Chapter 6 THERMOCHEMISTRY West Valley High School AP Chemistry Mr. Mata.

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Presentation transcript:

Chapter 6 THERMOCHEMISTRY West Valley High School AP Chemistry Mr. Mata

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 A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane! WORK IS NOT A STATE FUNCTION

 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

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

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 joule = calories

A Bomb Calorimeter

A Cheaper Calorimeter

Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·K) Water (liquid)4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) Carbon (graphite,solid) Iron (solid) Copper (solid) Mercury (liquid) Lead (solid) Gold (solid) 0.129

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

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.”

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 Step #3: Multiply reaction #3 by 2 2H 2 + O 2  2 H 2 O 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 #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