Ch. 7 Thermochemistry

thermochemistry

the study of the relationship between chemistry and energy

energy

the capacity to do work

work

force acting through a distance

heat

the flow of energy due to a temperature difference

kinetic energy

motion

thermal energy

temperature

potential energy

position or composition

chemical energy

atom and molecular composition

law of conservation of energy

energy can neither be created nor destroyed. it can only be transferred or transformed

system

the objects under study

open system

can exchange mass and energy

closed system

allows energy transfer but not mass

isolated system

does not transfer mass or energy

surroundings

everything outside of the system

joule

kg x m^2/s^2

calorie

the amount of energy required to raise the temperature of one gram of water by one degree Celsius (1 cal= 4.184 J)

Calorie (kilocalorie)

1000 calories

kWh

kilowatt-hour: electric bill units and is equivalent to 3.60 x 10^6 J

First Law

the total energy of the universe is considered constant

state function

value depends only on the state of the system, not how the system arrived at that state

internal energy

the sum of the kinetic and potential energies of all particles that compose the system

E = Eproducts - Ereactants

chemical reaction

E = Efinal - Einitial

a change in internal energy

energy must be conserved, when energy is exchanged

Esystem = - Esurroundings

out of system

Esys is -

into surroundings

Esurr is +

Into system

Esys is +

from the surroundings

Esurr is -

a system can exchange energy with its surroundings through what

heat and work

Energy = heat + work

q(heat)

\+system gains thermal energy

\-system loses thermal energy

w(work)

\+work done on the system

\-work done by the system

Energy

\+energy flow into the system

\-energy flows out of the system

ice melts

heat

\+into system

metal cylinder rolled up a ramp

work

\+into system

steam burns the skin

heat

\-out of system

the potato shoots out of the cannon and emits heat to the surroundings. If the burning of the fuel performs 855 J of work on the potato and produces 1422 J of heat, what is the energy for the burning of the fuel?

E = q+w

heat is released by the system -1422 J

work done on the potato by the system -855 J

\-1422 J + (-855) = -2277 J

a cylinder and piston system is warmed by an external flame. the contents of the cylinder expand, doing work on the surroundings by pushing the piston outward against the external pressure. if the system absorbs 559 J of heat and does 488 J of work during the expansion, what is the value for the energy?

E = q + w

heat absorbed +559 J

work done on surroundings -488 J

=71 J

thermal equilibrium

thermal energy exchange between a system and its surroundings caused by a temperature difference to reach

temperature is a measure of

thermal energy

heat capacity (C)

the amount of heat required to raise the temperature of a substance by 1 C

Specific heat capacity (Cs)

amount of heat required to raise the temperature of one gram of a substance by 1 C (J/gxC)

molar heat capacity

amount of heat required to raise the temperature of one mole of a substance by 1 C (J/mol x C)

a copper penny is found in the snow. how much heat is absorbed by the penny as it warms from -8.0 C to 37 C?

penny mass = 3.10 grams

Scu=0.385 J/gxC

q = m x s x T

(3.10)(0.385)(37-(-8.0))

53\.7 J

the mass of 4.7 g absorbs 57.2 J of heat as the rock temperature rises from 25 C to 57 C. Determine the specific heat capacity and decide if the rock is gold or not.

sAu= 0.128 J/g x C

q= m x s x T

57\.2 J=4.7(s)(57-25)

57\.2 J=150.4s

0\.380J/g x C=s

Not gold

a 55.0 g aluminum block initially at 27.5 C absorbs 725 J of heat. what is the final temperature of the aluminum?

sAl=0.903 J/g x C

\

q=m x s x T

725 J = 55.0(0.903)(Tf-27.5)

725 =49.7 (Tf-27.5)

725 = 49.7Tf - 136.7

2092 = 49.7Tf

Tf=42.1 C

a large water jug and a rock of equal mass are heated to 38 C (100 F). Which one will stay warmer longer?

the water, because it will release more heat as it cools

thermal energy transfer

heat transfer occurs between two objects of different temperature. If thermal isolation is assumed, the heat lost by one equals the heat gained by the other

qsys=-qsurr

a 32.5 g cube of Al initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium?

qsys=-qsurr

msT=-(msT)

32\.5(0.903)(Tf-45.8)=-(105.3)(4.18)(Tf-15.4)

29\.3(Tf-45.8)=-(440.2)(Tf-15.4)

29\.3Tf-1342=-(440.2Tf-6779)

29\.3Tf-1342=-440.2Tf+6779

469\.5Tf=8121

Tf=17.3 C

a block of copper of unknown mass has an initial temperature of 65.4 C. The copper is immersed in a beaker containing 95.7 g of water at 22.7 C. When the two substances reach thermal equilibrium, the final temperature is 24.2 C. What is the mass of the copper block?

sCu=0.385

qsys=-qsurr

qCu=-qH2O

msT=-(msT)

m(0.385)(24.2-65.4)=-(95.7)(4.18)(24.2-22.7)

\-15.9m=-600

m=37.8 g

substances A and B, initially at different temperatures, come into contact and reach equilibrium. the mass of A is twice that of B. specific heat capacity of B is twice that of A. which statement is true?

the Tf is exactly midway between the Ti of A and B

pressure-volume work

occurs when the force is caused by a volume change against external pressure

w=-PextV

\

a cylinder within a piston expands from a volume of 1.00 L to a volume of 2.00 L against an external pressure of 1.00 atm. How much work (in J) was done by the expansion?

w=-(1.00)(2.00-1.00)=-1.00 L x atm (101.3 J/L x atm)=-101 J

to inflate a balloon, some pressure-volume work on the surroundings must be done. if a balloon is inflated from a volume of 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work (in J) is done?

(101.3 J = 1 L x atm)

w = -PextV

V=Vf-Vi

w=-(1.00 atm)(1.85L-0.100L)

w=-(1.00)(1.75L)

w=-1.75 atm

\-1.75 L x atm (101.3 J/1 L x atm)= -177 J

a cylinder equipped with a piston expands against an external pressure of 1.58 atm. if the initial volume is 0.485 L and the final volume is 1.245 L , how much work in J is done?

w=-PextV

V=Vf=Vi

101\.3 J = 1 L x atm

V=1.245-0.485

V=0.76 L

w=-1.58 atm(0.76 J)

w=-1.20 L x atm

\-1.20 L x atm(101.3J/1 L x atm)=-122 J

when fuel is burned in a cylinder equipped with a piston, the volume expands from 0.255 L to 1.45 L against an external pressure of 1.02 atm. In addition, 875 J is emitted as heat. what is the energy for the burning of the fuel?

V=1.45-0.255

V=1.195

w=-1.02(1.195)

w=-1.22(101.3/1)=-124 J

E=q + w

E=-875 J + -124 J

E=-999J

Calorimetry

measure of heat change in a chemical reaction

Bomb calorimeter

measures the energy for constant volume combustion reactions

the heat gained by the calorimeter will be the amount of heat given off by the reaction

qcal=-qrxn

when 1.010 g of sucrose (C12H22O11) undergoes combustion in a bomb calorimeter, the temperature rose from 24.92 C to 28.33 C. Determine the energy for the reaction in kJ/mol. The heat capacity of the bomb calorimeter is 4.90 kJ/C.

qcal=Ccal x T

qcal=(4.90)(28.33-24.92)

=4.90(3.41)

=16.7 kJ so qrxn=-16.7

1\.010 g sucrose(1 mol/342.34)=2.950 10^-3 mol sucrose

E=(-16.7/2.950 x 10^-3)=-5.66 x 10^3 kJ/mol

when 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rose from 25.87 C to 38.13 C. determine the energy for the reaction in kJ/mol hexane. the heat capacity of the bomb calorimeter is 5.73 kJ/C

qcal=573(38.13-25.87)

qcal=70.2 kJ so qrxn=-70.2

1\.550(1 mol hexane/86.2g)=0.0180 mol C6H14

E=-70.2/0.0180 mol

E=-3.90 x 10^3 kJ/mol

the combustion of toluene (C7H8) has a E=-3.91 x 10^3 kJ/mol. when 1.55 g undergoes combustion in a bomb calorimeter, the temperature rose from 23.12 C to 37.57 C. what is the heat capacity of the calorimeter?

1\.55 g (1 mol/92.15 g C7H8)=0.01682 mol C7H8

\-3.91 x 10^3=qrxn/0.01682

\-65.8=qrxn

so qcal =65.8

65\.8 =Ccal (37.57-23.12)

65\.8=14.45Ccal

Ccal=4.55 kJ/C

enthalpy (H)

the sum of its internal energy and the product of its pressure and volume

H=E+ PV

Lighters are usually fueled by butane (C4H10), when 1 mol of butane burns at constant pressure, it produces 2658 kJ of heat and does 3 kJ of work. What are the values for H and E for the combustion of one mole of butane?

H=-2658 kJ; E=-2661 kJ

endothermic

a positive H indicates heat flow into the system

exothermic

a negative H indicates heat flow out of the system

sweat evaporating from skin

endothermic (+)

water freezing in a freezer

exothermic (-)

wood burning in a fire e

exothermic (-)

an ice cube melting

endothermic (+)

nail polish remover quickly evaporating after it is accidently spilled on the skin

endothermic (+)

gasoline burning within the cylinder of an automobile engine

exothermic (-)

an endothermic reaction occurs in a flask. what happens to the temperature of the flask?

the temperature falls

2 A → AA and Hrxn = -51.0 J

What is the heat associated with the reaction of 6 moles of A?

\-153 J

a propane tank in a home barbeque contains 13.2 kg C3H8. calculate the heat (in kJ) associated with the complete combustion of all the propane in the tank.

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Hrxn=-2044 kJ

13\.2 kg(1000g/1 kg)(1 mol C3H8/44.11 g C3H8)(-2044 kJ/1 mol C3H8)= -6.12 x 10^5 kJ

Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3

4 NH3 + 5 O2 → 4 NO + 6 H2O

Hrxn=-906 kJ

155g(1 mol/17.04g)(-906kJ/4 mol NH3)= -2.06 x 10^3 kJ or -2060

what mass of butane is necessary to produce 1.55 x 10^3 kJ of heat. what mass of CO2 is produced?

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

Hrxn=-2658 kJ

\-1.55 x 10^3 kJ (1 mol C4H10/-2658 kJ)(58.14 /1 mol C4H10)=33.9 C4H10

33\.9 g C4H10(1 mol/58.14)(4 mol CO2/1mol C4H10)(44.01g/1 mol CO2)= 103 g CO2

a calorimeter is used to measure what

the enthalpy of aqueous reactions

qsoln=(msoln)(Csoln)(T)

an insulated cup prevents heat from escaping

therefore: qrxn=-qsoln

if 0.158 g of Mg is reacted with sufficient HCl (100.0 mL0 for complete reaction, the temperature of the solution rises from 25.6 C to 32.8 C. Determine Hrxn

(1.00 g/mL for solution density and Cs, soln=4.18 J/g x C

qrxn=-qsol

Hrxn=qrxn/mol Mg

\-(100g)(4.18)(32.8-25.6)

=-(100)(4.18)(7.2)

=-3010 J

moles of Mg = 0.158g Mg(1 mol/24.31g)=6.50 x 10^-3 mol

Hrxn=-3010J/6.50 x 10^-3 mol=-4.63 x 10^5 J/mol

AgNO3 + HCl → AgCl + HNO3

when 50.0 mL of 0.100 M AgNO3 is combined with 50.0 mL of 0.100 M HCl the temperature changes from 23.40 C to 24.21 C. Calculate Hrxn

Hsol=-Hrxn

qsol=(100)(4.184)(24.21-23.40)

=338.9 J

so qxrn=-338.9 J

q = msT

molxrn=0.500L(0.100mol/1L)= 0.00500 mol

Hrxn=qrxn/mol=(-338.9/0.00500)=-6.78 x 10^4 J

the same reaction, with the same amount of reactant, is conducted in a bomb and coffee-cup calorimeter. In one of the calorimeters, qrxn=-12.5 kJ and in the other qrxn=-11.8 kJ. which value was obtained from the bomb calorimeter?

\-12.5 kJ

reaction enthalpy

1. if the reaction is multiplied by some amount, so is the Hrxn
2. if the reaction is reversed, Hrxn changes sign
3. if a reaction is expressed as a series of steps, the Hrxn is the sum of the steps values, Hess’s Law

Hess’s law

if a reaction is expressed as a series of steps, the Hrxn is the sum of the steps values

2 A + B → C

Hrxn=122 J

What is the Hrxn for the reaction 2C→ 4A + 2B?

\-244 J

3 C + 4 H2 → C3H8

Use these reactions with known H’s

a. C3H8 + 5O2 → 3 CO2 + 4 H2O H=-2043 kJ

b. C + O2 → CO2 H=-393.5 kJ

c. 2H2 + O2 → 2 H2O H=-483.6 kJ

a1. 3 CO2 + 4 H2O → C3H8 + 5 O2 H=2043 kJ

b1. 3 C + 3 O2 → 3 CO2 H=-1180 kJ

c1. 4 H2 + 2 O2 → 4 H2O H=-967.2 kJ

\
3 C + 4 H2 → C3H8 H=-104 kJ

N2O + NO2 → 3 NO

a. 2 NO + O2 → 2 NO2 H=-113.1 kJ

b. N2 + O2 → 2 NO H=182.6 kJ

c. 2 N2O → 2 N2 + O2 H=-163.2 kJ

a1. NO2 → NO + 1/2 O2 H=56.55 kJ

b1. N2 + O2 → 2 NO H=182.6 kJ

c1. N2O → N2 + 1/2 O2 H=-81.6

N2O + NO2 → 3 NO H=157.6 kJ

3 H2 + O3 → 3 H2O

a. 2 H2 + O2 → 2 H2O H=-483.6 kJ

b. 3 O2 → 2 O3 H=285.4 kJ

a1. 3 H2 + 3/2 O2 → 3 H2O H=-725.4

b1. O3 → 3/2 O2 H= -142.7

=-868.1

Standard state

a. for a gas: pure gas at 1 atm

b. liquid or solid: pure substance in its most stable form at 25 C

c. substance in solution: substance in solution of exactly 1 M

Standard enthalpy change (H)

the change in enthalpy when all reactants and products are in their standard states

standard enthalpy of formation (Hf)

a. pure compound: the change in enthalpy when 1 mole of the compound forms from its constituent elements

b. pure element: Hf=0

Pure compound

the change in enthalpy when 1 mole of the compound forms from its constituent elements

pure element

Hf=0

4 NH3 + 5 O2 → 4 NO + 6 H2O

Hf NH3=-45.9

Hf O2=0

Hf NO= 91.3

Hf H2O=-241.8

sum of products - sum of reactants

4(91.3)+6(-241.8) - 4(-45.9) + 5(0.0)

=-1085.6 -(-183.6)

=-902.0 kJ

2 Al + Fe2O3 → Al2O3 + 2 Fe + heat

Hf Al=0.0

Hf Fe2O3 =-824.2

Hf Al2O3 = -1675.7

Hf Fe=0.0

Hrxn= (-1675.7) + 2(0.0) - (2(0.0)+-824.2)

=-851.5 kJ

a city of 100,000 people uses approximately 1.0 x 10^11 kJ of energy per day. Suppose all of that energy comes from the combustion of liquid octane. Use the standard enthalpies of formation to determine Hrxn for teh combustion of octane and then determine the number of kilograms necessary to provide this amount of energy.

C8H18 + 25/2 O2 → 9 H2O + 8 CO2

Hf C8H18 = -250.1

Hf O2= 0.0

Hf H2O = -241.8

Hf CO2 = -393.5

Hrxn= 9(-241.8) + 8(-393.3) -(-250.1+25/2(0.0))

=-5324.2 -(-250.1)

=-5074.1

A city of 100,000 people uses approximately 1.0 x 10^11 kJ of energy per day. Determine the number of kilograms necessary to provide this amount of energy

C8H18 + 25/2 O2 → 9 H2O + 8 CO2

Hrxn=-5074.1 kJ

\

1\.0 x 10^11 (1mol/-5074.1 kJ)(114.26g/1 mol)(1 kg/1000 mol)= 2.3 x 10^6 kg C8H18

4 Fe + 3 O2 → 2 Fe2O3

Calculate Hrxn for the reaction and compute the heat produced from 15.0 g of iron powder

Hf Fe= 0.0

Hf O2 = 0.0

Hf Fe2O3 = -824.2 kJ

Hrxn= 2(-824.2)-4(0)+3(0)

=-1648.4 kJ

15\.0 g ( 1mol/55.85g)(-1648.4 kJ/4 mol)= -111 kJ

Energy consumption

most U.S. energy comes from the combustion of fossil fuels

originate from ancient plant and animal life and are a nonrenewable energy source

the combustion of fossil fuels produces by products that are harmful to the environment

fossil fuels

finite resource

* oil prone to spills

tens to hundreds of millions of years to make

* coal is the dirtiest

remains of ancient forests

air pollution

sulfur oxides (SOx)

* related to burning coal
* forms acid rain SO + H2O → H2SO4

Carbon monoxide (CO)

* forms from incomplete combustion
* respiratory toxin

Nitrogen Oxides (NOx)

* form high temperature combustion
* eye and lung irritant

Ozone (O3)


1. nitrous oxides and unburned V.O.C.’s react with sunlight
2. photochemical smog
3. not the same as stratospheric ozone