THERMOCHEMISTRY

Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object

Definitions #2 Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

State Functions State Functions depend ONLY on the present state of the system ENERGY IS A STATE FUNCTION WORK IS NOT A STATE FUNCTION WHY NOT???

 E = q + w  E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings

Work, Pressure, and Volume Expansion Compression +  V (increase) -  V (decrease) - w results+ w results E system decreases Work has been done by the system on the surroundings E system increases Work has been done on the system by the surroundings 1 L atm = J

Energy Change in Chemical Processes EndothermicExothermic

Energy Change in Chemical Processes Endothermic: Exothermic: Reactions in which energy flows into the system as the reaction proceeds. Reactions in which energy flows out of the system as the reaction proceeds. + q system - q surroundings - q system + q surroundings

Endothermic Reactions

Exothermic Reactions

Enthalpy – flow of heat represents the heat transferred per amount. at constant pressure,  H =  E + P  V The sign of  H indicates if system is gaining or losing energy +  H indicates system is endothermic indicates system is exothermic  H

Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1  C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1  F

The Joule The unit of heat used in modern thermochemistry is the Joule 1 calorie = joules

A Bomb Calorimeter

A Cheaper Calorimeter (aka – coffee cup calorimeter)

Heat Capacity Heat absorbed/change in temperature Cal / °C J / °C Cal / K J / K

Specific Heat Capacity C = Specific Heat Capacity Heat absorbed/ Mass*change in temperature Cal / g °C J / g °C Cal / g K J / g K

Molar Heat Capacity C = Molar Heat Capacity Heat absorbed/ mole*change in temperature Cal / mol °C J / mol °C Cal / mol K J / mol K

Calculations Involving Specific Heat C or s = Specific Heat Capacity q = Heat lost or gained  T = Temperature change OR

Calculations Involving Molar Heat Capacity C = Molar Heat Capacity q = Heat lost or gained  T = Temperature change OR

Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius Lead (solid) Mercury (liquid) Copper (solid) Carbon (graphite,solid) Aluminum (solid) Gold (solid) Iron (solid) 2.01 Water (vapor) 2.09 Water (solid) 2.44 Ethanol (liquid) Water (liquid) Specific Heat (J/g·K) Substance

Solutions to Problem 31,34,45,47,61,67

Cal Prob Solutions

Hess’s Law of bunny love

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.” Rules: If the reaction is reversed, the sign of ∆H is changed If the reaction is multiplied, so is ∆ H

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

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 CH 4  C + 2H kJ Step #2: Keep reaction #2 unchanged, because CO 2 belongs on the product side C + O 2  CO kJ

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 CH 4  C + 2H kJ C + O 2  CO kJ 2H 2 + O 2  2 H 2 O kJ Step #3: Multiply reaction #2 by 2

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 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

Do #75 on page 289

Calculation of Heat of Reaction Calculate  H for the combustion of methane, CH 4 : CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l)  H rxn =  n  H f (products) -  n  H f (reactants)  H rxn = [1( kJ) + 2( kJ)] – [1(-74.80kJ) + 2(0)]  H rxn = kJ kJ/molH2OH2O kJ/mol CH 4  H f CO kJ/mol O2O2 0 kJ/mol Substance CO 2 (g) + 2H 2 O (g)CH 4 (g) + 2O 2 (g)

Do #80 on page 289

Do #83 on page 289

The standard enthalpy of the combustion of glucose is MJ at 298K. Given the following enthalpies of formation, calculate ΔH f ° for glucose CO 2 (g) : kJ/mol H 2 O (l) : kJ/mol