Chemical Reactions / Thermochemistry / Gases

Chemical Reactions / Thermochemistry / GasesH Advanced ChemistryUnit 2

1Objective #4 Lewis vs. Bronsted-Lowery Definitions of Acids and BasesThe traditional Arrhenius definition for acids and bases is a rather narrow definition. In this definition, acids must contain ionizable hydrogen to be an acid and ionizable hydroxide to be a base. For example: HCl and NaOH. A broader definition, the Bronsted - Lowery concept defines an acid as a species that donates a proton and defines a base as a species that accepts a proton.Example:H2S + CH3NH2 HS-1 + CH3NH3+1Acid Base Conjugate Conjugate Base Acid All Arrhenius acids are also Bronsted-Lowery acids but some Bronsted-Lowery bases are not Arrhenius bases. Another broader definition for acids and bases is the Lewis concept. In this system, it is recognized that in order for a Bronsted-Lowery base to be a proton acceptor, it must also be able to donate an electron pair.Example:H+ + :NH3 > NH4+1In this system, an acid is defined as a species that can accept an electron pair.Example::NH3 + BF3 > NH3BF3 Lewis base Lewis acidObjective #5 Oxidation Numbers and Redox Reactions(Review of Oxidation Numbers+1 -1Na Cl +1 +6 -2Na2 S O4+2 +7 -2Mg (Cl O4)2+4 +6 -2Pb (Cr2 O7)2Writing Molecular and Net Ionic Equations for Displacement Reactions*a displacement reaction occurs when an ion displaced due to a change in oxidation number*example I: (oxidation state assignments)0 +2 +5 -2 +2 +5 -2Fe + Ni (N O3)2 > Fe (N O3)2 +0Ni (molecular equation)(ionic equation) Fe + Ni+2 + 2NO3-1 > Fe+2 + 2NO3-1 + Ni(net ionic equation)Fe + Ni+2 > Fe+2 + Ni(spectator ions)NO3-1

*example II: (oxidation state assignments)(molecular equation)0 +2 +6 -2 +2 +6 -2 0Mg + Co S O4 > Mg S O4 + Co

(ionic equation):Mg + Co+2 + SO4-2 > Mg+2 + SO4-2 + Co(net ionic equation):Mg + Co+2 > Mg+2 + Co(spectator ions):SO4-2Objectives #6-9 Working with Energy and the First Law of ThermodynamicsExothermic vs. Endothermic Reactions*exothermic reactions involve the loss of energy to the surroundings while endothermic reactions involve the gain of energy from the surroundings*the change in energy of a chemical process can be found by comparing the initial value of the energy of the reactants to the final value of the energy of the productsE = Efinal - Einitial12Objectives #6-9 Working with Energy and the First Law of Thermodynamics*if the E has a positive value, then the products of a reaction have a higher energy than the reactants, energy has been gained from the surroundings, and the sign of the energy value will be expressed as a positive value*if the E has a negative value, then the products of a reaction have a lower energy than the reactants, energy has been released to the surroundings, and the sign of the energy value will be expressed as a negative value*energy diagram for endothermic vs. exothermic13Endothermic vs. Exothermic Graphs

Objectives #6-9 Working with Energy and the First Law of Thermodynamics*when energy is exchanged with the surroundings, it can be in the form of heat or work*the change in this energy exchange can be found using the equation:E = q + wwhere q represents the heat transferred and w represents the work done on the reaction system or by the reaction system*when heat is added or work is done on the reaction system, the total energy of the system increases*the total energy of the system can also decrease if heat is lost to the surroundings and / or work is done on the surroundings15Objectives #6-9 Working with Energy and the First Law of Thermodynamics*the following sign conventions are used to designate the q, w, and t:

Thermodynamic QuantityIf PositiveIf Negativeqreaction gains heatreaction loses heatwwork done on reaction work done by reaction Enet gain of energy by reactionnet loss of energy by reaction 16If a reaction that produces a solid from various gaseous reactants loses 1200 J of heat to the surroundings and the external atmospheric pressure causes 500 J of work to be done on a collapsing piston, calculate the change in the internal energy of this reaction system.E = q + w = -1200 J + 500 J = -700 JCalculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.E = q + w = +140 J 85 J = +55 JObjectives #6-9 Working with Energy and the First Law of Thermodynamics*a related equation involves enthalpy which relates the heat flow of a process to its internal energy and work which is the result of pressure and volume:H = E + PV*If a reaction is performed at constant pressure, the product of PV will be very small as a result. Therefore, the enthalpy equals the heat gained or lost in a reaction. Enthalpy, H, is an easier concept to deal with than looking at the work involved in a chemical process*If heat is gained by a reaction system, its H value will be positive and the reaction is classified as endothermic*If heat is released by a reaction system, its H value will be negative and the reaction is classified as exothermic19*Examples: Determine if each of the following processes would have a positive or negative sign for the H value:Melting ice +HBurning methane gas -HOxidation of magnesium in oxygen -H

Objective #10 Interpreting Potential Energy Diagrams for Endothermic and Exothermic Reactions*potential energy diagrams can pictorially illustrate the energy changes as reactants are converted into products and the reverse reaction as well

21Potential Energy Diagrams:Exothermic vs. Endothermic

*diagram illustrating exothermic reaction:2H2 + O2 > 2H2OH < 0 (exothermic) Reactants at higher energy state*diagram illustrating exothermic reaction:H2O(l) > H2O(g)H > 0 (endothermic) Products at higher energy state*diagram illustrating a forward and reverse reaction (example is endothermic)N2 + 3H2 0Objective #10 Interpreting Potential Energy Diagrams for Endothermic and Exothermic Reactions*general conventions:Enthalpy depends on the amount of material used and produced in the reaction; for example if burning one mole of methane produces 890 kJ of heat then burning two moles will result in the production of 1780 kJ of heatThe amount of heat generated in the forward reaction will equal the amount of heat required in the reverse reaction and vice versa but the signs will be reversed.3. Enthalpy changes depend on the states of the materials involved in the reaction. For example forming gaseous water would take up more energy than forming liquid water and therefore the final energy output to the surroundings would be less

Objective #11 Problems Involving Energy Stoichiometry*if the H is known for a reaction, the mole ratios of the reactants and products included in the reaction can be used to determine the amount of heat evolved or required in a reaction(examples)

26Calculate the amount of heat released when 5.00 g of methane burns in oxygen gas:CH4 + 2O2 CO2 + 2H2Owhere the H = -890 kJ5.00 g CH4 X 1 mole CH4 / 16.0 g X-890 kJ / 1 mole CH4 = -278 kJCalculate the amount of heat released when 5.00 g of hydrogen peroxide decomposes.2H2O2 > 2H2O + O2 where the H = 196 kJ5.00 g H2O2 X 1 mole H2O2 / 34.0 g X196 kJ / 2 mole H2O2 = 14.4 kJObjectives #12-13 Problems Involving Calorimetry(bomb calorimeter)

Example of a Laboratory Calorimeter

Examples of Calorimeter ProblemsHow much heat is needed to warm 250. g of water from 22oC to 98oC if the specific heat of liquid water is 4.18 J/g.K.Q = mct = (250. g) (4.18 J/g. K) (76 K) = 79400 JCalculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0oC and the specific heat of the rocks is .082 J/g.K.Q = mct = (50000 g) (.082 J/g.K) (12 K) = 49200 JWhat temperature change would the rocks undergo if they absorbed 450 kJ of heat?Q = mctQ / mc = t450 kJ / (.082 J/g.K) (50000 g) = t110 K = tWhen 4.00 g of CH6N2 is combusted in a bomb calorimeter (in other words it goes BOOM!), the temperature of the calorimeter increases from 25.00 oC to 39.50 oC. If the heat capacity of the calorimeter is 7.794 kJ/oC determine the heat of reaction per mole of CH6N2.(so we need to find kJ of heat per mole of material)

Find heat of rx. from change in temp and specific heat of calorimeter:t = 39.50oC 25.00oC = 14.50oCqrxn = -Ccal X t = -(7.794 kJ/oC) (14.50oC) = -113.0 kJ*the heat produced came from the 4.00 g of material -113.0 kJ / 4.00 g CH6N2 X 46.1 g CH6N2/ 1 mole = -1.30 X 103 kJ/moleWhen 50.0 ml of .100 M AgNO3 and 50.0 ml of .100 M HCl are mixed in a constant pressure calorimeter, the temperature of the mixture increases from 22.20oC to 23.11oC. Calculate the H for this reaction in kJ/mole AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g.C.

*once again we need heat per mole of materialQ = mct = (100.0 g) (4.18 J/goC) (.91oC) = 380 J = .380 kJMoles AgNO3 = (.100 M) (.0500 L) = .00500 moles.380 kJ / .00500 moles = -76.0 kJ / mole AgNO3Germain Henri Hess1802-1850

Objectives #14-15 Hesss Law*if a reaction is carried out in a series of steps, the H for the reaction will equal the sum of the enthalpy changes for the individual steps40Objectives #14-15 Hesss Law

Objectives #14-15 Hesss Law*Hesss Law makes it possible to calculate the energy changes for a reaction indirectly when it is not possible to measure the energy changes directly for a particular reaction*Hesss Law problems often involve manipulating the step by step reactions that can lead to the overall reaction desired; these manipulations can include:reversing a reactionmultiplying or dividing a reaction by some factor canceling like termsWhen these changes are made, the sign and the of the energy term must also be adjusted(examples)

*For example, lets say we want to calculate the energy change for the following reaction:C(s) + 1/2O2(g) CO(g) given the following thermochemical data:C(s) + O2(g) CO2(g) H1 = -393.5 kJ2. CO(g) + 1/2O2(g) CO2(g) H2 = -283.0 kJStep I Rx.#1 has the carbon on the correct reactant side and so we can just recopy that reaction as is:C(s) + O2(g) CO2(g) H1 = -393.5 kJStep II Rx.#2 needs to be reversed so that the carbon monoxide appears on the product side as shown in the desired equation:CO2(g) CO(g) + 1/2O2(g)

since the reaction is reversed, the sign on the energy term needs to switch as wellH = +283.0 kJStep III The carbon dioxide molecules are on opposite sides of the equation and so they can now be canceled out and the remaining substances will now match the desired equation.C(s) + 1/2O2(g) CO(g) Step IV Add the energy terms to obtain the final energy value desired.H = +283.0 kJ -110.5 kJ = -110.5 kJ

Example II:Calculate the H for the reaction:1. 2C(s) + 2O2(g) 2CO2(g) H = -787 kJH2(g) + 1/2O2(g) H2O(l) H = -285.8 kJ2CO2(g) + H2O(l) C2H2(g) + 5/2O2(g)H = 1299.6 kJ---------------------------------------------2C(s) + H2(g) C2H2(g) H = 226.8 kJExample III:Calculate the H for the reaction:NO(g)+ O3(g)NO2(g) + O2(g) H=-198.9 kJO(g) 1/2O2(g) H = -247.5 kJ3/2O2(g) O3(g) H = +142.3 kJ---------------------------------------------NO(g) + O(g) NO2(g) H = -304.1 kJExample IVCalculate the H for:SO3(g) SO2(g) + 1/2O2(g) H = +98 kJS(s) + 3/2O2(g) SO3(g) H = -395 kJ----------------------------------------------S(s) + O2(g) SO2(g) H = -297 kJObjectives #14-15 Hesss LawPart II Using Enthalpies of Formation to Calculate Enthalpies of Reaction*standard enthalpies of formation can also be used to determine the H for a reaction*these enthalpy values represent the energy change that results when 1 mole of a compound is formed from its elements in their standard states at 25oC; when H values are determined in this manner, they are referred to as standard enthalpy change and are denoted by the symbol H050Objectives #14-15 Hesss Law*For example, the reaction:2Na(s) + 1/2O2(g) Na2O(g)would represent an equation showing a standard enthalpy of formation because all of the reaction constituents are in their standard states (sodium is a metal, oxygen is a gas, sodium oxide is a solid) at 25oC and the reaction is showing the formation of 1 mole of sodium oxide being formed*Example #2: Would the following reaction show the standard enthalpy of formation for potassium chloride? If not, make the appropriate corrections:2K(l) + Cl2(g) 2KCl(s)

51*Example #2: Would the following reaction show the standard enthalpy of formation for potassium chloride? If not, make the appropriate corrections:2K(l) + Cl2(g) 2KCl(s)(divide by 2)K(s) + 1/2Cl2(g) KCl(s)

Objectives #14-15 Hesss Law*Some conventions to keep in mind when using standard enthalpy of formation values to calculate reaction enthalpies:The standard enthalpy of formation for an element is defined as zero because it is already formedBe sure to reverse the sign of a standard enthalpy of formation value if the substance involved is a reactant in the reaction (formation means products!)If more than 1 mole of a substance is indicated from the reaction stoichiometry be sure to adjust the Hf0 value as appropriateThe final value for the standard heat of reaction will be equal to the sum of the heats of the products sum of the heat of the reactants(examples)53Example ICalculate the standard reaction enthalpy for the following reaction:CH4(g) + 2O2(g) CO2(g) + 2H2O(l)Horxn = +74.8 kJ -0 kJ -395.5 kJ -2(285.8 kJ) = -892.3 kJ Example IICalculate the standard reaction enthalpy for the following reaction:Horxn = -98 kJ O kJ -12(393.5 kJ) -1715 kJ = -6535 kJ

Objectives #18-19 Applications of the Ideal Gas Law*the ideal gas law; PV=nRT can be manipulated so that the density of a gas can be determined*since density equals mass/volume, we can rearrange the equation in terms of moles/volume n/v = p/RT*multiplying both sides by the molar mass will convert the moles to grams which will form the unit grams/L which are units of density for a gasD = PM/RT(examples)

56Example I:Calculate the density of carbon tetrachloride vapor at 714 torr and 125oC.D = PM/RT = (714 torr / 760 torr) (154 g) / (.0821 L.atm/mole.K) (398 K) = 4.43 g/LExample II:The molar mass of the atmosphere on Titan, a moon of Saturn, is 28.6 g/mol. Calculate the density of its atmosphere if the temperature is 95 K and the pressure is 1.6 atm.D = PM / RT = (1.6 atm.) (28.6 g/mol) / (.0821) (95 K)= 5.87 g/L*the density equation can also be solved for molar mass as follows:M = dRT/P(examples)

*Example ICalculate the molar mass of a gas having a mass of 2.889 g and a volume of 936 ml when the pressure is 735 torr at 31oC.M = dRT/P= (2.889 g / .936 L) (.0821) (304 K) / (735 torr / 760 torr) = 79.6 g

*Example IICalculate the molar mass of dry air if it has a density of 1.17 g/L at 21oC and 740.0 torr of pressure.M = (1.17 g/L) (.0821) (294 K) / (740.0 torr / 760 torr) = 29.0 gObjective #21 Daltons Law of Partial Pressures / Gas Stoichiometry / Raoults LawWhen gases are collected over water, the vapor pressure of water must be taken into consideration. According to Daltons Law:Pgas = Ptotal PwaterThe vapor pressure of water varies with increasing temperature and can be found on p.1058 of the text. In a related topic, the vapor pressure of a volatile solvent can be affected by the presence of a nonvolatile solute.According to Raoults law, adding a non-volatile solute to a pure volatile solvent will decrease its vapor pressure. For example the vapor pressure of pure water at 20oC is 17.5 torr from the vapor pressure chart. If by adding sugar to the water, the mole fraction of the water is now only .80 moles. The new decreased vapor pressure can be found by:Psolution = (.80) (17.5 torr) = 14 torrExample of gas stoichiometry involving the collection of gas over water:Calculate the volume of hydrogen produced at 1140 torr and 19oC by the reaction of 26.5 g of calcium metal with excess water. The vapor pressure of water is 16.48 torr.Ca + 2H2O H2 + Ca(OH)2*find moles of gas first:26.5 g Ca X 1 mole Ca / 40.1 g Ca X 1 mole H2 / 1mole Ca = .661 moles H2*find pressure of gas:1140 torr 16.48 torr = 1124 torr*find volume of gas using PV=nRTV = (.661 moles) (.0821) (292K) / = (1124 torr / 760 torr) = 10.7 L Objective #22 Real vs Ideal Gases*according to the kinetic theory of gases, ideal gas particles have no attractive forces and the individual gas particles occupy no volume66Objective #23 Real vs Ideal Gases

Objective #23 Real vs Ideal Gases*however under conditions of high pressure and low temperature, real gas properties deviate from these ideal conditions set forth in the Kinetic Theory; attractive forces increase as the gas particles are brought closer together and their motion is slowed down*due to these attractions between gas particles, the force that the particles hit the sides of a container become less forceful and therefore the pressure is decreased below its ideal value

Objective #23 Real vs Ideal Gases

The Effect of Temperature and Pressure on the Behavior of Nitrogen Gas

Johannes van der Waals 1837-1923

Objective #23 Real vs Ideal Gases*because many real world situations involve gases at high pressures, the van der Waals equation can be used to correct the pressure that would be calculated just using the ideal gas law:

Moles vs Pressure of Oxygen Gas(pink line is using ideal gas law and purple line is using van der Waals equation)

Objective #23 Real vs Ideal GasesP = nrt/V-nb n2a/V2*the constant a is specific for each type of gas and its magnitude depends on the attractive forces of the gas particles and the constant b, which is also specific for each type of gas, depends on the volume of the gas particles*the a constant, along with the number of moles of gas present and the volume the gas occupies corrects for molecular attraction*the b constant along with the moles of the gas correct for the volume of gas molecules74van der Waals Constants for Gas Molecules

Example:1.000 mole of Cl2 gas would be expected to occupy a volume of 22.4 L at 273 K if acting as an ideal gas. Determine its true pressure using the van der Waals equation. The a value for chlorine gas is 6.49 L2atm/mole2 and the b value is .0562 L/mole.*utilize equation:P = [nrt/V-nb] [n2a/V2][nrt/V nb] =P = (1.000 mole) (.0821) (273K) /[22.40 L (1.000 mole) (.0562 L / mole)]P = 22.41 / [22.40 - .0562] = 22.41 / 22.34 = 1.003 atm [n2a/V2] =[(1.000 mole)2 (6.49 L2atm / mole2)] / (22.40 L)26.49 / 501.76 = .0129 atmPreal = 1.003 atm - .0129 atm = .9901 atmPreal = .9901 atm vs. Pideal 1.000 atm