THERMOCHEMISTRY

3 MAIN PROCESSES THAT WILL ALTER THE ENERGY OF A CHEMICAL SYSTEM Heating and cooling Phase changes Chemical reactions

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

Depend ONLY on the present state of the system State Properties (State Functions) The state of a system is described by its composition, temperature and pressure. 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 WHY NOT???

SYSTEM and SURROUNDINGS OPEN VS CLOSED SYSTEMS THERMODYNAMIC FUNCTIONS = 3 PARTS

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

E = q + w Try This! Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. E = q + w ΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ The system has gained 17.0 kJ of energy!

Work, Pressure, and Volume Gas Expansion Compression +V (increase) - V (decrease) -w results +w results Esystem decreases Esystem increases Work has been done by the system on the surroundings Work has been done on the system by the surroundings

Work, Pressure, and Volume Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. W = -15 atm x 18 L = -270 L x atm (gas is expanding , therefore it is doing work on its surroundings!) Energy is flowing out of the gas, so W is a negative quantity

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

Endothermic Reactions

Exothermic Reactions

TIME TO REVIEW CHAPTER 5 EXAM THE FOLLOWING INDIVIDUALS WILL MOVE TO THE CENTER OF THEIR RESPECTIVE WORKSTATIONS: JEREMY RANJEEKA KUSH SINDY PRIYANKA SHYAMA PLEASE REMAIN AT YOUR WORKSTATION DURING THIS CHAPTER 6 MULTIPLE CHOICE REVIEW SESSION.

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

4.18 joule = 1 calorie The Joule The unit of heat used in modern thermochemistry is the Joule 4.18 joule = 1 calorie

A Cheaper Calorimeter

A Bomb Calorimeter

Heat Capacity C heat absorbed C = Increase in temperature Not enough!…………..When a substance is heated, the energy required will depend on ??? AMOUNT OF SUBSTANCE Specific Heat Capacity – J/oC x g or J/K x g Molar Heat Capacity – J/oC x mol or J/K x mol

Calculations Involving Specific Heat OR c = Specific Heat Capacity q = Heat lost or gained T = Temperature change m = mass of the solution

ΔH What is this????? ENTHALPY

Let’s Test Our “Calorimetry “Acumen” Suppose we mix 50.0 mL of 1.0 M HCl at 25oC with 50.0 mL of 1.0 M NaOH also at 25oC in a calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.9oC. (Assume that the solution can be treated as if it were pure water with a density of 1.0 g/mL). How much energy is released?

Answer is………….??? 2.9 x 103 J

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 Δ H

Hints for Using Hess’s Law work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products.

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

Hess’s Law Example Problem CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

Hess’s Law Example Problem CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ Step #3: Multiply reaction #3 by 2

Hess’s Law Example Problem CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ CH4 + 2O2  CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

Two forms of carbon are graphite and diamond. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), calculate H for the conversion of graphite to diamond: Cgraphite(s) Cdiamond(s) The combustion reactions are: Cgraphite(s) + O2(g) CO2(g) H = -394 kJ Cdiamond(s) + O2(g) CO2(g) H = -396 kJ Cgraphite(s) + O2(g) CO2(g) H = -394 kJ CO2(g) Cdiamond(s) + O2(g) ) H = -(-396 kJ) Cgraphite(s) Cdiamond(s) ) H = 2 kJ

6.4 Standard Enthalpies of Formation The standard enthalpy of formation (ΔHof) of a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard state The above degree symbol indicates that the corresponding process has been carried out under standard conditions. The standard state for a substance is a precisely defined reference state. Conventional Definitions of Standard States for a compound or element are located on pg. 246 in your textbook.

Key Concepts When Doing Enthalpy Calculations: When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer. Elements in their standard states are not included in the ΔH reaction calculations. That is ΔH of for an element in its standard state is zero.

ΔH o reaction = Σnp ΔH of(products) - Σnr ΔH of(reactants) The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: ΔH o reaction = Σnp ΔH of(products) - Σnr ΔH of(reactants)

Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O Hrxn =  Hf(products) -   Hf(reactants)     Substance Hf   CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ O Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

Specific Heat Capacity The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·C) Water (liquid) 4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid)