17-1 CHEM 102, Fall 15 LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office:...

17-1CHEM 102, Fall 15 LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: upali@latech.edu

Office: CTH 311

Phone 257-4941

Office Hours: M.W &F, 8:00-9:00 & 11:00-12:00 and Tu ,Th 8:00 - 10:00 am. am. or by appointment

Test Dates

Chemistry 102 Fall 2015

Sept. 29, 2015 (Test 1): Chapter 13

Oct. 27, 2015 (Test 2): Chapter 14 &15

Nove. 17, 2015 (Test 3): Chapter 16 &17

November 19, 2015 (Make-up test) comprehensive:

Chapters 13-17

Chapter 6. Thermochemistry

6.1 Chemical Hand Warmers 231

6.2 The Nature of Energy: Key Definitions 232

6.3 The First Law of Thermodynamics: There Is No Free Lunch 234

6.4 Quantifying Heat and Work 240

6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry246

6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249

6.7 Constant-Pressure Calorimetry: Measuring 253

6.8 Relationships Involving 255

6.9 Determining Enthalpies of Reaction from Standard Enthalpies

of Formation 257

6.1 0 Energy Use and the Environment 263

Chapter 17. Free Energy and Thermodynamics17.1 Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 769

17.2 Spontaneous and Nonspontaneous Processes 771

17.3 Entropy and the Second Law of Thermodynamics 773

17.4 Heat Transfer and Changes in the Entropy of the Surroundings 780

17.5 Gibbs Free Energy 784

17.6 Entropy Changes in Chemical Reactions: Calculating 788

17.7 Free Energy Changes in Chemical Reactions: Calculating 792

17.8 Free Energy Changes for Nonstandard States: The Relationship

between and 798

17.9 Free Energy and Equilibrium: Relating to the Equilibrium Constant

Chapter 6. Thermochemistry

6.1 Chemical Hand Warmers 231

6.2 The Nature of Energy: Key Definitions 232

6.3 The First Law of Thermodynamics: There Is No Free Lunch 234

6.4 Quantifying Heat and Work 240

6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry 246

6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249

6.7 Constant-Pressure Calorimetry: Measuring 253

6.8 Relationships Involving 255

6.9 Determining Enthalpies of Reaction from Standard Enthalpies

of Formation 257

6.1 0 Energy Use and the Environment 263

Method 1: Calculate DH for the reaction:

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

Other reactions:

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

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

H2(g) +1/2O2(g) -----> H2O(g); DH = -242 kJ

SO2(g) ------> S(s) + O2(g); DH1 = 297 kJ - 1

H2(g) + S(s) + 2O2(g) ------> H2SO4(l); DH2 = -814 kJ - 2

H2O(g) ----->H2(g) + 1/2 O2(g) ; DH3 = +242 kJ - 3

______________________________________

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

DH = DH1+ DH2+ DH3

DH = +297 - 814 + 242

DH = -275 kJ

Calculate DH for the reaction

1) Calculate entropy change for the reaction:

2 C(s) + 1/2 O2(g) + 3 H2(g) --> C2H6O(l); ∆H = ? (ANS 1493.2 kJ/mol)Given the following thermochemical equations:

C2H6O(l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O(l); ∆H = - 1366.9 kJ/mol

1/2 O2(g) + H2(g) ----> H2O(l); ∆H = -285.8 kJ/mol

C(s) + O2(g) ----> CO2(g); ∆H = -393.3 kJ/mol

Calculate Heat (Enthalpy) of Combustion: 2nd method C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; DHo = ?

DHf (C7H16) = -198.8 kJ/mol

DHf (CO2) = -393.5 kJ/mol

DHf (H2O) = -285.9 kJ/mol

DHf O2(g) = 0 (zero)

What method?DHo = S n DHf

o products – S n DHfo reactants

n = stoichiometric coefficients2nd method

DH = [Sn ( DHof) Products] - [Sn (DHo

f) reactants]

DH = [ 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

Calculate DH for the reaction

Why is DHof of elements is zero?

DHof, Heat formations are for compounds

Note: DHof of elements is zero

2) Calculate enthalpy change given the

∆Hfo[SO2(g)] = -297 kJ/mole and

∆Hfo [SO3(g)] = -396 kJ/mole

2SO2 (g) + O2 (g) -----> 2 SO3(g); ∆H= ? ANS -198 kJ/mole)

What is relation of DH of a reaction to covalent bond energy?

DH = S½bonds broken½- S ½bonds formed½

How do you calculate bond energy from DH?

How do you calculate DH from bond energy?

3) Use the table of bond energies to find the ∆Ho for the reaction:

H2(g) + Br2(g) 2 HBr(g);

H-H = 436 kJ, Br-Br= 193 kJ, H-Br = 366 kJ

Example.

Calculate the DSo

rxn at 25 o

C for the following reaction.

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

Substance So

(J/K.mol)

CH4 (g) 186.2

O2 (g) 205.03

CO2 (g) 213.64

H2O (g) 188.72

Calculation of standard entropy changes

Calculate the DS for the following reactions using D So

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

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

D So [N2O(g)] = 220 J/K mole; D S[ H2O(l)] = 70 J/K mole

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

DSo 496 205 514

DSo = S DSo (products) - S DS o(reactants)

DSo = [514] - [496 + 205]

DSo = 514 - 701

DSo = -187 J/K mole

2 H2(g) + O2(g) 2 H2O(liq)

DSo = 2 So (H2O) - [2 So (H2) + So (O2)]

DSo = 2 mol (69.9 J/K•mol) – [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]

DSo = -326.9 J/K

There is a decrease in S because 3 mol of gas give 2 mol of liquid.

Calculating DS for a Reaction Based on Hess’s Law second method:

(products) - So

(reactants)

Based on Hess’s Law second method:

(products) - So

(reactants)

4) Calculate the ∆S for the following reaction using:

a) 2SO2 (g) + O2 (g) ----> 2SO3(g)

So [SO2(g)] = 248 J/K mole ;

So [O2(g)] = 205 J/K mole;

So [SO3(g)] = 257 J/K mole

The sign of DG indicates whether a reaction will occur spontaneously.

+ Not spontaneous

0 At equilibrium

- Spontaneous

The fact that the effect of DS will vary as a function of temperature is important.

This can result in changing the sign of DG.

Free energy, DG

Free energy change that results when one mole of a substance if formed from

its elements will all substances in their standard states.

DG values can then be calculated from:

= S npDGfo

products – S nrDGfo

reactants

Standard free energy of formation, DGfo

Substance DGfo

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) -371.08

All have units of kJ/mol and are for 25 oC

Standard free energy of formation

How do you calculate DG

There are two ways to calculate DG

for chemical reactions.

i) DG = DH - TDS.

ii) DGo = S DGof (products) - S DG o

f (reactants)

Calculating DGorxn

Method (a) : From tables of thermodynamic data we find

DHorxn = +25.7 kJ

DSorxn = +108.7 J/K or +0.1087 kJ/K

DGorxn = +25.7 kJ - (298 K)(+0.1087 J/K)

= -6.7 kJReaction is product-favored in spite of positive DHo

Reaction is “entropy driven”

NH4NO3(s) + heat NH4NO3(aq)

Calculating DGorxn

Combustion of carbon

C(graphite) + O2(g) --> CO2(g)

Method (b) :

DGorxn = DGf

o(CO2) - [DGfo(graph) + DGf

o(O2)]

DGorxn = -394.4 kJ - [ 0 + 0]

Note that free energy of formation of an element in its standard state is 0.

DGorxn = -394.4 kJ

Reaction is product-favored

rxn = S DGfo

(products) - S DGfo

(reactants)DGo

rxn = S DGfo

(products) - S DGfo

(reactants)

We can calculate DGo

values from DHo and DSo

values at a constant temperature

and pressure.

Example.

Determine DGo

for the following reaction at 25o

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

, kJ/mol 0.00 0.00 -46.11

, J/K.mol 191.50 130.68 192.3

Calculation of DGo

Predict the spontaneity of the following processes from DH and DS at various temperatures.

a)DH = 30 kJ, DS = 6 kJ, T = 300 Kb)DH = 15 kJ,DS = -45 kJ,T = 200 K

a) DH = 30 kJ DS = 6 kJ T = 300 K DG = DHsys-TDSsys or DG = DH - TDS.DH = 30 kJDS = 6 kJ T = 300 K DG = 30 kJ - (300 x 6 kJ) = 30 -1800 kJDG = -1770 kJ

b) DH = 15 kJ DS = -45 kJ T = 200 KDG = DHsys-TDSsys or DG = DH - TDS.DH = 15 kJDS = -45 kJ T = 200 K DG = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJDG = 15 + 9000 kJ = 9015 kJ

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

a) ∆H = 30 kJ ∆S = 6 kJ T = 300 K

b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K

6) Calculate the ∆Go for the following chemical reactions using given ∆Ho values, ∆So calculated above and the equation ∆G = ∆H - T∆S.2SO2 (g) + O2 (g) > 2 SO 3(g) ; ∆Go=∆Ho = -198 kJ/mole; ∆So = -187 J/K mole; T = 298 K

system ∆Ho

system ∆So

system T

7) Which of the following condition applies to a particular chemical reaction, the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K. This reaction is

∆G system ∆H system ∆S system T

a) Negative always Negative (exothermic) Positive Yes

a) Negative at low T Positive at high T

Negative (exothermic)

Negative ∆G =- ,at low T; ∆G= +, at high T

a) Positive at low T Negative at high T

Positive (endothermic)

Positive ∆G = + ,at low T; ∆G= -, at high T

a) Positive always Positive (endothermic)

Negative∆G= +, at any T

Effect of Temperature on Reaction Spontaneity

DGo = DHo - TDSo

8) At what temperature a particular chemical reaction, with the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K becomes

a) Equilibrium:

b) Spontaneous:

How do you calculate DG at different T and P

DG = DGo + RT ln Q

Q = reaction quotient

at equilibrium DG = 00 = DGo + RT ln K

DGo = - RT ln K

If you know DGo you could calculate K

Concentrations, Free Energy, and the Equilibrium Constant

Equilibrium Constant and Free Energy

DG = DGo + RT ln Q

Q = reaction quotient

0 = DGo + RT ln Keq

DGo = - RT ln Keq

9) Calculate the non standard ∆G for the following equilibrium reaction and predict the direction of the change using the equation:

∆G= ∆Go + RT ln Q

Given ∆Gfo[NH3(g)] = -17 kJ/mole

N2 (g) + 3H2(g) → 2NH3(g); ∆G=? at 300K, PN2= 300, PNH3 = 75 and PH2 = 300

10) The Ka expression for the dissociation of acetic acid in water is based on the following equilibrium at 25°C:

HC2H3O2(l) + H2O ⇄ H+(aq) + C2H3O2 -(aq)

What is ∆G° if Ka=1.8 x 10-5?

Calculate the DG value for the following reactions using: D Go = S D Go

f (products) - S D Gof (reactants)

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

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

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

N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ?DGf

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

f (products) - 3 DGof (reactants)

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

Free Energy and Temperature

2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)

DHorxn = +467.9 kJ DSo

rxn = +560.3 J/K

DGorxn = +300.8 kJ

Reaction is reactant-favored at 298 K

At what T does DGorxn change from (+) to (-)?

Set DGorxn = 0 = DHo

rxn - TDSorxn

K 835.1 = kJ/K 0.5603

kJ 467.9 =

Keq is related to reaction favorability and so to Gorxn.

The larger the (-) value of DGorxn the larger the value

DGorxn = - RT lnK

where R = 8.31 J/K•mol

Thermodynamics and Keq

For gases, the equilibrium constant for a reaction can be related to DGo

= -RT lnK

For our earlier example,

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

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

ln K = =

ln K = 13.27; K = 5.8 x 105

-32.91 kJ

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

Free energy and equilibrium

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

o[ NH3(g) ] = -17 kJ/mole

N2 (g) + 3 H2 (g) 2 NH3 (g); D G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300

N2 (g) + 3 H2 (g) 2 NH3 (g); DG = ?

To calculate DGo

Using DGo = S DGof (products) - S DGo

f (reactants)

DGfo[ N2(g)] = 0 kJ/mole; DGf

o[ H2(g)] = 0

kJ/mole; DGfo[ NH3(g)] = -17 kJ/mole

Notice elements have DGfo = 0.00 similar to DHf

N2 (g) + 3 H2 (g) 2 NH3 (g); DG = ?DGf

o 0 0 2 x (-17) 0 0 -34 DGo = S DGo

f (products) - S DGof (reactants)

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

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

K = _________ pN2 p3

Q = _________ ; pN2 p3

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 DGo

DG = DGo + RT ln Q

DGo= -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

DG = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln

6.94 x 10-7)

DG = -34 + 2.49 ln 6.94 x 10-7

DG = -34 + 2.49 x (-14.18)

DG = -34 -35.37

DG = -69.37 kJ/mole

Calculate K (from G0)

N2O4 --->2 NO2 DGorxn = +4.8 kJ

DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K

rxn = - RT lnK

1.94- = K)J/K)(298 (8.31

J 4800 - = ln K

K = 0.14

DGorxn > 0 : K < 1

DGorxn < 0 : K > 1

K = 0.14

DGorxn > 0 : K < 1

DGorxn < 0 : K > 1

Thermodynamics and Keq

Concentrations, Free Energy, and the Equilibrium Constant

The Influence of Temperature on Vapor Pressure

H2O(l) => H2O(g)

Keq = pwater vapor

pwater vapor = Keq = e- G'/RT

DG as a Function of theExtent of the Reaction

DG as a Function of theExtent of the Reactionwhen there is Mixing

Maximum WorkDG = wsystem = - wmax

(work done on the surroundings)

Coupled ReactionsHow to do a reaction that is not

thermodynamically favorable?

Find a reaction that offset the (+) DG

Thermite Reaction

Fe2O3(s) => 2Fe(s) + 3/2O2(g)

2Al(s) + 3/2O2(g) Al2O3(s)

ADP and ATP

Acetyl Coenzyme A

Gibbs Free Energy and Nutrients

Photosynthesis: Harnessing Light Energy

Using Electricity for reactions with (+) DG: Electrolysis

Non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current

Electrolysis

2NaCl(l) => 2Na(s) + 2Cl2(g)

17-1 CHEM 102, Fall 15 LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office:...

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