Thermochemisty (Enthalpy) and Hess’s Law Chapter 10 Sections

Enthalpy Change in enthalpy (ΔH p = q p ): the amount of heat exchanged when heat exchange occurs under conditions of constant pressure Enthalpy is a state function –State function is defined as a property that changes independent of pathway ΔH is independent of the path taken

At constant pressure, 890kJ of energy per mole of CH 4 is produced as heat Q p = ΔH p = -890 kJ exothermic If 5.8 g of methane is burned, how much heat will be released? 5.8g CH 4 x 1mol CH 4 = 0.36 mol CH g CH mol CH 4 x -890 kJ = -320 kJ mol CH 4

Calorimeter Calorimetry the process of measuring heat flow in the form of temperature changes in a system Calorimeter an insulated device used to measure temperature changes

Hess’s Law –ΔH for a multi-step process is the sum of ΔH for the individual steps –Hess’s law allows us to calculate ΔH for new reactions from previously determined values Germain Henri Hess

Altitude San Francisco, CA 0 ft Salt Lake City, UT 4,266 ft Denver, CO 5,280 ft Flight 1 Flight 2 Flight 1: Δ Alt = Flight 2: Δ Alt = SL City-Denver Δ Alt = What is the change in altitude (Δ Alt) between Salt Lake City and Denver? ? 4266 ft 1014 ft 5280 ft

Hess’s Law Applying Hess’s Law –When reversing a reaction, change the sign of ΔH. –When multiplying a reaction by a new coefficient, multiply ΔH

Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Calculate the enthalpy change for a partial combustion of 1 mol of graphite to diamond C(s, graphite)  C(s, diamond) ΔH = ? Given: C(s, graphite) + O 2 (g)  CO 2 (g) ΔH = KJ C(s, diamond) + O 2 (g)  CO 2 (g) ΔH = KJ Answer: C(s, graphite) + O 2 (g)  CO 2 (g) ΔH = kJ CO 2 (g)  C(s, diamond) + O 2 (g) ΔH = kJ C(s, graphite)  C(s, diamond) ΔH = +1.9 kJ Hess’s Law

Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH kJ C + O 2  CO kJ H 2 + ½ O 2  H 2 O kJ Step #1: CH 4 must appear on the reactant side, so we reverse reaction #1 and change the sign on  H. CH 4  C + 2H kJ

Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH kJ C + O 2  CO kJ H 2 + ½ O 2  H 2 O kJ CH 4  C + 2H kJ Step #2: Keep reaction #2 unchanged, because CO 2 belongs on the product side C + O 2  CO kJ

Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH kJ C + O 2  CO kJ H 2 + ½ O 2  H 2 O kJ CH 4  C + 2H kJ C + O 2  CO kJ Step #3: Multiply reaction #2 by 2 2H 2 + O 2  2 H 2 O kJ

Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O Reaction  H o C + 2H 2  CH kJ C + O 2  CO kJ H 2 + ½ O 2  H 2 O kJ CH 4  C + 2H kJ C + O 2  CO kJ 2H 2 + O 2  2 H 2 O kJ Step #4: Sum up reaction and  H CH 4 + 2O 2  CO 2 + 2H 2 O kJ

Calculation of Heat of Reaction Calculate  H for the combustion of methane, CH 4 : CH 4 + 2O 2  CO 2 + 2H 2 O  H rxn =   H f (products) -   H f (reactants) Substance  H f CH kJ O2O2 0 kJ CO kJ H2OH2O kJ  H rxn = [ kJ + 2( kJ)] – [-74.80kJ]  H rxn = kJ