

3. If several equations, when summed up, create a new equation, you can also add the enthalpy values of those component equations to get the enthalpy value for the new equation

For example, let’s say you want to calculate the enthalpy change for the following reaction:

4NH_{3}(*g*) + 5O_{2}(*g*) → 4NO(*g*) + 6H_{2}O(*g*)

Given:

N |
∆ |

N |
∆ |

2H |
∆ |

First, we want to make sure all of the equations have the products and reactants on the correct side. A quick glance shows us the NH_{3} should end up on the left but is given to us on the right, so the second equation needs to be flipped. All other species appear to be on the side they need to be on, so we’ll leave the other two reactions alone.

Next, we’re going to see what coefficients we need to get to. The NO needs to have a 4 but is only 2 in equation 1, so we’ll multiply that equation by two. The NH_{3} also needs to have a 4 but is only 2 in equation 2, so we’ll also multiply that equation by two. Finally, the H_{2}O needs to have a 6 but only has a 2 in equation three, so the last equation gets multiplied by three. We now have:

2N |
∆ |

4NH |
∆ |

6H |
∆ |

Fortunately, the two O_{2} on the products side add up to 5, and both the N_{2} and H_{2} cancel out completely, giving us the final equation that we wanted. So, our final value for the enthalpy will be 361.2 + 183.6 − 1451 = −906 kJ/mol.