1. Hess’ Law

2. Many reactions can occur by many alternative routes. Hess' Law states: The enthalpy change for a reaction depends only on the energy of the reactants and final products. The enthalpy change in a reaction does not depend on the reaction pathway.

3. Example The combustion of ethane can occur in a single step [T] OR by a series of steps 1,2 and 3. The overall enthalpy change is the same for both T and the sum of steps1,2 and 3

4. Using the Energy Level Diagram 1) What is ∆H° T ? 2) Find the sum of the energy changes for steps 1, 2, and 3.

5. Using the Energy Level Diagram 1) What is ∆H° T ? -1560.4 kJ 2) Find the sum of the energy changes for steps 1, 2, and 3. ∆ H° 1 + ∆ H° 2 + ∆ H° 3 = 136.2 + (-1410.8) + (-285.8) = -1560.4 kJ

6. STRATEGY FOR SOLVING HESS' LAW PROBLEMS Compare the given equations with the target (overall) equation. Manipulate the given equations so that they add to give the target equation. Remember to do the same manipulation to the  H's for each of the given equations.

7. Possible Manipulations: A) Reverse a given equation so that the products become reactants and the reactants become the products. When you reverse an equation, you need to change the sign on the  H for that equation.

8. Possible Manipulations: B) Multiply ALL the coefficients in a given equation by the same number (integer or fraction) to obtain the same coefficients in the target equation. When you multiply an equation, you need to multiply the  H by the same number.

9. After completing the necessary manipulations: Collect all the reactants on the left side and all the products on the right side. Cancel substances that appear on both sides. Be careful… pay attention to the states (s, l, g, or aq). If they are different states, you cannot cancel them.

10. Ensure that you have obtained the target equation. Add the manipulated  H's for the combined equations to obtain the target  H.

11. Example Use Hess’ Law to find the  H for this reaction using the given equations: 4 NH 3 (g) + 7 O 2 (g)  4 NO 2 (g) + 6H 2 O (g) Given: 1) 2 N 2 (g) + 4 O 2 (g)  4 NO 2 (g)  H 1 = 8.1 kcal 2) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (g)  H 2 = -57.8 kcal 3) 2 N 2 (g) + 6 H 2 (g)  4 NH 3 (g)  H 3 = -11.0 kcal