chapter 7: energy and chemical change chemistry: the molecular nature of matter, 6e...

Chapter 7: Energy and

Chemical Change

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

2

Thermochemistry

Study of energies given off by or absorbed by reactions.

Thermodynamics Study of heat transfer or heat flowEnergy (E )

Ability to do work or to transfer heat.

Kinetic Energy (KE) Energy of motion KE = ½mv 2


Potential Energy (PE) Stored energy Exists in natural attractions

and repulsions Gravity Positive and negative charges Springs

Chemical Energy PE possessed by chemicals Stored in chemical bonds Breaking bonds requires energy Forming bonds releases energy

3


4

Your Turn!Which of the following is not a form of kinetic energy?

A. A pencil rolls across a desk

B. A pencil is sharpened

C. A pencil is heated

D. A pencil rests on a desk

E. A pencil falls to the floor


Factors Affecting Potential EnergyIncrease Potential Energy

Pull apart objects that attract each other A book is attracted to the earth

by gravity North and south poles of magnets Positive and negative charges

Push together objects that repel each other Spring compressed Same poles on two magnets Two like charges

5


Decrease Potential Energy Objects that attract each other come

together Book falls North and south poles of two magnets Positive and negative charges

Objects that repel each other move apart Spring released North poles on two

magnets Two like charges

6

Factors Affecting Potential Energy


Your Turn!Which of the following represents a decrease in the potential energy of the system?

A. A book is raised six feet above the floor.

B. A ball rolls downhill.

C. Two electrons come close together.

D. A spring is stretched completely.

E. Two atomic nuclei approach each other.

7


Law of Conservation of Energy Energy can neither be created nor

destroyed Can only be

converted fromone form to another

Total energy of universe is constant

8

Total Energy

Potential Energy

Kinetic Energy

= +


9

Temperature vs. Heat Temperature

Proportional to average kinetic energy of object’s particles

Higher average kinetic energy means Higher temperature Faster moving molecules

Heat Total amount of energy transferred between

objects Heat transfer is caused by a temperature

difference Always passes spontaneously from warmer

objects to colder objects Transfers until both are the same temperature


10

Heat Transfer Hot and cold objects placed in

contact Molecules in hot object moving

faster KE transfers from hotter to

colder object A decrease in average KE of hotter

object An increase in average KE of colder

object Over time

Average KEs of both objects becomes the same Temperature of both becomes the same

hot cold


11

Units of EnergyJoule (J) KE possessed by 2 kg object moving at

speed of 1 m/s.

If calculated value is greater than 1000 J, use kilojoules (kJ)

1 kJ = 1000 J

2

s 1m 1

kg 221

J1

2

2

smkg 1

J1


12

Units of EnergyA calorie (cal) Energy needed to raise the

temperature of 1 g H2O by 1 °C 1 cal = 4.184 J (exactly) 1 kcal = 1000 cal 1 kcal = 4.184 kJ

A nutritional Calorie (Cal) note capital C 1 Cal = 1000 cal = 1 kcal 1 kcal = 4.184 kJ


Your Turn!Which is a unit of energy?

A. Pascal

B. Newton

C. Joule

D. Watt

E. Ampere

13


14

Heat Pour hot coffee into cold cup

Heat flows from hot coffee to cold cup Faster coffee molecules bump into wall of cup Transfer kinetic energy Eventually, the cup and the coffee reach the

same temperature

Thermal Equilibrium When both cup and coffee reach same

average kinetic energy and same temperature Energy transferred through heat comes

from object’s internal energy


Internal Energy (E ) Sum of energies of all particles in system

E = total energy of system

E = potential + kinetic = PE + KE

Change in Internal Energy

E = Efinal – Einitial means change final – initial What we can actually measure

Want to know change in E associated with given process

15


16

E, Change in Internal Energy For reaction: reactants products E = Eproducts – Ereactants

Can use to do something useful Work Heat

If system absorbs energy during reaction Energy coming into system has a positive sign

(+) Final energy > initial energyEx. Photosynthesis or charging battery As system absorbs energy

Increase potential energy Available for later use


Kinetic Molecular Theory Kinetic Molecular Theory tells us

Temperature Related to average kinetic energy of

particles in object

Internal energy Related to average total molecular kinetic

energy Includes molecular potential energy

Average kinetic energy Implies distribution of kinetic energies

among molecules in object17


18

Temperature and Average Kinetic Energy

In a large collection of gas molecules Wide distribution of kinetic energy (KE) Small number with KE = 0

Collisions can momentarily stop a molecule’s motion

Very small number with very high KE Most molecules intermediate KEs Collisions tend to average kinetic

energies Result is a distribution of energies


19

Distribution of Kinetic Energy

1: lower temperature2: higher temperature

At higher temperature, distribution shifts to higher kinetic energy


20

Kinetic Energy DistributionTemperature

Average KE of all atoms and molecules in object

Average speed of particles

Kelvin temperature of sample

T (K) Avg KE = ½ mvavg2

At higher temperature Most molecules moving at higher average speed

Cold object = Small average KE Hot object = Large average KE

Note: At 0 K KE = 0 so v = 0


21

Kinetic Theory: Liquids and Solids Atoms and molecules in liquids and

solids also constantly moving Particles of solids jiggle and vibrate in

place Distributions of KEs of particles in gas,

liquid and solid are the same at same temperatures

At same temperature, gas, liquid, and solid have Same average kinetic energy But very different potential energy


Your Turn!Which statement about kinetic energy (KE) is

true?A.Atoms and molecules in gases, liquids and solids

possess KE since they are in constant motion.

B.At the same temperature, gases, liquids and solids all have different KE distributions.

C.Molecules in gases are in constant motion, while molecules in liquids and solids are not.

D.Molecules in gases and liquids are in constant motion, while molecules in solids are not.

E.As the temperature increases, molecules move more slowly.

22


E, Change in Internal Energy E = Eproducts – Ereactants

Energy change can appear entirely as heat

Can measure heat

Can’t measure Eproduct or Ereactant

Importantly, we can measure E

Energy of system depends only on its current condition

DOES NOT depend on: How system got it

What energy the system might have sometime in future

23


State of Object or System Complete list of properties that specify

object’s current condition

For Chemistry Defined by physical properties

Chemical composition Substances Number of moles

Pressure Temperature Volume

24


25

State Functions Any property that only depends on

object’s current state or condition Independence from method, path or

mechanism by which change occurs is important feature of all state functions

Some State functions, E, P, t, and V : Internal energy E Pressure P Temperature t Volume V


26

State of an object If tc = 25 °C, tells us all we need to know

Don’t need to know how system got to that temperature, just that this is where it currently is

If temperature increases to 35 °C, then change in temperature is simply: t = tfinal – tinitial

Don’t need to know how this occurred, just need to know initial and final values

What does t tell us? Change in average KE of particles in object Change in object’s total KE Heat energy


27

Defining the SystemSystem What we are interested in studying

Reaction in beaker

Surroundings Everything else

Room in which reaction is run

Boundary Separation between system and surroundings

Visible Ex. Walls of beaker Invisible Ex. Line separating warm and cold

fronts


28

Three Types of SystemsOpen System

Open to atmosphere

Gain or lose mass and energy across boundary

Most reactions done in open systems

Closed System Not open to atmosphere

Energy can cross boundary, but mass cannot

Open system

Closed system


29

Three Types of SystemsIsolated System

No energy or matter can cross boundary

Energy and mass are constant

Ex. Thermos bottle

Adiabatic Process Process that occurs in isolated

system Process where neither energy

nor matter crosses the system/surrounding boundary

Isolated system


Your Turn!A closed system can __________

A.include the surroundings.

B.absorb energy and mass.

C.not change its temperature.

D.not absorb or lose energy and mass.

E.absorb or lose energy, but not mass.

30


31

Heat (q) Cannot measure heat directly Heat (q) gained or lost by an object

Directly proportional to temperature change (t) it undergoes

Adding heat, increases temperature

Removing heat, decreases temperature

Measure changes in temperature to quantify amount of heat transferred

q = C × t C = heat capacity


32

Heat Capacity (C ) Amount of heat (q) required to raise

temperature of object by 1 °C

Heat Exchanged = Heat Capacity × t

q = C × t Units for C = J/°C or J°C –1

Extensive property Depends on two factors

1. Sample size or amount (mass) Doubling amount doubles heat capacity

2. Identity of substance Water vs. iron


33

Learning Check: Heat CapacityA cup of water is used in an experiment.

Its heat capacity is known to be 720 J/ ˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚ C?

q = 3.1 × 103 J


Learning Check: Heat CapacityIf it requires 4.184 J to raise the

temperature of 1.00 g of water by 1.00 °C, calculate the heat capacity of 1.00 g of water.

34

tq

C

4.18 J/°C


Your Turn!What is the heat capacity of 300. g of an object if it requires 2510. J to raise the temperature of the object by 2.00˚ C?

A. 4.18 J/˚ C

B. 418 J/˚ C

C. 837 J/˚ C

D. 1.26 × 103 J/˚ C

E. 2.51 × 103 J/°C

35

1255 J/°C


36

Specific Heat (s) Amount of Heat Energy needed to raise

temperature of 1 g substance by 1 °C

C = s × m or Intensive property

Ratio of two extensive properties

Units J/(g °C) or J g1 °C1

Unique to each substance Large specific heat means substance

releases large amount of heat as it cools

mC

s


Learning Check Calculate the specific heat of water if it the

heat capacity of 100. g of water is 418 J/°C.

What is the specific heat of water if heat capacity of 1.00 g of water is 4.18 J/°C?

Thus, heat capacity is independent of amount of substance

37

g .100C J/418

s

g 00.1

C J/18.4 s

4.18 J/g °C

4.18 J/g °C


Your Turn!The specific heat of silver 0.235 J g–1 °C–1. What is the heat capacity of a 100. g sample of silver?

A. 0.235 J/°C

B. 2.35 J/°C

C. 23.5 J/°C

D. 235 J/°C

E. 2.35 × 103 J/°C

38


39


40

Using Specific HeatHeat Exchanged = (Specific Heat mass)

t

q = s m t

Units = J/(g °C) g °C = J Substances with high specific heats resist

changes in temperature when heat is applied Water has unusually high specific heat

Important to body (~60% water) Used to cushion temperature changes

Why coastal temperatures are different from inland temperatures


41

Learning Check: Specific HeatCalculate the specific heat of a metal if it

takes 235 J to raise the temperature of a 32.91 g sample by 2.53 °C.


Your Turn!The specific heat of copper metal is 0.385 J/(g ˚C). How many J of heat are necessary to raise the temperature of a 1.42 kg block of copper from 25.0 ˚C to 88.5 ˚C?

A. 547 J

B. 1.37 × 104 J

C. 3.47 × 104 J

D. 34.7 J

E. 4.74 × 104 J

42


Direction of Heat Flow Heat is the energy transferred between two

objects Heat lost by one object has the same

magnitude as heat gained by other object Sign of q indicates direction of heat flow

Heat is gained, q is positive (+) Heat is lost, q is negative (–)

q1 = –q2

Ex. A piece of warm iron is placed into beaker of cool water. Iron loses 10.0 J of heat, water gains 10.0 J of heatqiron = –10.0 J qwater = +10.0 J

43


44

Your Turn!A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is false?A. The water heatsB. The skillet coolsC. The heat transfer for the skillet has a

negative (–) signD. The heat transfer for the skillet is the same as the heat transfer for the water


45

Ex. 1 Using Heat CapacityA ball bearing at 260˚ C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚ C?

Heat capacity of the cup of water = 1046 J / ˚Cqlost by ball bearing = –qgained by water

1. Determine temperature change of watert water = (37.3 ˚ C – 25.0 ˚C) = 12.3 ˚C

qwater = Cwater twater = 1046 J/˚C 12.3 ˚C2. Determine how much heat gained by water

= 12.87 ×103 J


Ex. 1 Using Heat Capacity (cont)

46

3. Determine how much heat ball bearing lostqball bearing = – qwater

4. Determine T change of ball bearing

t ball bearing = (37.3 ˚C – 260 ˚C)5. Calculate C of ball bearing

= –12.87 × 103 J

= –222.7 ˚ C

= 57.8 J/˚ C

A ball bearing at 260 ˚C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚C? C of the cup of water = 1046 J /˚ C


47

Ex. 2 Specific Heat CalculationHow much heat energy must you lose from

a 250. mL cup of coffee for the temperature to fall from 65.0 ˚C to a 37.0 ˚C? (Assume density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J g1 ˚C1)q = s m t

t = 37.0 – 65.0 ˚ C = – 28.0 ˚C

q = 4.18 J g1 ˚ C1250. mL1.00 g/mL(– 28.0 ˚ C)q = (–29.3 103 J) = –29.3 kJ


Ex. 3 Using Specific HeatIf a 38.6 g piece of gold absorbs 297 J of heat, what will the final temperature of the gold be if the initial temperature is 24.5 ˚C? The specific heat of gold is 0.129 J g–1 ˚C–1.

Need to find tfinal t = tf – ti

First use q = s m t to calculate t

Next calculate tfinal

59.6 °C = tf – 24.5 °C

tf = 59.6 °C + 24.5 °C 48

= 59.6 ˚ C

= 84.1 °C


Your Turn!What is the heat capacity of a container if 100. g of water (s = 4.18 J/g °C) at 100.˚ C are added to 100. g of water at 25 ˚C in the container and the final temperature is 61˚ C?§ 35 J/˚C§ 4.12 × 103 J/˚ C§ 21 J/˚C§ 4.53 × 103 J/˚C§ 50. J/˚C


Your Turn! - Solution What is the heat capacity of a container if 100. g of water (s = 4.18 J/g ˚C) at 100. ˚C are added to 100. g of water at 25 ˚C in the container and the final temperature is 61˚C?qlost by hot water = m × t × s

=(100. g)(61 °C – 100.˚C)(4.18 J/g ˚C) = –16,302 J

qgained by cold water = (100. g)(61 °C – 25 ˚C)(4.18J/g ˚C)

= 15,048 J qlost by system = 15,048 J + (–16,302 J) = –1254 J

qcontainer = –q lost by system = +1254 J= 35 J/°C


51

Chemical Bonds and EnergyChemical bond Attractive forces that bind

Atoms to each other in molecules, or Ions to each other in ionic compounds Give rise to compound’s potential energy

Chemical energy Potential energy stored in chemical bonds

Chemical reactions Generally involve both breaking and

making chemical bonds


Chemical ReactionsForming Bonds

Atoms that are attracted to each other are moved closer together

Decrease the potential energy of reacting system

Releases energy

Breaking Bonds Atoms that are attracted to each other are

forced apart Increase the potential energy of reacting

system Requires energy

52


53

Exothermic Reaction Reaction where products have less

chemical energy than reactants Some chemical heat energy converted to

kinetic energy Reaction releases heat energy to

surroundings Heat leaves the system; q is negative ( – ) Heat energy is a product Reaction gets warmer, temperature increases

Ex. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat


54

Endothermic Reaction Reaction where products have more chemical

energy than reactants Some kinetic energy converted to chemical energy Reaction absorbs heat from surroundings Heat added to system; q is positive (+) Heat energy is a reactant Reaction becomes colder,

temperature decreases

Ex. Photosynthesis

6CO2(g) + 6H2O(g) + solar energy C6H12O6(s) + 6O2(g)


Bond Strength Measure of how much energy is needed to

break bond or how much energy is released when bond is formed.

Larger amount of energy equals a stronger bond Weak bonds require less energy to break than

strong bonds Key to understanding reaction energies

Ex. If reaction has Weak bonds in reactants and Stronger bonds in products Heat released

55


Why Fuels Release Heat

Methane and oxygen have weaker bonds Water and carbon dioxide have stronger bonds

56


Your Turn!Chemical energy is

A. the kinetic energy resulting from violent decomposition of energetic chemicals.

B. the heat energy associated with combustion reactions.

C. the electrical energy produced by fuel cells.

D. the potential energy which resides in chemical bonds.

E. the energy living plants receive from solar radiation.

57


Heat of Reaction Amount of heat absorbed or released in

chemical reaction Determined by measuring temperature

change they cause in surroundings

Calorimeter Instrument used to measure temperature

changes Container of known heat capacity Use results to calculate heat of reaction

Calorimetry Science of using calorimeter to determine

heats of reaction58


59

Heats of Reaction Calorimeter design not standard

Depends on Type of reaction Precision desired

Usually measure heat of reaction under one of two sets of conditions Constant volume, qV

Closed, rigid container Constant pressure, qP

Open to atmosphere


60

What is Pressure? Amount of force acting on unit area

Atmospheric Pressure Pressure exerted by Earth’s atmosphere by

virtue of its weight. ~14.7 lb/in2

Container open to atmosphere Under constant P conditions P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar

areaforce

Pressure


Comparing qV and qP

Difference between qV and qP can be significant

Reactions involving large volume changes, Consumption or production of gas

Consider gas phase reaction in cylinder immersed in bucket of water Reaction vessel is cylinder topped by piston Piston can be locked in place with pin Cylinder immersed in insulated bucket containing

weighed amount of water Calorimeter consists of piston, cylinder, bucket,

and water61


Comparing qV and qP Heat capacity of calorimeter = 8.101 kJ/°C Reaction run twice, identical amounts of

reactants

Run 1: qV - Constant Volume Same reaction run once at constant

volume and once at constant pressure Pin locked; ti = 24.00 C; tf = 28.91 C

qCal = Ct

= 8.101 J/C (28.91 – 24.00)C = 39.8 kJqV = – qCal = –39.8 kJ

62


63

Comparing qV and qP

Run 2: qP

Run at atmospheric pressure

Pin unlocked

ti = 27.32 C; tf = 31.54 C

Heat absorbed by calorimeter is

qCal = Ct

= 8.101 J/C (31.54 27.32)C

= 34.2 kJ

qP = – qCal = –34.2 kJ


Comparing qV and qP

qV = -39.8 kJ

qP = -34.2 kJ

System (reacting mixture) expands, pushes against atmosphere, does work Uses up some energy that would otherwise be

heat

Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ

Expansion work or pressure volume work Minus sign means energy leaving system

64


Work ConventionWork = –P × V P = opposing pressure against which

piston pushes V = change in volume of gas during

expansion V = Vfinal – Vinitial

For Expansion Since Vfinal > Vinitial

V must be positive So expansion work is negative Work done by system

65


Your Turn!Calculate the work associated with the expansion of a gas from 152.0 L to 189.0 L at a constant pressure of 17.0 atm.

A. 629 L atm

B. –629 L atm

C. –315 L atm

D. 171 L atm

E. 315 L atm

66

Work = –P × V

w = –17.0 atm × 37.0 L

V = 189.0 L – 152.0 L


Your Turn!A chemical reaction took place in a 6 liter cylindrical enclosure fitted with a piston. Over the course of the reaction, the system underwent a volume change from 0.400 liters to 3.20 liters. Which statement below is always true?A.Work was performed on the system.

B.Work was performed by the system.

C.The internal energy of the system increased.

D.The internal energy of the system decreased.

E.The internal energy of the system remained unchanged.

67


68

First Law of Thermodynamics In an isolated system, the change in

internal energy (E) is constant:

E = Ef – Ei = 0

Can’t measure internal energy of anything Can measure changes in energy

E is state function

E = heat + work

E = q + w

E = heat input + work input


69

First Law of Thermodynamics Energy of system may be transferred as

heat or work, but not lost or gained. If we monitor heat transfers (q) of all

materials involved and all work processes, can predict that their sum will be zero Some energy transfers will be positive, gain

in energy

Some energy transfers will be negative, a loss in energy

By monitoring surroundings, we can predict what is happening to system


70

First Law of Thermodynamics E = q + w

q is (+) Heat absorbed by system (IN)

q is (–) Heat released by system (OUT)

w is (+) Work done on system (IN)

w is (–) Work done by system (OUT)

Endothermic reaction E = + Exothermic reaction E = –


71

E is Independent of Pathq and w

NOT path independent

NOT state functions

Depend on how change takes place


72

Discharge of Car BatteryPath a

Short out with wrench

All energy converted to heat, no work

E = q (w = 0)

Path b Run motor

Energy converted to work and little heat

E = w + q (w >> q)

E is same for each path Partitioning between two paths differs


Your Turn!A gas releases 3.0 J of heat and then performs 12.2 J of work. What is the change in internal energy of the gas?

A. –15.2 J

B. 15.2 J

C. –9.2 J

D. 9.2 J

E. 3.0 J

73

E = q + w

E = – 3.0 J + (–12.2 J)


Your Turn!Which of the following is not an expression for the First Law of Thermodynamics?

A.Energy is conserved

B.Energy is neither created nor destroyed

C.The energy of the universe is constant

D.Energy can be converted from work to heat

E.The energy of the universe is increasing

74


75

Bomb Calorimeter (Constant V) Apparatus for

measuring E in reactions at constant volume

Vessel in center with rigid walls

Heavily insulated vat Water bath No heat escapes

E = qv


76

Ex. 4 Calorimeter ProblemWhen 1.000 g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from 22.000 ˚C to 26.049 ˚C. a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is 9.032 kJ/˚C.

= 36.57 kJqreleased by oil = – qcalorimeter = – 36.57 kJ

qabsorbed by calorimeter = Ct = 9.032 kJ/°C × 4.049 ˚C

t = 26.049 ˚ C – 22.000 ˚C = 4.049 ˚C

kcal 1Cal 1

kJ 4.184kcal 1

g 0001kJ 5736

cal/g) (in

.

.oilq

–8.740 Cal/g oil


77

Ex. 4 Calorimeter Problem (cont)b) Olive oil is almost pure glyceryl trioleate,

C57H104O6. The equation for its combustion is

C57H104O6(l) + 80O2(g) 57CO2(g) + 52H2O

What is E for the combustion of one mole of glyceryl trioleate (MM = 885.4 g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate.

610457

610457

610457 OHC mol 1OHC g 4.885

OHC g 000.1kJ 57.36

E = qV = –3.238 × 104 kJ/mol oil


Your Turn!A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 3.74×10–3 mol sample of ethylene was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for one mole of ethylene.

A. –5.29 kJ/mol

B. 5.29 kJ/mol

C. –148 kJ/mol

D. –1410 kJ/mol

E. 1410 kJ/mol

78

= 2.47 kJ/K × 2.14 K = 5.286 kJqethylene = – qcal = – 5.286 kJ

qcal = Ct


79

Enthalpy (H) Heat of reaction at constant Pressure (qP)

H = E + PV Similar to E, but for systems at constant P Now have PV work + heat transfer H = state function At constant pressure

H = E + PV = (qP + w) + PV

If only work is P–V work, w = – P V

H = (qP + w) – w = qP


80

Enthalpy Change (H)H is a state function

H = Hfinal – Hinitial

H = Hproducts – Hreactants

Significance of sign of H

Endothermic reaction System absorbs energy from surroundings H positive

Exothermic reaction System loses energy to surroundings H negative


81

Coffee Cup Calorimeter Simple Measures qP

Open to atmosphere Constant P

Let heat be exchanged between reaction and water, and measure change in temperature Very little heat lost

Calculate heat of reaction qP = Ct


82

Ex. 5 Coffee Cup CalorimetryNaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = 4.184 J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(aq)

qabsorbed by solution = mass s t

massHCl = 50.0 mL 1.02 g/mL = 51.0 g

massNaOH = 50.0 mL 1.04 g/mL = 52.0 g

massfinal solution = 51.0 g + 52.0 g = 103.0 g

t = (32.2 – 25.5) °C = 6.7 °C


83

Ex. 5 Coffee Cup Calorimetryqcal = 103.0 g 4.184 J g–1 °C–1 6.7 °C = 2890 J

Rounds to qcal = 2.9 103 J = 2.9 kJqrxn = –qcalorimeter = –2.9 kJ

= 0.0500 mol HCl

Heat evolved per mol HCl =

= -58 kJ/mol

soln HClL 1 HCl mol 1

soln HClL 0500.0


Ex. 5: Coffee Cup CalorimetryWhen 50.0 mL of 0.987 M H2SO4 is added to 25.0

mL of 2.00 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution increases to 33.9 °C. Calculate H in kJ/mole of limiting reactant. Assume: specific heat of the solution is 4.184 J/g°C, density is 1.00 g/mL, and the calorimeter absorbs a negligible amount of heat.

84

2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(aq)

Write balanced equation

qsoln = 75.0 g × (33.9 – 25.0)°C × 4.184 J/g°C = 2.8×103 J

Determine heat absorbed by calorimeter

masssoln = (25.0 mL + 50.0 mL) × 1.00 g/mL = 75.0 g


Ex. 6 Determine Limiting Reagent

= 0.04935 mol H2SO4 present

= 0.0500 mol NaOH present NaOH is

limiting

= 0.0987 mol NaOH needed

SOHL 1 SOH mol .9870

SOHmL 1000 SOHL 1

SOHmL 50.042

42

42

4242

NaOHL 1 NaOH mol 2

NaOHmL 1000NaOHL 1

NaOHmL 25.0

= –56 kJ/mol J1000kJ 1

NaOH mol 0.0500 J102.8 3

H


86

Your Turn!A 43.29 g sample of solid is transferred from boiling water (t = 99.8 ˚ C) to 152 g water at 22.5 ˚C in a coffee cup. The temperature of the water rose to 24.3 ˚C. Calculate the specific heat of the solid.

A. –1.1 × 103 J g–1 ˚C–1

B. 1.1 × 103 J g–1 ˚C–1

C. 1.0 J g–1 ˚C–1

D. 0.35 J g–1 ˚C–1

E. 0.25 J g–1 ˚C–1

q = m × s × t

= 1.1 × 103 J

qsample = – qwater = – 1.1 × 103 J

C99.8) – (24.3g 43.29

J1011 3

.s


87

Enthalpy Changes in Chemical Reactions

Focus on systems Endothermic

Reactants + heat products

Exothermic Reactants products + heat

Want convenient way to use enthalpies to calculate reaction enthalpies

Need way to tabulate enthalpies of reactions


The Standard State A standard state specifies all the necessary

parameters to describe a system. Generally this includes the pressure, temperature, and amount and state of the substances involved.

Standard state in thermochemistry Pressure = 1 atmosphere Temperature = 25 °C = 298 K Amount of substance = 1 mol (for formation

reactions and phase transitions) Amount of substance = moles in an equation

(balanced with the smallest whole number coefficients)

88


Thermodynamic Quantities E and H are state functions and are also

extensive properties E and H are measurable changes but still

extensive properties. Often used where n is not standard, or

specified

E ° and H ° are standard changes and intensive properties

Units of kJ /mol for formation reactions and phase changes (e.g. H °f or H °vap)

Units of kJ for balanced chemical equations (H °reaction)

89


90

H in Chemical Reactions Standard Conditions for H 's

25 °C and 1 atm and 1 mole

Standard Heat of Reaction (H ° ) Enthalpy change for reaction at 1 atm and 25 °C

Example:

N2(g) + 3H2(g) 2 NH3(g)

1.000 mol 3.000 mol 2.000 mol When N2 and H2 react to form NH3 at 25 °C and

1 atm 92.38 kJ released H= –92.38 kJ


91

Thermochemical Equation Write H immediately after equation

N2(g) + 3H2(g) 2NH3(g) H = –92.38 kJ

Must give physical states of products and reactants

H different for different states

CH4(g) + 2O2(g) CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ

Difference is equal to the energy to vaporize water


92

Thermochemical Equation Write H immediately after equation

N2(g) + 3H2(g) 2NH3(g) H= –92.38 kJ

Assumes coefficients is the number of moles 92.38 kJ released when 2 moles of NH3 formed

If 10 mole of NH3 formed

5N2(g) + 15H2(g) 10NH3(g) H= –461.9 kJ

H° = (5 × –92.38 kJ) = – 461.9 kJ Can have fractional coefficients

Fraction of mole, NOT fraction of molecule

½N2(g) + 3/2H2(g) NH3(g) H°rxn = –46.19 kJ


93

State Matters!C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)

ΔH °rxn= –2043 kJ

C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )

ΔH °rxn = –2219 kJ

Note: there is difference in energy because states do not match

If H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol

4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol

Or –2219 kJ + 176 kJ = –2043 kJ


94

Consider the following reaction:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

ΔE ° = –2511 kJ

The reactants (acetylene and oxygen) have 2511 kJ more energy than products. How many kJ are released for 1 mol C2H2?

Learning Check:

–1,256 kJ


95

Learning Check: Given the equation below, how many kJ are required for 44 g CO2 (MM = 44.01 g/mol)?

6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g) ΔH˚reaction = 2816 kJ

If 100. kJ are provided, what mass of CO2 can be converted to glucose?

9.38 g

470 kJ22

22 CO mol 6

kJ 2816CO g 44.01

CO mol 1CO g 44

2

22

CO mol 1 CO g 44.0

kJ 2816CO mol 6

kJ 100


Your Turn!Based on the reaction

CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g)

H˚reaction = – 434 kJ/mol CH4

What energy change occurs when 1.2 moles of methane reacts?

A. – 3.6 × 102 kJ

B. +5.2 × 102 kJ

C. – 4.3 × 102 kJ

D. +3.6 × 102 kJ

E. – 5.2 × 102 kJ

96

H = –434 kJ/mol × 1.2 mol

H = –520.8 kJ = 5.2 × 102 kJ


97

Running Thermochemical Equations in Reverse

Consider

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H°reaction = – 802.3 kJ

Reverse thermochemical equation Must change sign of H

CO2(g) + 2H2O(g) CH4(g) + 2O2(g)

H°reaction = 802.3 kJ


98

Reverse Thermochemical Equation, Changes sign of H Makes sense:

Get energy out when form products

Must put energy in to go back to reactants

Consequence of Law of Conservation of Energy Like mathematical equation

If you know H ° for reaction, you also know H ° for the reverse


99

Multiple Paths; Same H ° Can often get from reactants to

products by several different paths

Should get same H ° Enthalpy is state function and path

independent Let’s see if this is true

Intermediate A

Products Reactants

Intermediate B


100

Ex. 7 Multiple Paths; Same H °Path a: Single step

C(s) + O2(g) CO2(g) H°rxn = –393.5 kJ

Path b: Two step Step 1: C(s) + ½O2(g) CO(g) H °rxn = –110.5 kJ

Step 2: CO(g) + ½O2(g) CO2(g) H °rxn = –283.0 kJ

Net Rxn: C(s) + O2(g) CO2(g) H °rxn = –393.5 kJ

Chemically and thermochemically, identical results

True for exothermic reaction or for endothermic reaction


101

Ex. 8 Multiple Paths; Same H

°rxnPath a: N2(g) + 2O2(g) 2NO2(g) H °rxn = 68 kJ

Path b:

Step 1: N2(g) + O2(g) 2NO(g) H °rxn = 180. kJ

Step 2: 2NO(g) + O2(g) 2NO2(g) H °rxn = –112 kJ

Net rxn: N2(g) + 2O2(g) 2NO2(g) H °rxn = 68 kJ

Hess’s Law of Heat Summation For any reaction that can be written into steps,

value of H °rxn for reactions = sum of H °rxn values of each individual step


Enthalpy Diagrams Graphical description of Hess’ Law

Vertical axis = enthalpy scale Horizontal line =various states of reactions Higher up = larger enthalpy Lower down = smaller enthalpy

102


Enthalpy Diagrams Use to measure Hrxn

Arrow down Hrxn = negative

Arrow up Hrxn = positive

Calculate cycle One step process =

sum of two step process

103

Ex. H2O2(l ) H2O(l ) + ½O2(g)

–286 kJ = –188 kJ + Hrxn

Hrxn = –286 kJ – (–188 kJ )

Hrxn = –98 kJ


104

Hess’s LawHess’s Law of Heat Summation Going from reactants to products Enthalpy change is same whether reaction

takes place in one step or many Chief Use

Calculation of H °rxn for reaction that can’t be measured directly

Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction


105

Rules for Manipulating Thermochemical Equations

1. When equation is reversed, sign of H°rxn must also be reversed.

2. If all coefficients of equation are multiplied or divided by same factor, value of H°rxn must likewise be multiplied or divided by that factor

3. Formulas canceled from both sides of equation must be for substance in same physical states


106

Strategy for Adding Reactions Together:

1. Choose most complex compound in equation for one-step path

2. Choose equation in multi-step path that contains that compound

3. Write equation down so that compound is on appropriate side of equation has appropriate coefficient for our reaction

4. Repeat steps 1 – 3 for next most complex compound, etc.


107

Strategy for Adding Reactions (Cont.)5. Choose equation that allows you to

cancel intermediates

multiply by appropriate coefficient

6. Add reactions together and cancel like terms

7. Add energies together, modifying enthalpy values in same way equation modified

If reversed equation, change sign on enthalpy

If doubled equation, double energy


108

Ex. 9 Calculate H°rxn for C (s, graphite) C (s, diamond)

Given C (s, gr) + O2(g) CO2(g) H°rxn = –394 kJ

C (s, dia) + O2(g) CO2(g) H°rxn = –396 kJ

To get desired equation, must reverse second equation and add resulting equations

C(s, gr) + O2(g) CO2(g) H°rxn = –394 kJ

CO2(g) C(s, dia) + O2(g) H°rxn = –(–396 kJ)

C(s, gr) + O2(g) + CO2(g) C(s, dia) + O2(g) + CO2(g)

H° = –394 kJ + 396 kJ = + 2 kJ

–1[ ]


109

Learning Check: Ex. 10

Calculate H °rxn for 2 C (s, gr) + H2(g) C2H2(g)

Given the following:

a. C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(l ) H °rxn = –

1299.6 kJ

b. C(s, gr) + O2(g) CO2(g) H °rxn = –393.5 kJ

c. H2(g) + ½O2(g) H2O(l ) H °rxn = –285.8 kJ


110

Ex.10 Calculate for 2C(s, gr) + H2(g) C2H2(g)

2CO2(g) + H2O(l ) C2H2(g) + 5/2O2(g) H°rxn = – (–1299.6 kJ) = +1299.6 kJ

2C(s, gr) + 2O2(g) 2CO2(g) H°rxn =(2 –393.5 kJ) = –787.0 kJ

H2(g) + ½O2(g) H2O(l ) H°rxn = –285.8 kJ

2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g) C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l )

2C(s, gr) + H2(g) C2H2(g) H°rxn = +226.8 kJ


Your Turn!Which of the following is a statement of Hess's Law?A.H for a reaction in the forward direction is equal to H for the reaction in the reverse direction.B.H for a reaction depends on the physical states of the reactants and products.C.If a reaction takes place in steps, H for the reaction will be the sum of Hs for the individual steps.D.If you multiply a reaction by a number, you multiply H by the same number.E.H for a reaction in the forward direction is equal in magnitude and opposite in sign to H for the reaction in the reverse direction.

111


Your Turn!Given the following data: C2H2(g) + O2(g) 2CO2(g) + H2O(l ) H rxn= –1300.

kJ C(s) + O2(g) CO2(g) Hrxn = –394 kJ

H2(g) + O2(g) H2O(l ) Hrxn = –286 kJ

Calculate for the reaction2C(s) + H2(g) C2H2(g)

A. 226 kJ

B. –1980 kJ

C. –620 kJ

D. –226 kJ

E. 620 kJ112

Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)


113

Tabulating H° values Need to Tabulate H° values Major problem is vast number of reactions Define standard reaction and tabulate

these Use Hess’s Law to calculate H° for any

other reaction

Standard Enthalpy of Formation, Hf° Amount of heat absorbed or evolved when one

mole of substance is formed at 1 atm (1 bar) and 25 °C (298 K) from elements in their standard states

Standard Heat of Formation


114

Standard State Most stable form and physical state of element

at 1 atm (1 bar) and 25 °C (298 K)

Element Standard state

O O2(g)

C C (s, gr)

H H2(g)

Al Al(s)

Ne Ne(g) See Appendix C in back of textbook and

Table 7.2

Note: All Hf° of elements in their standard states = 0

Forming element from itself.


115

Uses of Standard Enthalpy (Heat) of Formation, Hf°

1. From definition of Hf°, can write balanced equations directly

Hf° of C2H5OH(l )

2C(s, gr) + 3H2(g) + ½O2(g) C2H5OH(l ) Hf° = –277.03

kJ/mol

Hf° of Fe2O3(s)

2Fe(s) + 3/2O2(g) Fe2O3(s) Hf° = –822.2 kJ/mol


Your Turn!What is the reaction that corresponds to the standard enthalpy of formation of NaHCO3(s ), Hf ° = – 947.7 kJ/mol?

A. Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr) NaHCO3(s)

B. Na+(g) + H+(g) + 3O2–(g) + C4+(g) NaHCO3(s)

C. Na+(aq) + H+(aq) + 3O2–(aq) + C4+(aq) NaHCO3(s)

D. NaHCO3(s) Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr)

E. Na+(aq) + HCO3–(aq) NaHCO3(s)

116


117

Using Hf°2. Way to apply Hess’s Law without needing to

manipulate thermochemical equations

Consider the reaction:aA + bB cC + dD

H°reaction = c × H°f(C) + d × H°f(D) – {a×H°f(A) +

b×H°f(B)} H°rxn has units of kJ because Coefficients heats of formation have units of mol

kJ/mol

H°reaction = –Sum of all H°f of all of the products

Sum of all H°f of all of the reactants

H°rxn has units of kJ H°f has units of kJ/mol


118

Ex. 11 Calculate H°rxn Using Hf°

Calculate H°rxn using Hf° data for the reactionSO3(g) SO2(g) + ½O2(g)

1. Multiply each Hf° (in kJ/mol) by the number of moles in the equation2. Add the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each product3. Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each reactant

)(SO)(O)(SO )(3f)(2f21)(2frxn ggg HHHH

H°rxn = 99 kJ

H°rxn has units of kJH°f has units of kJ/mol


119

Learning CheckCalculate H°rxn using Hf° for the reaction

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l )

H°rxn = [136 – 1715.4 + 184] kJ

H°rxn = –1395 kJ


120

Check Using Hess’s Law4 ×[NH3(g) ½N2(g) + 3/2H2(g)] – 4 × Hf°(NH3,

g)

7 ×[ O2(g) O2(g) ] –7 × Hf°(O2, g)

4 ×[ O2(g) + ½ N2(g) NO2(g)] 4 × Hf°(NO2, g)

6 ×[ H2(g) + ½ O2(g) H2O(l ) ] 6 × Hf°(H2O, l)

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l )Same as before


121

Other Calculations Don’t always want to know H°rxn

Can use Hess’s Law and H°rxn to calculate Hf° for compound where not known

Ex. Given the following data, what is the value of Hf°(C2H3O2

–, aq)?

Na+(aq) + C2H3O2–(aq) + 3H2O(l )

NaC2H3O2·3H2O(s)

H°rxn = –19.7 kJ/mol

Naaq) H f= –239.7 kJ/mol

NaC2H3O2•3H2O(s) H f= 710.4 kJ/mol

H2O(l) H f= 285.9 kJ/mol


122

Ex. 13 cont.H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+,

aq) – Hf° (C2H3O2–, aq) – 3Hf° (H2O, l )

Rearranging

Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) –

Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l)

Hf°(C2H3O2–, aq) =

–710.4 kJ/mol – (–239.7kJ/mol) – (–19.7 kJ/mol) – 3(–285.9 kJ/mol)

= +406.7 kJ/mol


123

Learning CheckCalculate H for this reaction using Hf° data.

2Fe(s) + 6H2O(l) 2Fe(OH)3(s) + 3H2(g)

Hf° 0 –285.8 –696.5 0

H°rxn = 2×Hf°(Fe(OH)3, s) + 3×Hf°(H2, g) – 2× Hf°(Fe, s) – 6×Hf°(H2O, l )

H°rxn = 2 mol× (– 696.5 kJ/mol) + 3×0 – 2×0 – 6 mol× (–285.8 kJ/mol)

H°rxn = –1393 kJ + 1714.8 kJ

H°rxn = 321.8 kJ


124

Learning CheckCalculate H°rxn for this reaction using Hf° data.

CO2(g) + 2H2O(l ) 2O2(g) + CH4(g)

Hf° –393.5 –285.8 0 – 74.8

H°rxn = 2×Hf°(O2, g) + Hf°(CH4, g) –Hf°(CO2, g) – 2× Hf°(H2O, l )

H°rxn = 2 × 0 + 1 mol × (–74.8 kJ/mol) – 1 mol × (–393.5 kJ/mol) – 2 mol × (–285.8 kJ/mol)

H°rxn = –74.8 kJ + 393.5 kJ + 571.6 kJ

H°rxn = 890.3 kJ

chapter 7: energy and chemical change chemistry: the molecular nature of matter, 6e...

Documents

energy energy

potential energy slide

potential energy objects

form of kinetic energy

e jespersenbradyhyslop

jespersenbradyhyslop

molecular nature of

bonds releases energy