10-4 Enthalpy (Section 10.6) And you

Enthalpy, symbolized by H, can be thought of as the potential energy stored in the bonds of molecules. Chemists use the change in enthalpy ∆H to measure the heat content of a system (when the pressure is constant). We define the “system” to be the chemicals and everything else is termed the “surroundings”. Applying the First Law of Thermodynamics (Conservation of Energy), any heat lost by the system will equal the heat gained by the surroundings (and vice versa). Exothermic (“exo” means released or “exits” and therm refers to heat)

Exothermic reactions characteristics Reactants: high E (H), less stable, weak bonds Products: low E (H), more stable, strong bonds System releases PE from bonds to KE of surroundings (which feel hot). ∆H = P – R = negative value (heat released)

Endothermic (“endo” means absorbed or “going in” and therm refers to heat)

Characteristics: Reactants: low E (H), more stable, strong bonds Products: high E (H), less stable, weak bonds System absorbs KE from surroundings as PE in the bonds. Surroundings will feel cold. ∆H = P – R = positive value (heat absorbed)

Bond Energies and ∆H: It requires energy to break the bonds of the reactants. It releases energy when new bonds of the products form. The difference between these two energies is the ∆H. Note, though, if: Energy absorbed to break reactants > Energy released forming products Endo ∆H = +  Energy absorbed to break reactants < Energy released forming products Exo ∆H = -

Example ½ H2 + ½ Cl2  HCl Think of bond energies as KE entering or leaving the system. 216 + 120  427 The change in KE = the change in PE  91 H = -91kJ/mol Exothermic

10-5 Enthalpy of Formation The enthalpy of formation, ∆Hf, is defined as the heat absorbed or released when making 1 mole of a compound from its elements (at 25oC and 1 atm = standard state). Note that the conditions are important! By convention, the Hf of any element at this temperature and pressure is zero. ex: O2 (g)

what does the balanced reaction look like? N2 (g) + 3H2 (g) = 2NH3 (g) Example reaction 1: Nitrogen gas reacts with hydrogen gas to form ammonia (NH3) what does the balanced reaction look like?  N2 (g) + 3H2 (g) = 2NH3 (g) How would you write this to show 1 mole of product? ½ N2 (g) + 3/2 H2 (g)  NH3 (g) ▲Hf = -46 kJ.mol What is more stable the reagents or the product? The product!!! E is given off – product has less H

Example reaction 2: Nitrogen gas reacts with oxygen gas to form nitrogen dioxide. Bal. Rxn. w/1 mole product??? ½ N2 (g) + O2 (g)  NO2 (g) ▲Hf = +33.9 KJ/mol What is more stable the reagents or the products? The reagents E goes into syst – prod. have greater H

Reaction #3: Aluminum solid reacts with oxygen gas to form aluminum oxide. ? rxn. w/1 mol. product 2Al (s) + 3/2 O2 (g)  Al2O3 (s) ▲H = -1676 KJ/mol What is more stable? The product

Summary Reaction Hf (kJ/mol) Stability of Product ½ N2(g) + 3/2 H2(g) → NH3 -46 stable ½ N2(g) + O2 → NO2 +34 unstable 2 Al + 3/2 O2 → Al2O3 -1676 very stable

10-6 Enthalpy of a Reaction H = ∑Hf (products) – ∑Hf (reactants) Hc = enthalpy of combustion ~ defined for the combustion of 1 mole of a fuel  CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) [-393.5 + 2 (-285.8)] – [-74.8 + 0 ] = -890.3 kJ/mol Burning fuels is always exothermic!!!