2Specific Heat Formula Q = Energy (heat) lost or gained cp = Specific HeatT = Temperature changem = Mass
3Two Types of Thermal Reactions Exothermic: Releases Thermal Energy (heat) sign is –Endothermic: Absorbs Thermal Energy (heat)sign is +
4Exothermic ProcessesProcesses in which energy is released as it proceeds, and surroundings become warmerReactants Products + energy
5Endothermic Processes Processes in which energy is absorbed as it proceeds, and surroundings become colderReactants + energy Products
6Enthalpy = HIn a chemical reaction, Enthalpy (H) is equal to the energy that flows as heat.(at a constant pressure)Example:When 1 mol of methane gas is burned it releases 890 kJ of energyCH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heatH = -890kj = exothermic reaction
7Solution: Molar mass of CH4 = 16.04 g / mol Calculate H for a process in which 5.8 g of methane are burned. H = -890kJ per mol CH4Rxn: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) +heatSolution:Molar mass of CH4 = g / mol5.8 g CH4 1 mol CH4 = 0.36 mol CH416.04 g CH40.36 mol CH kJ = -320 kJ1 mol CH4
8Solution: Molar mass of SO2 = 64.07 g / mol Calculate H when 12.8 g of sulphur dioxide reacts with excess oxygen to form sulphur trioxide. H = kJ per mol SO2Solution:Molar mass of SO2 = g / mol12.8 g SO2 1 mol SO2 = mol SO264.07 g SO2mol SO kj = kJ1 mol SO2
9Hydrogen peroxide decomposes according to the following thermochemical reaction: H2O2(l) → H2O(l) + 1/2 O2(g)ΔH = kJCalculate the change in enthalpy (ΔH) when 4.00 grams of hydrogen peroxide decomposes.Calculate the change in enthalpy (ΔH) when grams of water are created…
10The reaction that occurs in hand warmers is: 4Fe(s) + 3O2(g) 2Fe2O3(s) + heatH of reaction = -1652kJHow much heat is released when 2.5 grams of Fe(s) is reacted with excess oxygen?
11Law of Concert Tickets!Miller Stitts LadyGaga $6002 Winters + Miller Justin Bieber + $300Stitt 2 Elvis Winter $500Lady Gaga Justin Bieber Elvis $_____ ???
12Law of Concert Tickets = Hess’s Law You can add KNOWN equations to solve UNKOWN equationsTwo Rules:FLIP (reverse) the equation – you must FLIP the sign (- +)MULTIPY / DIVIDE equation – must Multiply / Divide the ΔH!
13Example #1: Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g)ΔH° = ??? kJGiven the following thermo chemical equations:C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l)ΔH° = kJC(s) + O2(g) ---> CO2(g)ΔH° = kJH2(g) + (1/2)O2(g) ---> H2O(l)ΔH° = kJa) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. kJ + (-787 kJ) + ( kJ) = kJ
15+234 + (+592) + (-1220) = -394 Example #2: Given the following data: Find the ΔH of the following reaction:C(s) + O2(g) ---> CO2(g)SrO(s) + CO2(g) ---> SrCO3(s)ΔH = -234 kJ2SrO(s) ---> 2Sr(s) + O2(g)ΔH = kJ2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g)ΔH = kJa) first equation - flip itb) second equation - divide by twoc) third equation - flip it, divide by two(+592) + (-1220) = -394
16+234 + (+592) + (-1220) = -394 Example #2: Given the following data: Find the ΔH of the following reaction:C(s) + O2(g) ---> CO2(g)SrO(s) + CO2(g) ---> SrCO3(s)ΔH = -234 kJ2SrO(s) ---> 2Sr(s) + O2(g)ΔH = kJ2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g)ΔH = kJa) first equation - flip itb) second equation - divide by twoc) third equation - flip it, divide by two(+592) + (-1220) = -394
17Hess's LawIn a reaction, the change in enthalpy (ΔH)is the same - regardless if the reaction occurs in a single step or in several steps.If a series of reactions are added together, the net change in ΔH is the sum of the enthalpy changes for each step.
18Rules for using Hess's Law If the reaction is multiplied (or divided) by some factor, Δ H must also be multiplied (or divided) by that same factor.If the reaction is reversed (flipped), the sign of Δ H must also be reversed.
19Water phase changes & Energy constantTemperature remains __________ during a phase change.
20Phase Change Diagram Processes occur by addition of energy Processes occur by removal of energy
21Phase DiagramRepresents phases as a function of temperature and pressure.Critical temperature: temperature above which the vapor can not be liquefied.Critical pressure: pressure required to liquefy AT the critical temperature.Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).
23Effect of Pressure on Boiling Point Boiling Point of Water at Various LocationsLocationFeet above sea levelPatm (kPa)Boiling Point (C)Top of Mt. Everest, Tibet29,0283270Top of Mt. Denali, Alaska20,32045.379Top of Mt. Whitney, California14,49457.385Leadville, Colorado10,1506889Top of Mt. Washington, N.H.6,29378.693Boulder, Colorado5,43081.394Madison, Wisconsin90097.399New York City, New York10101.3100Death Valley, California-282102.6100.3
24Phase DiagramRepresents phases as a function of temperature and pressure.Critical temperature: temperature above which the vapor can not be liquefied.Critical pressure: pressure required to liquefy AT the critical temperature.Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm).
30Reaction PathwayShows the change in energy during a chemical reaction
31Exothermic Reaction 2H2(l) + O2(l) 2H2O(g) + energy reaction that releases energyproducts have lower PE than reactantsenergy released2H2(l) + O2(l) 2H2O(g) + energy
32Endothermic Reaction 2Al2O3 + energy 4Al + 3O2 reaction that absorbs energyreactants have lower PE than productsenergy absorbed2Al2O3 + energy 4Al + 3O2
33Latent Heat of Phase Change Molar Heat of FusionThe energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.Molar Heat of SolidificationThe energy that must be removed in order to convert one mole of liquid to solid at its freezing point.
34Latent Heat of Phase Change #2 Molar Heat of VaporizationThe energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.Molar Heat of CondensationThe energy that must be removed in order to convert one mole of gas to liquid at its condensation point.
35Latent Heat – Sample Problem Problem: The molar heat of fusion of water is kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?Massof iceMolarMass ofwaterHeatoffusion
36Heat of SolutionThe Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.SubstanceHeat of Solution(kJ/mol)NaOH-44.51NH4NO3+25.69KNO3+34.89HCl-74.84
37Energy is the capacity to do work Thermal energy is the energy associated with the random motion of atoms and moleculesChemical energy is the energy stored within the bonds of chemical substancesNuclear energy is the energy stored within the collection of neutrons and protons in the atomElectrical energy is the energy associated with the flow of electronsPotential energy is the energy available by virtue of an object’s position6.1
38Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study.SURROUNDINGSSYSTEMopenclosedisolatedExchange:mass & energyenergynothing6.2
392H2 (g) + O2 (g) 2H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2 (g) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.2H2 (g) + O2 (g) H2O (l) + energyH2O (g) H2O (l) + energyEndothermic process is any process in which heat has to be supplied to the system from the surroundings.energy + H2O (s) H2O (l)energy + 2HgO (s) Hg (l) + O2 (g)6.2
40DH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.DH = H (products) – H (reactants)DH = heat given off or absorbed during a reaction at constant pressureHproducts < HreactantsHproducts > HreactantsDH < 0DH > 06.3
41Thermochemical Equations Is DH negative or positive?System absorbs heatEndothermicDH > 06.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.H2O (s) H2O (l)DH = 6.01 kJ6.3
42Thermochemical Equations Is DH negative or positive?System gives off heatExothermicDH < 0890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DH = kJ6.3
43Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substanceH2O (s) H2O (l)DH = 6.01 kJIf you reverse a reaction, the sign of DH changesH2O (l) H2O (s)DH = kJIf you multiply both sides of the equation by a factor n, then DH must change by the same factor n.2H2O (s) H2O (l)DH = 2 x 6.01 = 12.0 kJ6.3
44Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations.H2O (s) H2O (l)DH = 6.01 kJH2O (l) H2O (g)DH = 44.0 kJHow much heat is evolved when 266 g of white phosphorus (P4) burn in air?P4 (s) + 5O2 (g) P4O10 (s) DH = kJ1 mol P4123.9 g P4x3013 kJ1 mol P4x266 g P4= 6470 kJ6.3
45Heat (q) absorbed or released: The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.C = msHeat (q) absorbed or released:q = msDtq = CDtDt = tfinal - tinitial6.4
46Dt = tfinal – tinitial = 50C – 940C = -890C How much heat is given off when an 869 g iron bar cools from 940C to 50C?s of Fe = J/g • 0CDt = tfinal – tinitial = 50C – 940C = -890Cq = msDt= 869 g x J/g • 0C x –890C= -34,000 J6.4