First Law of Thermodynamics

Enthalpy -H, Hf

Second Law of Thermodynamics

Third Law of Thermodynamics

Entropy S, Free Energy G

Spontaneity

Chapter 17. Thermodynamics: Spontaniety, Entropy and Free Energy

Thermochemistry

Heat changes during chemical reactions Thermochemical equation. eg.

H2 (g) + O2 (g) ---> 2H2O(l) H =- 256 kJ; is called the enthalpy of reaction.

if H is + reaction is called endothermic

if H is - reaction is called exothermic

Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

Why is it necessary to divide Universe into System and Surrounding

Universe = System + Surrounding

What is the internal energy What is the internal energy change (change (U) of a system?U) of a system?

U is associated with changes in atoms, molecules and subatomic particles

Etotal = Eke + E pe + U

U = heat (q) + w (work) U = q + w U = q -P V; w =- P V

What forms of energy are found in the Universe? mechanical thermal electrical nuclear mass: E = mc2

others yet to discover

What is 1st Law of Thermodynamics Eenergy is conserved in the universe

All forms of energy are inter-convertible and conserved

Energy is neither created nor destroyed.

What exactly is H?

Heat measured at constant pressure qp

Chemical reactions exposed to atmosphere and are held at a constant pressure.

Volume of materials or gases produced can change. ie: work = -PV

U = qp + w; U = qp -PV

qp = U + PV; w = -PV

H = U + PV; qp = H(enthalpy )

How do you measure U?

Heat measured at constant volume qv

Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -PV= 0 U = qv + wqv = U + o; w = 0U = qv = U(internal energy )

What is Hess's Law of Summation of Heat?

To heat of reaction for new reactions. Two methods? 1st method: new H is calculated by

adding Hs of other reactions. 2nd method: Where Hf ( H of

formation) of reactants and products are used to calculate H of a reaction.

Method 1: Calculate H for the reaction:

SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) H = ?

Other reactions: SO2(g) ------> S(s) + O2(g) H = 297kJ H2SO4(l)------> H2(g) + S(s) + 2O2(g)

H = 814 kJH2(g) +1/2O2(g) -----> H2O(g) H = -242 kJ

SO2(g) ------> S(s) + O2(g);H1 = 297 kJ - 1

H2(g) + S(s) + 2O2(g) ------> H2SO4(l) H2 = -814 kJ - 2

H2O(g) ----->H2(g) + 1/2 O2(g) H3 = +242 kJ - 3

______________________________________ SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) ªH = H1 + H2 + H3

ªH = +297 - 814 + 242 ªH = -275 kJ

Calculate Heat (enthalpy) of Combustion: 2nd method

C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; H = ?

Hf (C7H16) = -198.8 kJ/mol

Hf (CO2) = -393.5 kJ/mol

Hf (H2O) = -285.9 kJ/mol

Hf O2(g) = 0 (zero)

What method? 2nd method

H = [n ( Hof) Products] -

[n (Hof) reactants]

H = [ 7(-393.5 + 8 (-285.9)] - [-198.8 + 11 (0)]

= [-2754.5 - 2287.2] - [-198.8] = -5041.7 + 198.8 = -4842.9 kJ = -4843 kJ

Why is Hof of elements is zero?

Hof, Heat formations are for compounds

Note: Hof of elements is zero

What is 2nd Law of Thermodynamics

Entropy (S) of the Universe increases during spontaneous process.

What is entropy Entropy (S) : A measure of randomness or

disorder of a system. Spontaneous Process: A process that occurs

without outside intervention.

Definitions The Universe: The sum of all parts under

consideration. System: Part of the Universe we are

interested in and a change is taking place. Surrounding: Part of the Universe we are

not interested in or can not observe.

Entropy S

Suniv = entropy of the Universe

Ssys = entropy of the System

Ssurr = entropy of the Surrounding

Suniv = Ssys+ Ssurr

Suniv= Ssys + Ssurr

Suniv Ssys Ssurr

+ + +

+ +(Ssys>Ssurr) -

+ - + (Ssurr>Ssys)

Standard entropies at 25oC Substance So (J/K.mol) Substance So

(J/K.mol)

C (diamond) 2.37 HBr (g) 198.59 C (graphite) 5.69 HCl (g)186.80 CaO (s) 39.75 HF (g) 193.67

CaCO3 (s) 92.9 HI (g) 206.33

C2H2 (g) 200.82 H2O (l) 69.91

C2H4 (g) 219.4 H2O (g) 188.72

C2H6 (g) 229.5 NaCl (s) 72.12

CH3OH (l) 127 O2 (g) 205.03

CH3OH (g) 238 SO2 (g) 248.12

CO (g) 197.91 SO3 (g) 256.72

Entropy Change Entropy (S) normally increase (+) for

the following changes: i) Solid ---> liquid (melting) ii) Liquid ---> gas iii) Solid ----> gas iv) Increase in temperature v) Increasing in pressure(constant

volume, and temperature) vi) Increase in volume ( at constant

temperature and pressure)

Free energy

Gibbs free energy (Gibbs free energy (GG)) can be used to describe the energy changes of a system.

As usual, it is the changes in free energy that are of interest - GG.

At constant temperature and pressure, G is equal to:

GG = = HH - - TTSS

T is the Kelvin temperature.

What is G ?

Free Energy

G = - TSuni.

G = H - TS.

How do you get:G = H - TS. suniv= ssys +ssurr

ssurr = -Hsys/T)

Suniv = Ssys -Hsys/T

Suniv x T = TSsys -Hsys : x T

-Suniv x T = -TSsys +Hsys : x -1

-Suniv x T = Hsys-TSsys

-Suniv x T = G;

G = Hsys-TSsys or G = H - TS.

What G means

If G is - for a change it will take place spontaneously If G is + for a change it will not take

place If G is 0 for a change it will be in

equilibrium

Free energy

The sign ofG indicates whether a reaction will occur spontaneously.

++ Not spontaneousNot spontaneous 0 0 At equilibriumAt equilibrium -- SpontaneousSpontaneous The fact that the effect of S will vary as a

function of temperature is important. This can result in changing the sign of G.

Predict G at different H, S, T

G = H - T S. - - all + + + all - - /+ + high/low + -/+ - low/high -

Predict the spontaneity of the following processes from H and S at various temperatures.

a)H = 30 kJ, S = 6 kJ, T = 300 Kb)H = 15 kJ,S = -45 kJ,T = 200 K

H = 30 kJ S = 6 kJ T = 300 K G = Hsys-TSsys or G = H - TS.H = 30 kJS = 6 kJ T = 300 K G = 30 kJ - (300 x 6 kJ) = 30 -1800 kJG = -1770 kJ

b) H = 15 kJ S = -45 kJ T = 200 KG = Hsys-TSsys or G = H - TS.H = 15 kJS = -45 kJ T = 200 K G = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJG = 15 + 9000 kJ = 9015 kJ

Calculation of Go

We can calculate Go values from Ho and So values at a constant temperature and pressure.

Example.Example. Determine Go for the following reaction at

25oC Equation N2 (g) + 3H2 (g) 2NH3 (g)

Hfo, kJ/mol 0.00 0.00 -46.11

So, J/K.mol 191.50 130.68 192.3

Qualitative prediction of S of Chemical Reactions Look for (l) or (s) --> (g) If all are gases: calculate n = n(gaseous prod.) - n(gaseous reac.) N2 (g) + 3 H2 (g) --------> 2 NH3 (g)

n = 2 - 4 = -2 If n is - S is negative (decrease in S) If n is + S is positive (increase in S)

Predict S!

2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g).

2 CO(g) + O2(g)-->2 CO2(g).

HCl(g) + NH3(g)-->NH4Cl(s).

H2(g) + Br2(l) --> 2 HBr(g).

Calculating S of reactions

Based on Hess’s Law second method:

So=So(prod.) - S o(react.)

Calculation of standard entropy changes

Example.Example. Calculate the So

rxn at 25 oC for the following reaction.

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

SubstanceSubstance SSoo (J/K (J/K..mol)mol) CH4 (g) 186.2

O2 (g) 205.03

CO2 (g) 213.64

H2O (g) 188.72

Calculate the S for the following reactions using: So

= 3 So (products) - 3 S o(reactants)a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K mole

b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l) So[ NH3(g)] = 193 J/K mole ; So [N2(g)] = 192 J/K mole; So [N2O(g)] = 220 J/K mole; S[ H2O(l)] = 70 J/K mole

a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K mole

So 496 205 514So

= 3 So (products) - 3 S o(reactants) So

= [514] - [496 + 205]So

= 514 - 701So

= -187 J/K mole

How do you calculate G

There are two ways to calculateG for chemical reactions. i) G = H - TS. ii) Go = Go

f (products)

- G of (reactants)

Standard free energy of formation

GGffoo

Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.

G values can then be calculated from: Go = npGf

o products - npGfo reactants

Standard free energy of formation Substance Gf

o Substance Gfo

C (diamond) 2.832 HBr (g) -53.43 CaO (s) -604.04 HF (g) -273.22

CaCO3 (s) -1128.84 HI (g) 1.30

C2H2 (g) 209 H2O (l) -237.18

C2H4 (g) 86.12 H2O (g) -228.59

C2H6 (g) -32.89 NaCl (s) -384.04

CH3OH (l) -166.3 O (g) 231.75

CH3OH (g) -161.9 SO2 (g) -300.19

CO (g) -137.27 SO3 (g) -371.08All have units of kJ/mol and are for 25 oC

Calculate the G value for the following reactions using:

Go = Gof (products) - Go

f (reactants)

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ? Gf

o[ N2O5 (g) ] = 134 kJ/mole ; Gfo [H2O(g)] = -237

kJ/mole; Gfo[ HNO3(l) ] = -81 kJ/mole

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ?Gf

o 1 x 134 1 x (-237) 2 (-81) 134 -237 -162 Go = Go

f (products) - 3 Gof (reactants)

Go = [-162] - [134 + (-237)]Go = -162 + 103Go = -59 kJ/mole The reaction have a negative G and the reaction is spontaneous or will take place as written.

How do you calculate G at different T and P G = Go + RT ln Q Q = reaction quotient at equilibrium G = = Go + RT ln K Go = - RT ln K If you know Go you could calculate K

Free energy and equilibrium For gases, the equilibrium constant for a reaction

can be related to Go by: GGoo = - = -RTRT ln lnKK For our earlier example,

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

At 25oC, Go was -32.91 kJ so K would be:

ln K = = =

ln K = 13.27; K = 5.8 x 105

Go

-RT-32.91 kJ

-(0.008315 kJ.K-1mol-1)(298.2K)

Calculate the G for the following equilibrium reaction and predict the direction of the change using the equation: G = Go + RT ln Q ; [Gf

o[ NH3(g) ] = -17 kJ/moleN2 (g) + 3 H2 (g) W 2 NH3 (g); G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300 N2 (g) + 3 H2 (g) W 2 NH3 (g); G = ?

To calculate Go

Using Go = Gof (products) - 3 Go

f (reactants)

Gfo[ N2(g) ] = 0 kJ/mole; Gf

o[ H2(g) ] = 0 kJ/mole; Gf

o[ NH3(g) ] = -17 kJ/moleNotice elements have Gf

o = 0.00 similar to Hfo

N2 (g) + 3 H2 (g) W 2 NH3 (g); G = ?Gf

o 0 0 2 x (-17) 0 0 -34 Go = Go

f (products) - 3 Gof (reactants)

Go = [-34] - [0 +0]Go = -34Go = -34 kJ/mole

To calculate QEquilibrium expression for the reaction in terms of partial pressure:N2 (g) + 3 H2 (g) W 2 NH3 (g) p2

NH3

K = _________ pN2 p3

H2

p2NH3

Q = _________ ; pN2 p3

H2

Q is when initial concentration is substituted into the equilibrium expression 752

Q = _________ ; p2NH3= 752; pN2 =300; p3

H2=3003

300 x 3003

Q = 6.94 x 10-7

To calculate Go

G = Go + RT ln QGo= -34 kJ/mole R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole T = 300 K Q= 6.94 x 10-7

G = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln 6.94 x 10-7)G = -34 + 2.49 ln 6.94 x 10-7

G = -34 + 2.49 x (-14.18)G = -34 -35.37G = -69.37 kJ/mole