Energy

Heat

Work

Heat Capacity

Enthalpy

1 © Prof. Zvi C. Koren 20.07.2010

Thermodynamics

E

Can a reaction occur??? (Is it spontaneous)???

= Thermo + Dynamics

Thermodynamics vs. Kinetics

Thermo

Kinetics

Diamond vs. Graphite

Thermodynamically unstable

Kinetically stableThermodynamically stable

C(di) C(gr)?2 © Prof. Zvi C. Koren 20.07.2010

(Note: Absolute E can never be determined by humans!)

Kinetic Energy = K.E. = energy of motion = ½ mv2

Potential Energy = P.E. = energy of position

= m·g·h, g = 9.81 m/s2, h = height

Energy

Energy Units

SI unit: Joule (J)

1 J = 1 kgm2/s2 (“kms”) [Note units of “mv2”]

= 1 VC

= 1 Pam3

= 107 erg,

1 erg = 1 gcm2/s2 (“cgs”)

6.2415 ×1011 eV

1 cal ≡ 4.184 J (exactly)

1 BTU = 1054.35 J

1 kWhr = 3.6 MJ

1 Latm ≡ 101.325 J (exactly)

1 erg = 6.2415 ×1011 eV

R = Gas Constant

= energy/molK

= 0.0821 Latm/molK

= 1.99 cal/molK

= 8.31 J/molK

3 © Prof. Zvi C. Koren 20.07.2010

E = q + w

Note the equation is NOT any of the following:

heat work

E is a state function

q and w are path functions

w: For example, compression or expansion

E = q + w

or

E = q + w

Popular Expression: “Law of Conservation of Energy”

Mathematical Expression (and more general):

(P1,V1,T1) (P2,V2,T2)

E1 E2

q

w

E = E2 - E1

The First Law of Thermodynamics

4 © Prof. Zvi C. Koren 20.07.2010

Work:

Energy transferred when an object is moved against an opposing force.

Note: Two conditions need to be met: movement and resistance.

Heat:

Energy transferred from a “hot” body (at Thigh) to a “cold” body (at Tlow)--------------------------------------------------------------------------------------------------------

Work and Heat

m

m

expansion

fopposingfopposing = weight = mg

w = – f • d = – mg • h

Also: p = f/A f = p·A

needed for direction

popposing

For piston of constant weight or if open to atmosphere: popposing = Pinternal

(piston)

gasSystem

Surroundings

w = – fopposing • x compression

= – pA • x = – p • V

w = – P • V

w = – fopposing • d = – fopposing • x

f = force

d = distance

WORK:

5 © Prof. Zvi C. Koren 20.07.2010

w causes ordered motion

Qualitative Differences Between q and w

q causes random motion

Work that the SYSTEM performs causes the molecules (or

atoms) in an object in the SURROUNDINGS to all move

in the same direction.

Heat flowing from SYSTEM to SURROUNDINGS

causes the molecules in the surroundings to move in

random directions.

6 © Prof. Zvi C. Koren 20.07.2010

Amount of heat (q) needed to raise a quantity of substance by 1 degree

For: 1 gram

“specific heat capacity”

Units of C cal (or J)/g•deg

For: 1 mole

“molar heat capacity”

Units of C cal (or J)/mol•deg

For example, for water:

Cspecific = 1 cal/g•deg, Cmolar = ?

Heat Capacity, C

7 © Prof. Zvi C. Koren 20.07.2010

Law of Dulong and Petit:Molar Heat Capacities of

Elemental Metals

3R

6.0 0.3 cal/moldeg

Specific Heat Capacities of Some Common Substances

C (J/gdeg)NameSubstance

0.902AluminumAl

So

lids

0.451IronFe

0.385CopperCu

0.128GoldAu

2.06IceH2O(s)

1.76Wood

0.88Concrete

0.84Glass

0.79Granite

4.70AmmoniaNH3(l)

Liq

uid

s

2.46EthanolC2H5OH(l)

4.184Water

(liquid)H2O(l)

1.88SteamH2O(g)

Gas

es

0.917OxygenO2(g)

1.04NitrogenN2(g)

Note:

C(liq) > C(solid)

Why?

(one of the highest)

8 © Prof. Zvi C. Koren 20.07.2010

Start with “C”. Recall its units:

qbathtub qcup

Why?

Let’s heat water or Cu from one temperature to another.

What are the factors that affect how much heat is needed in each case?

If: q = + (endothermic), q = – (exothermic)

(Note the overall units)

Heating Substances

9 © Prof. Zvi C. Koren 20.07.2010

C m t C n t

q =

Problems involving 7 parameters in 1 equation

Heat Balanceמאזן חום בתהליכי העברת חום

Σqi = 0

qhot + qcold = 0

qhot = – qcold

Heat Transfer

For example:

A hot iron rod is placed in cold water.

Eventually, everything comes to thermal equilibrium

Recall: q = Cmt

10 © Prof. Zvi C. Koren 20.07.2010

(assuming no heat loss to surroundings)

© Prof. Zvi C. Koren

Tem

p.

time

Heating Curve

s

l

Fusion: s l

g

Vaporization: l v

Hfus

(H2O: 333 J/g)

Hvap

(H2O: 2260 J/g)

Tbp

Tmp

For example: Calculate the heat (“thermal energy”) required for the

following process: 10.0 g, (ice, -10oC) (steam, 120oC)

(Cooling Curve)

Condensation: v l, Hcond = ?

Solidification (or Crystallization): l s, Hsolid (or cryst) =

For H2O: C(s) = 2.06 J/g·deg

C(l) = 4.184 J/g·deg = 1.0 cal/ g·deg

C(g) = 1.88 J/g·deg

Heating Involving Phase Changes (Physical processes)

11 © Prof. Zvi C. Koren 20.07.2010

[Constant Pressure: P = f/A; f = w = mg]

s s l l l v g

vaporization

fusion

100

0

Tem

p.

(0C

)

Time (min)

100 600

Heating/Cooling Curve for Water.

1 mol water is heated from –100C to 1100C.

A constant heating rate of 100 J/min is assumed.

Pistonm

12 © Prof. Zvi C. Koren 20.07.2010

Calculation of the Heats Involved With Each Step

in the Heating/Cooling Curve

vaporization

fusion

100

0

Tem

p.

(0C

)

Time (min)

100 600

Value for H2ONameSymbol

1.00 cal/g·deg

4.18 J/g ·deg

specific heat capacity

(of liquid)C(l)

18.00 cal/mol·deg

75.2 J/mol ·deg

molar heat capacity

(of liquid)

333 J/gheat of fusionΔHfus

2250 J/gheat of vaporizationΔHvap

)( lC“C-bar”

13 © Prof. Zvi C. Koren 20.07.2010

Find the values of C(s) and C(g) of H2O

Enthalpy H E + PV

(a convenient definition for H)

(Note: Absolute H can never be determined. Why?

H = E + (PV)

= q + w + (PV)

= q - PV + PV, [P]

H = qP

--------------------------------------------------------------------------------------------------

Note:

For constant volume processes, V = 0:

w = –PV = 0

And

E = qV

open to

atmosphere:

Pinternal = pexternal = constant

qP

Rxn.

pexternal

Pinternal

Rxn.qV

V is constant

(not P)

Enthalpy

14 © Prof. Zvi C. Koren 20.07.2010

Recall,

H E + PV

Hrxn = Erxn + (PV)rxn

(PV)rxn = (PV)products - (PV)reactants

= (PV)products(s,l,g) - (PV)reactants(s,l,g)

But, for a given quantity,

Vgas >> Vsolid,liquid

PVgas >> PVsolid,liquid

(PV)products(g) - (PV)reactants(g)

Assume ideal gases: PV = nRT,

(nRT)products(g) - (nRT)reactants(g)

= RT· ng

Hrxn Erxn + RT· ng

For example, consider the following rxn.:

2A(s) + B(l) + 4C(g) 2D(g) + E(g)

Hrxn Erxn + RT· ngHrxn Erxn + RT· (3-4)

Hrxn Erxn - RT

H = qP

E = qV

15 © Prof. Zvi C. Koren 20.07.2010

Calculating Heat of Reaction, Hrxn:Energy of Reaction, Erxn

If a rxn is made up of other rxns, then the heats are summed.

Why? Because H (like E) is a state function.HA

HB HC HD HE

HA = HB + HC + HD + HE

[Recall Born-Haber Cycle]

Calculating Heat of Reaction, Hrxn:Hess’s Law of Heat Summation

Problem:

Find H for rxn (1), (1) A + 2D C, H1 = ?

From the following data: (2) A + 2B 5C, H2 = 50 kJ

(3) B 2C + D, H3 = -75 kJ

Solution: Because rxn(1) = rxn(2) – 2 • rxn(3),

H1 = H2 – 2 • H3= 50 – 2 • (-75) = 200 kJ.

By the way, of course it’s the same for E:

E1 = E2 - 2 • E3But recall: K1 = K2 / K3

2

(1) (2) (3)

16 © Prof. Zvi C. Koren 20.07.2010

For example:

Reactants Products

The Standard State

standard state of a substance (at a specific T)

= most stable state of the substance at 1 atm (or 1 bar) at that T.

For example, the standard state for nitrogen:

At 25oC: N2 (diatomic) and a gas;

At 2,000,000oC: N (monatomic) and a gas, probably even N+;

At -270oC: crystalline (solid)

At other temps., between Tmp and Tbp, the liquid is most stable

Another example, the standard state for carbon:

At 25oC: graphite (solid); At 1 atm, graphite is more stable than diamond.

At 2,000,000oC: C (monatomic) and a gas;

(continued)

17 © Prof. Zvi C. Koren 20.07.2010

Calculating Heat of Reaction, Hrxn:Heats (or Enthalpies) of Formation

(continued)

“Formation”

Formation is a rxn where:

1 mole of a compound is formed from its elements in their standard

state (most stable form at 1 atm)

For example, formation of CH4:

C(s,gr) + 2H2(g) CH4(g), Hf (CH4) (measured from qP of the rxn)

Hf = standard heat (or enthalpy) of formationo

o

18 © Prof. Zvi C. Koren 20.07.2010

Note: (any property) ≡ final – initial

Hrxn ≡ Hfinal – Hinitial = ΣHproducts – ΣHreactants

So, in effect:

Hf (CH4) = H(CH4) – H(C) – 2H(H2) o

So, the “enthalpy of formation” of a compound is in effect the “relative

enthalpy” of that compound, that is, its enthalpy relative to the enthalpies of

the elements from which it is composed.

(continued)

Consider the following combustion reaction:

C6H6(l) + 7½ O2(g) 6 CO2(g) + 3H2O(l)

Recall: (any property) ≡ final – initial

Hrxn ≡ Hfinal – Hinitial = ΣHproducts – ΣHreactants

Hcomb rxn = 6•H(CO2) + 3•H(H2O) – H(C6H6) – 7½ H(O2)

(This last equation is correct but not useful, because we can never know absolute H.

So, we must use relative enthalpies, that is enthalpies of formation, Hf.)

So, build the overall reaction from a series of “formation reactions” (in the Hess-way):

6 • (C + O2 CO2), 6•Hf(CO2)

3 • (H2 + ½ O2 H2O), 3•Hf(H2O)

(6 C + 3 H2 C6H6), Hf(C6H6)

C6H6(l) + 7½ O2(g) 6 CO2(g) + 3H2O(l), Hcomb rxn = ?

Hcomb = 6•Hf(CO2) + 3•Hf(H2O) – Hf(C6H6)o o o o

We can generalize that for any reaction:

Hrxn = Hf - Hfo o o

P R

Hf (element) ≡ 0o

(continued)

19 © Prof. Zvi C. Koren 20.07.2010

20 © Prof. Zvi C. Koren 20.07.2010

For every experiment use the same overall calorimeter mass.

Calibrate Calorimeter: Use a weighed mass of substance with known ΔHcomb.

Calorimetry

Measuring heats of combustion reactions

O2

ignitionwires

water

“Bomb Calorimeter”

(constant V)

stirrer

thermometer

21 © Prof. Zvi C. Koren 20.07.2010

Heat Capacity of Calorimeter

(or “Calorimeter Constant”),

Ccalorimeter,

in “energy”/deg:

(continued)

Measuring Heats of Reaction (qV

) with a “Bomb Calorimeter”:

3 parts: Reaction, Bomb Apparatus, Water

Heat Balance Equation: Σqi = 0 qrxn + qbomb + qwater = 0

– qrxn = qbomb + qwater

= Cbomb· tbomb + (C·m·t)water

[Recall: units of Cbomb = “energy”/deg]

For example, consider the following combustion of octane:

C8H18(l) + 12.5 O2(g) 8 CO2(g) + 9 H2O(l)

1.0 g of octane burns in a constant-volume calorimeter (Cbomb = 837 J/deg) containing

1.20 kg of water. The temperature rises from 25.00oC to 33.20oC. Calculate: (a) the

heat of comb., qV

, for this quantity of octane, (b) Hcomb for 1 mole of octane.

Answer to (a):

–qrxn = qbomb + qwater

= Cbomb · tbomb + (C · m · t)water = [Cbomb + (C·m)water]t; t tf – ti

= [(837 J/deg) + (4.184 J/g·deg) (1200 g)] (8.20oC) = 48.1 kJ

qV

= – 48.1 kJ

Answer to (b): First calculate # of moles: 1.0 g (1 mol/114 g) = 0.0088 mol

qV

= – 48.1 kJ / 0.0088 mol Ecomb = qV

= – 5,500 kJ/mol

qP

= Hcomb = Ecomb + RT(ng)

= -5.5x106 J/mol + (8.31 J/mol·K)(298 K)(– 4.5) = – 5.5x106 J/mol

22 © Prof. Zvi C. Koren 20.07.2010

(continued)