Hess’s Law Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy Changes.

Hess’s Law the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps.

Combustion of methane ∆H1 shows the magnitude of the single step reaction. ∆H2 and ∆H3 when combined are equivalent to ∆H1, demonstrating Hess’s Law.

Most reactions occur in more than one step or could hypothetically occur in more than one step. We focus on the OVERALL reaction based on the balanced chemical equation.

Goal: Achieve the Target equation by manipulating several sub-equations. The enthalpy change of the target equation will be the sum total of the enthalpy changes for each sub-equation.

Rule #1 If you reverse an equation the ∆H must switch signs. Na (s) + ½ Cl2(g) NaCl (s) ∆H = -411 kJ NaCl (s)  Na (s) + ½ Cl2(g) ∆H = 411 kJ

Rule #2 If you need to multiply or divide any part of the equation, the whole equation must follow the same operation, including ∆H. Na (s) + ½ Cl2(g) NaCl (s) ∆H = -411 kJ 2 Na (s) + Cl2(g) 2 NaCl (s) ∆H = -822 kJ

Practice A + B  AB ∆H1 = 20 kJ AB + B  AB2 ∆H2 = 50 kJ What is the ∆H for the overall reaction A + 2 B  AB2 ∆H3 = ? ∆H3 = 70 kJ

For example, suppose you are given the following data: Could you use these data to obtain the enthalpy change for the following reaction? 3

If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

Example Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ Calculate DHº for this reaction

Practice:

Practice: