Energy and Chemical Reactions Chapter 6. Energy - Thermodynamics Basic Principles Basic Principles...

Energy and Energy and Chemical Chemical ReactionsReactions

Chapter 6Chapter 6

EnergyEnergy - - ThermodynamicsThermodynamics Basic PrinciplesBasic Principles

Specific Heat Capacity and Heat Specific Heat Capacity and Heat TransferTransfer

Energy and Changes of StateEnergy and Changes of State First Law of ThermodynamicsFirst Law of Thermodynamics Enthalpy Changes of Chemical Enthalpy Changes of Chemical ReactionsReactions

CalorimetryCalorimetry Hess’s LawHess’s Law Standard Enthalpies of FormationStandard Enthalpies of Formation Product or Reactant Favored ReactionsProduct or Reactant Favored Reactions

Basic PrinciplesBasic Principles EnergyEnergy - the capacity to do - the capacity to do workwork

Kinetic EnergyKinetic Energy Thermal energy - particles in motionThermal energy - particles in motion Mechanical energy- macroscopic Mechanical energy- macroscopic objects in motionobjects in motion

Electric Energy - electrons moving Electric Energy - electrons moving through a conductorthrough a conductor

Sound compression and expansion of Sound compression and expansion of spaces between particlesspaces between particles

Basic PrinciplesBasic Principles EnergyEnergy - the capacity to do work - the capacity to do work Potential EnergyPotential Energy

Chemical potential energy - attractions among Chemical potential energy - attractions among electrons and atomic nuclei. Rearranging electrons and atomic nuclei. Rearranging electrons and nuclei changes the potential electrons and nuclei changes the potential energyenergy

Gravitational energy - ball held above the Gravitational energy - ball held above the floorfloor

hypertextbook.com/physics/matter/hypertextbook.com/physics/matter/energy-chemicalenergy-chemical

Electrostatic energy - positive and Electrostatic energy - positive and negative ions a small distance apart.negative ions a small distance apart.

Potential Energy Potential Energy can be converted to can be converted to Kinetic EnergyKinetic Energy

Conservation of EnergyConservation of Energy

The First Law of ThermodynamicsThe First Law of Thermodynamics The law of conservation of energyThe law of conservation of energy The total energy of the universe is The total energy of the universe is constantconstant

Energy can be transferred from one Energy can be transferred from one form to another, but must be form to another, but must be conserved.conserved.

All standard heat engines (steam, gasoline, diesel) work by supplying All standard heat engines (steam, gasoline, diesel) work by supplying heat to a gas, the gas then expands in a cylinder and pushes a piston to heat to a gas, the gas then expands in a cylinder and pushes a piston to do its work. The catch is that the heat and/or the gas must somehow do its work. The catch is that the heat and/or the gas must somehow then be dumped out of the cylinder to get ready for the next cycle.then be dumped out of the cylinder to get ready for the next cycle.QuickTime™ and a

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Temperature and HeatTemperature and Heat

Heat is not the same as Heat is not the same as temperaturetemperature

The more thermal energy a The more thermal energy a substance has the greater motion substance has the greater motion of its atoms and moleculesof its atoms and molecules

The total thermal energy in an The total thermal energy in an object is the sum of the object is the sum of the individual energies of all atoms, individual energies of all atoms, molecules or ions in that object.molecules or ions in that object.

Systems and Systems and surroundingssurroundings

A A System - the object or collection System - the object or collection of objects being studiedof objects being studied

The Surroundings - everything The Surroundings - everything outside the system that the exchange outside the system that the exchange energy with a system.energy with a system.

ApplicationEnergy efficiency ApplicationEnergy efficiency Combustion engine10-50%Combustion engine10-50%[4][4] Electric motors 30-60% (small ones < Electric motors 30-60% (small ones < 10W); 50-90 (middle ones between 10-10W); 50-90 (middle ones between 10-200W); 70-99.99% above 200W200W); 70-99.99% above 200W

Household refrigerators low end systems Household refrigerators low end systems ~ 20%; high end systems ~ 40-50%~ 20%; high end systems ~ 40-50%

Incandescent bulbs5-10%Incandescent bulbs5-10%

Directionality of Heat Directionality of Heat Transfer:Thermal Transfer:Thermal

EquilibriumEquilibrium Heat transfer always occurs from a Heat transfer always occurs from a hotter object to cooler object. hotter object to cooler object. (directionality of heat transfer)(directionality of heat transfer)

Transfer of heat continues until both Transfer of heat continues until both objects are at the same temperature objects are at the same temperature (thermal equilibrium)(thermal equilibrium)

The quantity of heat lost by a hotter The quantity of heat lost by a hotter object and the quantity of heat gained object and the quantity of heat gained by a cooler object are numerically by a cooler object are numerically equal (law of conservation of energy)equal (law of conservation of energy)

Directionality of Heat Directionality of Heat Transfer:Thermal Transfer:Thermal

EquilibriumEquilibrium Exothermic process heat is Exothermic process heat is transferred from the system to transferred from the system to the surroundings.the surroundings.

Symbol for heat q, q Symbol for heat q, q sys sys > 0> 0

Endothermic process heat is Endothermic process heat is transferred from the transferred from the surroundings to the system.surroundings to the system.

Symbol for heat q, q Symbol for heat q, q sys sys < 0< 0

Energy UnitsEnergy Units JouleJoule is the SI unit of thermal energy is the SI unit of thermal energy

related to mechanical energy related to mechanical energy 1 J = 1 kg 1 J = 1 kg . .

mm22 / s / s22

kilojoule kilojoule (kJ) is 1000 J(kJ) is 1000 J

11 calorie calorie the amount of energythe amount of energy required required to raise 1 gram of water 1 degree C to raise 1 gram of water 1 degree C (from 14.5 (from 14.5 ooC - 15.5 C - 15.5 o o C).C).

1 kilocalorie 1 kilocalorie equals 1000 calories equals 1000 calories

4.184 Joule = 1 calorie4.184 Joule = 1 calorie

Specific Heat Capacity Specific Heat Capacity and Heat Transferand Heat Transfer

The quantity of heatThe quantity of heat transferred transferred to or from an object when its to or from an object when its temperature changes depends on temperature changes depends on three things:three things: the the quantity quantity of materialof material the the size size of the temperature of the temperature changechange

the the identity identity of the material of the material gaining gaining

or losing heator losing heat

Specific Heat Capacity Specific Heat Capacity ( C ) ( C ) or ( S.H.)or ( S.H.) Specific Heat CapacitySpecific Heat Capacity

the quantity of heat required the quantity of heat required to raise the temperature of 1 gram to raise the temperature of 1 gram of a substance by one kelvin.of a substance by one kelvin.

Units are Units are J / g J / g .. K K The quantity of heat transferred The quantity of heat transferred is described by the equationis described by the equation

q = C q = C .. m m .. T T


q = C q = C .. m m .. T T

T = TT = Tfinalfinal - T - Tinitialinitial

T can be positive or negativeT can be positive or negative

when when T > 0 then q> 0 exothermic T > 0 then q> 0 exothermic when when T < 0 then q < 0 endothermic T < 0 then q < 0 endothermic

C has the units J / g C has the units J / g .. K K q will have units of Joule q will have units of Joule

Units for T Units for T and and Specific Heat Specific Heat CapacityCapacity T, Celsius = 100T, Celsius = 100ooC - 0C - 0ooC = 100C = 100o o

CC T, Kelvin = 373 K - 273 K = T, Kelvin = 373 K - 273 K = 100 C100 CUnits of specific Heat Capacity J / Units of specific Heat Capacity J / mol mol . . KK(Water 4.184 J/g (Water 4.184 J/g . . K) (18.02 g/mol) K) (18.02 g/mol) = 75.40 J/g = 75.40 J/g . . KK



In an experiment it was determined that In an experiment it was determined that 59.8 J was required to change the 59.8 J was required to change the temperature of 25.0 g of ethylene temperature of 25.0 g of ethylene glycol (a compound used as antifreeze glycol (a compound used as antifreeze in automobile engines) by 1.00 K. in automobile engines) by 1.00 K. Calculate the specific heat capacity of Calculate the specific heat capacity of ethylene glycol from these data.ethylene glycol from these data.

Use Use q = C q = C .. m m .. T TSolve for CSolve for C

Specific Heat Capacity Specific Heat Capacity and Heat Transferand Heat Transfer

• The specific heat capacity of a The specific heat capacity of a substance is determined substance is determined experimentally experimentally by accurately by accurately measuring temperature changes that measuring temperature changes that occur when heat is transferred from occur when heat is transferred from the substance to a known quantity of the substance to a known quantity of water (whose heat capacity is known)water (whose heat capacity is known)

Specific Heat Capacity of a Specific Heat Capacity of a MetalMetal

• Assumptions the water and metal will end Assumptions the water and metal will end up at the same temperature Tup at the same temperature Tfinalfinal

• Assume no heat transferred to Assume no heat transferred to surroundingssurroundings

• Heat transferred from the metal to the Heat transferred from the metal to the water, qwater, qmetalmetal, has a negative value because , has a negative value because the temp of metal dropped conversely qthe temp of metal dropped conversely qwaterwater has a positive value has a positive value

• The values of qThe values of qmetal metal andand qqwaterwater are are numerically equal but of opposite sign numerically equal but of opposite sign so so • - q- qmetal metal = = qqwater water

• qqmetal metal + + qqwaterwater = 0 = 0

Specific Heat Capacity of a Specific Heat Capacity of a MetalMetal

qqwaterwater + q + qmetalmetal = 0 = 0

CCwater water X m X mwaterwater X (T X (Tfinal final - T- Tinitial, waterinitial, water) + ) +

CCmetal metal X m X mmetalmetal X (T X (Tfinal final - T- Tinitial, metalinitial, metal) = ) = 00

A 15.5 g piece of chromium, heated to 100.0 A 15.5 g piece of chromium, heated to 100.0 ooC, is dropped into 55.5 g of water at C, is dropped into 55.5 g of water at 16.5 16.5 ooC C

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Energy and Changes of StateEnergy and Changes of State

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Heat is required to invigorate Heat is required to invigorate molecules to break bonds and molecules to break bonds and separate from each other. separate from each other.

The heat required to convert a The heat required to convert a substance from a solid substance from a solid at its at its melting point melting point to liquid is called to liquid is called the the heat of fusion.heat of fusion.

The heat required to convert The heat required to convert liquid liquid at its boiling pointat its boiling point to a to a gas called the gas called the heat of heat of vaporization.vaporization.


Heat of Fusion = 333 J / g or Heat of Fusion = 333 J / g or 6.00 kJ / mol 6.00 kJ / mol

Heat of vaporization = 2256 Heat of vaporization = 2256 J /gJ /g

Note the the units for the Note the the units for the heats have only 2 values, J and heats have only 2 values, J and g because there is no g because there is no T T


Heat of Fusion = 333 J / g or Heat of Fusion = 333 J / g or 6.00 kJ / mol6.00 kJ / mol

Heat of vaporization = 2256 J /gHeat of vaporization = 2256 J /g


What is the minimum amount of ice What is the minimum amount of ice at 0at 0ooC that must be added to the C that must be added to the contents of a can of diet cola 340 contents of a can of diet cola 340 mL to cool it from 20.5mL to cool it from 20.5ooC ? Assume C ? Assume the Specific Heat Capacity and the Specific Heat Capacity and density of diet cola are te same as density of diet cola are te same as for water and that no heat is for water and that no heat is gained or lost to the surroundings.gained or lost to the surroundings.

The First Law of The First Law of ThermodynamicsThermodynamics

If a system does work on its If a system does work on its surroundings, energy must be expended surroundings, energy must be expended and the energy content of the system and the energy content of the system will decreasewill decrease

If work is done by the surrounding on If work is done by the surrounding on a system, the energy content of the a system, the energy content of the system increases.system increases.

Work done will change the system’s Work done will change the system’s energy content.energy content.

The First Law of The First Law of ThermodynamicsThermodynamics

E is a measurable quantityE is a measurable quantity E : measure the heat transferred E : measure the heat transferred and the work done to or by the and the work done to or by the systemsystem

EnthalpyEnthalpy Most experiment is a chemical laboratory Most experiment is a chemical laboratory are carried out in vessels open to the are carried out in vessels open to the atmosphere so heat is measured under the atmosphere so heat is measured under the condition of constant pressurecondition of constant pressure

Enthalpy Enthalpy H, H, is the difference is the difference between the final and initial between the final and initial enthalpy content. enthalpy content. Negative Negative H H specify that energy is specify that energy is transferred from the system to the transferred from the system to the surroundingssurroundings

Positive Positive H H specify that energy is specify that energy is transferred from the surroundings to transferred from the surroundings to the systemthe system

State FunctionState Function A State Function is a quantity A State Function is a quantity that accompany chemical or that accompany chemical or physical changes do not depend physical changes do not depend on which path is chosen in going on which path is chosen in going from the initial state to the from the initial state to the final state.final state.

Analogy of state function. You Analogy of state function. You now have $25 in your savings now have $25 in your savings account. You could have put in account. You could have put in exactly $25 or you could have exactly $25 or you could have put in $100 and removed $75.put in $100 and removed $75.

Enthalpy Changes for Chemical Enthalpy Changes for Chemical ReactionsReactions

Enthalpy changes are specific to the Enthalpy changes are specific to the reactants and products and their amounts. reactants and products and their amounts. Both the identity of reactants and products Both the identity of reactants and products and their states (s,l,g,aq) are importantand their states (s,l,g,aq) are important

H has a negative value if heat is evolved H has a negative value if heat is evolved (an exothermic reaction). (an exothermic reaction). H has a positive H has a positive value if heat is required (an endothermic value if heat is required (an endothermic reaction).reaction).

Values of Values of H are numerically that same, but H are numerically that same, but opposite in sign for chemical reactions opposite in sign for chemical reactions that are the reverse of each other.that are the reverse of each other.

The enthalpy change depends on the molar The enthalpy change depends on the molar amounts of reactants and products. If the amounts of reactants and products. If the amounts are doubled, the enthalpy change is amounts are doubled, the enthalpy change is doubled. doubled.

Enthalpy Changes for Chemical Enthalpy Changes for Chemical ReactionsReactions

A + B -> C A + B -> C H = + 25 kJ H = + 25 kJ C -> A+ B C -> A+ B H= - 25 kJH= - 25 kJ

2A + 2B -> 2 C 2A + 2B -> 2 C H = 2 (+ 25 kJ)H = 2 (+ 25 kJ)

Sucrose CSucrose C1212HH2222OO1111 is burned releasing 5645 kJ per is burned releasing 5645 kJ per mol. What is the enthalpy change for burning 5 mol. What is the enthalpy change for burning 5 g of sucrose?g of sucrose?

When ethane (CWhen ethane (C22HH22) gas is burned according to ) gas is burned according to the following equation, the enthalpy change of the following equation, the enthalpy change of -2857.3 kJ is measured. Calculate the value of -2857.3 kJ is measured. Calculate the value of H for 15 g of CH for 15 g of C22HH66

2 C2 C22HH6(g) 6(g) + 7 O+ 7 O2(g) 2(g) -> 4 CO -> 4 CO2(g)2(g) + 6 H + 6 H22OO(g)(g)

CalorimetryCalorimetry

The heat transferred in a The heat transferred in a chemical or physical process is chemical or physical process is determined by an experimental determined by an experimental techniques call techniques call calorimeter.calorimeter. Constant pressure calorimeterConstant pressure calorimeter

(open to the air)(open to the air) Constant volume calorimeterConstant volume calorimeter

Coffee Cup CalorimetryCoffee Cup Calorimetry

Coffee Cup CalorimetryCoffee Cup Calorimetry

Insulated container with measure Insulated container with measure amount of wateramount of water

qqrxn rxn + q+ qsolution solution = = 0 0

Suppose you place 0.500 g of magnesium chips Suppose you place 0.500 g of magnesium chips in a coffee-cup calorimeter and then add in a coffee-cup calorimeter and then add 100.0 mL of 1.00 M HCl. The temperature 100.0 mL of 1.00 M HCl. The temperature of the solution increased from 22.2 of the solution increased from 22.2 ooC to C to 44.8 44.8 ooC. What is the enthalpy change for C. What is the enthalpy change for the reaction er mole of Mg? Assume the reaction er mole of Mg? Assume specific heat capacity of solution is 4.20 specific heat capacity of solution is 4.20 J/ gJ/ g..K and the density of HCl is 1.00 g/mLK and the density of HCl is 1.00 g/mL

Bomb CalorimetryBomb Calorimetry


Reaction occurs in a sealed Reaction occurs in a sealed metal container inside a metal container inside a calorimeter with watercalorimeter with water

qqrxn rxn + q+ qbomb bomb ++

qqwaterwater = 0 = 0

notenote: : qqbomb bomb has units (Chas units (Cbombbomb) ( ) ( T) T)

notenote: : qqwaterwater has units (C has units (Cbombbomb)( )( T)T)(m(mwaterwater) )


qqrxn rxn + q+ qbomb bomb ++ qqwaterwater = 0 = 0

A 1.00 g of sucrose (CA 1.00 g of sucrose (C1212HH2222OO1212) is burned in ) is burned in a bomb calorimeter. The temperature of a bomb calorimeter. The temperature of 150 g of water in the calorimeter rises 150 g of water in the calorimeter rises from 25.00from 25.00ooC to 27.32C to 27.32ooC. The heat C. The heat capacity of the bomb is 837 J/K and the capacity of the bomb is 837 J/K and the specific heat capaccity of water is 4.20 specific heat capaccity of water is 4.20 J/K . Calculate the J/K . Calculate the

(a) the heat evolved per gram of sucrose(a) the heat evolved per gram of sucrose

(b) the heat evolved per mole of sucrose. (b) the heat evolved per mole of sucrose.

Hess’s LawHess’s Law If a reaction is the sum of tow If a reaction is the sum of tow or more other reactions, or more other reactions, H for H for the overall process is the sum the overall process is the sum of of H values of those reactions.H values of those reactions.

A + B -> CA + B -> C H = H = -25kJ-25kJ

C + D -> EC + D -> E H = H = +70kJ+70kJ

A + B + D -> EA + B + D -> E H = +50 kJH = +50 kJ

Hess’s LawHess’s Law

Hess’s LawHess’s Law

Standard Molar Enthalpies of Standard Molar Enthalpies of FormationFormation

HHoof f standard molar enthalpies of standard molar enthalpies of

formationformation

the enthalpy change for the the enthalpy change for the formation of formation of

1 mol of a compound directly 1 mol of a compound directly from its component elements in from its component elements in their standard statestheir standard states


HHoof f standard molar enthalpies of standard molar enthalpies of formationformation

-the standard enthalpy of formation for an -the standard enthalpy of formation for an element in its standard state is zeroelement in its standard state is zero

-values of compounds in solution refer to -values of compounds in solution refer to the enthalpy change for the formation of 1 M the enthalpy change for the formation of 1 M solution of the compound from the elements solution of the compound from the elements making up the compounds plus th enthalpy making up the compounds plus th enthalpy change occurring when the substance change occurring when the substance dissolves in water.dissolves in water.

mostmost HHoof f are negative, indicating that are negative, indicating that

formation of most compounds from elements is formation of most compounds from elements is exothermic.exothermic.

HHoof f can be used to compare thermal can be used to compare thermal

stabilities of related compounds.stabilities of related compounds.


Enthalpy Change for a Enthalpy Change for a ReactionReaction HHoo

rxn rxn = = HHoof f (products)](products)]

HHoof f

(reactants)](reactants)]

note note HHoorxn rxn is calculated from is calculated from

HHoof f

Calculate the Calculate the HHoorxn rxn for the combustion for the combustion

of benzene (Cof benzene (C66HH6(l)6(l)) Given ) Given HHoof f

[[CC66HH6(l)6(l)] = + 48.95 kJ/mol] = + 48.95 kJ/mol

Enthalpy Change for a Enthalpy Change for a ReactionReaction

HHoorxn rxn = = HHoo

f f (products)](products)] HHoof f

(reactants)](reactants)]

In general In general

if if HHoorxn rxn < 0,< 0, the reaction is the reaction is

product-favoredproduct-favored

if if HHoorxn rxn > 0,> 0, the reaction is the reaction is

reactant-favoredreactant-favored

Energy ResourcesEnergy Resources

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Homework Questions Ch 6Homework Questions Ch 617. What quantity of heat is required to raise the temperature of 50.00 mL of water from 25.52 to 28.75 oC? The density of water at this temperature is 0.997 g/mL.

20. A 45.5-g sample of copper at 99.8 °C is dropped into a beaker containing 152 g of water at 18.5 oC. When thermal equilibrium is reached, what is the final temperature?

23. When 108 g of water at a temperature of 22.5 °C is mixed with 65.1 g of water at an unknown temperature, the final temperature of the resulting mixture is 47.9 oC. What was the initial temperature of the second sample of water?

Homework Questions Ch 6Homework Questions Ch 625. A 237-g piece of molybdenum, initially at 100.0 oC, is

dropped into 244 g of water at 10.0o C. When the system comes to thermal equilibrium, the temperature is 15.3 oC. What is the specific heat capacity of molybdenum?

29. Chloromethane, CH3Cl, is used as a topical anesthetic. What quantity of heat must be absorbed to convert 92.5 g of liquid to a vapor at its boiling point, - 24.09 oC? The heat of vaporization of CH3Cl is 21.40 kJ/mol.

30. Ethanol, C2H5OH, boils at 78.29 oC. What quantity of heat energy (in joules) is required to raise the temperature of 1.00 kg of ethanol from 20.0 °C to the boiling point and then change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is 2.44J / g- K, and its enthalpy of vaporization is 855 J/g.)

Homework Questions Ch 6Homework Questions Ch 636. Isooctane (2,2,4-trimethylpentane), one of many

hydrocarbons that make up gasoline, burns in air to give water and carbon dioxide.

2 C8H 18(l) + 25 O 2(g) -> 16 CO 2 (g) + 18 H2O ( l) H rxn = -10,922 kJ

If you burn 1.00 L of isooctane (density = 0.69 g/mL), what quantity of heat is evolved?

39. You mix 125 mL of 0.250 M CsOH with 50.0 mL of 0.625 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 21.50oC before mixing to 24.40 oC after the reaction.

CsOH(aq) + HF(aq) ----+ CsF(aq) + H20( l)

What is the enthalpy of reaction per mole of CsOH? Assume the densities of the solutions are all 1.00 g/mL and the specific heats of the solutions are 4.2J/g.K.

Homework Questions Ch 6Homework Questions Ch 642. Adding 5.44 g NH4N03(s) to 150.0 g of water in a coffee cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6 to 16.2 °C. Calculate the enthalpy change for dissolving NH4N0 3(s) in water, in kilo joules per mole. Assume that the solution (whose mass is 155.4 g) has a specific heat capacity of 4.2J/g°K.

44. Sulfur (2.56 g) is burned in a bomb calorimeter with excess 02(g). The temperature increases from 21.25 to 26.72 oC. The bomb has a heat capacity of 923 J/K, and the calorimeter contains 815 g of water. Calculate the heat evolved, per mole of S02 formed, for the reaction

S8(8) + 8 O2(g) -> 8 SO2(g))

Homework Questions Ch 6Homework Questions Ch 645. You can find the amount of heat evolved in the combustion of

carbon by carrying out the reaction in a combustion calorimeter. Suppose you burn 0.300 g of C(graphite) in an excess of O 2(g) to give CO 2(g). C(graphite) + 0 2(g) -> CO 2(g)

The temperature of the calorimeter, which contains 775 g of water, increases from 25.00 to 27.38 oC. The heat capacity of the bomb is 893 J/K. What quantity of heat is evolved per mole of carbon?

51. The enthalpies of the following reactions can be measured. C2H 4(g) + 3 O 2(g) -> 2 CO 2(g) + 2 H20( l) Ho = -1411.1 kJ

C2H5OH(l) + 3 O 2(g) -> 2 CO 2(g) + 3 H20( l) Ho = -1367.5 kJ

(a) Use these values and Hess's law to determine the enthalpy change for the reaction

(b) Draw an energy level diagram that shows the relationship between the energy quantities involved in this problem.

Homework Questions Ch 6Homework Questions Ch 652. Enthalpy changes for the following reactions can be

determined experimentally. N 2(g) + 3 H 2(g) -> 2 NH3(g) Ho = -91.8 kJ

4 NH 3(g) + 5 O 2(g) -> 4 NO (g) + 6 H20( g) Ho = -906.2 kJ

H 2(g) + 1/2 O 2(g) -> H20( g) Ho = -241.8 kJ

Use these values to determine the enthalpy change for the formation of NO(g) from the elements (an enthalpy that cannot be measured directly because the reaction is reactant-favored) . 1/2 N 2(g) + 1/2 O 2(g) -> NO(g) Ho = ?

56. (a) Write a balanced chemical equation for the formation of 1 mol of Cr2O 3(S) from Cr and O2 in their standard states and find the value for H; for Cr2O 3(S) in Appendix L.

(b) What is the standard enthalpy change if 2.4 g of chromium is oxidized to Cr2O 3(S) ?

Homework Questions Ch 6Homework Questions Ch 661. The Romans used calcium oxide (CaO) to produce a very

strong mortar in stone structures. The CaO was mixed with water to give Ca(OH)2' which reacted slowly with CO2 in the air to give CaC03.

(a) Calculate the standard enthalpy change for this reaction.

(b) What quantity of heat is evolved or absorbed if 1.00 kg of Ca(OH)2 reacts with a stoichiometric amount of CO2?

62. The standard enthalpy of formation of solid barium oxide, BaO, is - 553.5 kJ/mol, and the enthalpy of formation of barium peroxide, BaO2' is -634.3 kJ/mol.

(a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic?

(b) Draw an energy level diagram that shows the relationship between the enthalpy of this reaction and the heats of formation of BaO(s) and BaO2 (S).

Homework Questions Ch 6Homework Questions Ch 666. Use your "chemical sense" and decide if each of the following reactions is product- or reactant-favored. Calculate Ho rxn in each case, and draw an energy level diagram like those in Figure 6.18. (a) The reaction of aluminum and chlorine to produce AlCl 3(s)

(b) The decomposition of mercury(II) oxide to produce liquid mercury and oxygen gas

Homework Questions Ch 6Homework Questions Ch 674. Three 45-g ice cubes at 0 oC are dropped into 5.00 X 102 mL of tea to make ice tea. The tea was initially at 20.0 DC; when thermal equilibrium was reached the final temperature was 0 oC. How much of the ice melted and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

77. A commercial product called "Instant Car Kooler" contains 10% by weight ethanol, C2H5OH, and 90% by weight water. You spray the "Kooler" inside an overheated car. It works because thermal energy of the air in the car will be used to evaporate some of the alcohol and water. If the air inside an average size car must lose 3.6 kJ of heat to drop the air temperature from 55 to 25 oC, what mass of the ethanol water mixture must evaporate to absorb this heat? (The enthalpy of vaporization for ethanol is 850J / g and for water it is 2260J / g.)

Homework Questions Ch 6Homework Questions Ch 683. The standard molar enthalpy of formation of CS 2(g) cannot be determined directly because the compound cannot be prepared by the reaction of carbon and sulfur. It can be calculated from other enthalpy changes, however. The following enthalpies can be measured. C(s) + O 2(g) -> CO 2(g) Ho = -393.5 kJ S(s) + O 2(g) -> S02(g) Ho = -296.8 kJ CS2(g) + 3 O 2(g) -> CO 2(g) + 2 S02(g) Ho = -1103.9 kJ (a) Modify these equations to give a new set of equations, which, when added together, give the equation for the formation of CS2(g) from C(s) and S(s) in their standard states. Assign enthalpy changes to each reaction. (b) Calculate the Ho;for CS2(g). (c) Draw an energy level diagram that shows how the various enthalpies in this problem are related. (d) Is the formation of CS2(g) from its elements product-

or reactant-favored?

Homework Questions Ch 6Homework Questions Ch 684. The meals-ready-to-eat (MREs) in the military can be heated on a flameless heater. The source of energy in the heater is Mg(s) + 2 H2O (l) -> Mg(OH)2(s) + H2(g)

Calculate the enthalpy change under standard conditions (in joules) for this reaction. What quantity of magnesium is needed to supply the heat required to warm 250 mL of water (d = 1.00 g/mL) from 25 to 85 oC?

Homework Questions Ch 6Homework Questions Ch 691. Prepare a graph of molar heat capacities for metals versus their atomic weight. Use the data in Table 6.1 and the values in the following table. Does any relation exist between specific heat capacity and atomic weight? Use this relation to predict the specific heat capacity of platinum. (The specific heat capacity for platinum is given in the literature as 0.133J/g·K.) How good is the agreement between the predicted and actual values?

Metal chromium lead silver tin titanium

Specific heat capacity (J/g· K) 0.450 0.127 0.236

0.227 0.522

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Energy and Chemical Reactions Chapter 6. Energy - Thermodynamics Basic Principles Basic Principles...

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