Change in Enthalpy Unit 11

Learning Objective You should be able to define and calculate the enthalpy of a reaction. You should be able to identify whether a reaction is endothermic or exothermic from a chemical reaction or graph.

Chemical Changes Chemical reactions involve the breaking of existing bonds (reactants) and the formation of new bonds (products). Breaking of bonds requires energy. Formation of bonds releases energy. CH4 + Cl2  CH3Cl + HCl + Cl-Cl H-Cl +

Enthalpy Measure of the heat content of the system The change in enthalpy for a reaction lets us know if energy is being taken in or if energy is being released.

Endothermic Reactions Reaction consumes energy Energy is listed on the reactant (left) side of the reaction ΔH is a positive value Products have more enthalpy (energy) than the reactants

Endothermic Reactions The endothermic reaction absorbs heat; energy is a reactant 6

Exothermic Reactions Reaction releases energy Energy is listed on the product (right) side of the reaction ΔH is a negative value Reactants have more enthalpy (heat) than the products

Exothermic Reactions The exothermic reaction releases heat; energy is a product 8

Changes in Enthalpy ΔH To measure and study energy changes in reactions, chemists defined a property called enthalpy (H) Enthalpy is a measure of heat energy of a system at constant pressure It can’t be measured directly, we measure the change in enthalpy. ΔHrxn FIX THIS SLIDE IT IS UGLY

2 NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) Find the energy required to react 1mole of NaHCO3

2 NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) 1.Find the energy required to produce ½ mole of CO2(g) ? 2.Find the energy required to produce 1/3 mole of H2O(g) ?

Enthalpy Question For the decomposition of hydrogen peroxide, it is known that: H2O2(l) → H2O(l) + 1/2 O2(g) ΔH = -98.2 kJ Using this information, determine ΔH for the reaction: 2 H2O(l) + O2(g) → 2 H2O2(l)

Find ΔH for each given the pair of equations CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ ½ Ca(OH)2(s)  ½ CaO(s) + ½ H2O(l)

Calculating Changes in Enthalpy Change in enthalpy is called heat of reaction ΔHrxn ΔHrxn = ΣHf(products) - ΣHf(reactants) “Δ Hf0” is called standard heat of formation - the change in enthalpy from the formation of 1 mole of a compound from its elements This information will come from a table “Δ” is the letter delta – it means “change” “Σ” is the letter sigma – it means “sum” “ o ” refers to standard temperature and pressure (STP) Units: kJ/mol FIX THIS SLIDE IT IS UGLY

Change in Enthalpy All elements in their standard states have an enthalpy of zero Because there is no change involved in their formation Endothermic When N2 and O2 combine to form NO2, the ΔH = +33.2 kJ/mol Exothermic When S and O2 combine to form SO3, the ΔH = -396 kJ/mol.

Calculate ΔH for the following reaction: 8 Al(s) + 3 Fe3O4(s) --> 4 Al2O3(s) + 9 Fe(s)

8 Al(s) + 3 Fe3O4(s) --> 4 Al2O3(s) + 9 Fe(s) ΔH = Σ ΔHf products - Σ ΔHf reactants ΔH = sum of ΔHf products – sum of ΔHf reactants : ΔH = (0) + 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s) + (0)

The values for ΔHf may be found in the Heats of Formation of Compounds table. Plugging in these numbers: ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ) ΔH = -3316.5 kJ

Enthalpy Change Problem Use standard heat of formation to calculate ΔHrxn for the combustion of methane. Then, decide if the reaction is exothermic or endothermic. Molecule Standard Heat of Formation ΔHf0 CH4 (g) -75 kJ/mole C2H2 (g) 227 kJ/mole C2H6 (g) -85 kJ/mole C6H12O6 (s) -1268 kJ/mole O2 (g) 0 kJ/mole H2 (g) CO2 (g) -394 kJ/mole H2O (l) -286 kJ/mole CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)