Thermochemistry Lesson # 4: Hess’s Law

Enthalpy in Chemical Reactions From experimental evidence we have determined that the change of enthalpy in chemical reactions is independent of the path taken. This means that whether a reaction occurs in one step or five, if the products are the same in the end, the enthalpy change will be the same overall.

Nitrogen Dioxide Example For example, we have determined that: N2 (g) + 2 O2 (g) → 2 NO2 (g) ΔH = 68 kJ. But we can do this reaction in two steps as well: N2 (g) + O2 (g) → 2 NO (g) ΔH = 180 kJ 2 NO (g) + O2 (g) → 2 NO2 (g) ΔH = -112 kJ If we add these two enthalpies together (ΔH = 180 kJ + ΔH = -112 kJ) we get the same value of 68 kJ.

Potential Energy Diagram

Hess’s Law The enthalpy change for the conversion of reactants to products is the same whether the conversion occurs in one step or several steps. Rules for using Hess’s Law: If you reverse a chemical reaction, you must also reverse the sign of ΔH. The magnitude of ΔH id directly proportional to the number of moles of reactants and products in a reaction. If the coefficients in a balanced equation are multiplied by a factor, the value of ΔH is multiplied by the same factor.

Example 1 Graphite and diamond are two forms of solid carbon. Use the data below to calculate the enthalpy change for the conversion of graphite into diamond. Cgraphite (s) → Cdiamond (s) GIVEN: Cgraphite (s) + O2 (g) → CO2 (g) ΔH = -394 kJ Cdiamond (s) + O2 (g) → CO2 (g) ΔH = -396 kJ

Example 2 Ethane gas dehydrogenates to ethene and hydrogen: C2H6 (g) → C2H4 (g) + H2 (g). Determine the change in enthalpy for this reaction using the following thermochemical equations: C2H6 (g) + 3.5 O2 (g) → 2 CO2 (g) + 3 H2O (l) ΔH = -1559 kJ C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ΔH = -1411 kJ 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = -572 kJ