chemistry 102(001) fall 2012

18-1CHEM 102, Spring 2012 LA TECH

CTH 328 10:00-11:15 am

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311 Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;

Tu, Th, F 8:00 - 10:00am..

Exams: 10:00-11:15 am, CTH 328.

September 27, 2012 (Test 1): Chapter 13

October 18, 2012 (Test 2): Chapter 14 &15

November 13, 2012 (Test 3): Chapter 16 &18

Optional Comprehensive Final Exam: November 15, 2012 :

Chapters 13, 14, 15, 16, 17, and 18

Chemistry 102(001) Fall 2012


Review of Chapter 6. Energy and Chemical Reactions 6.1 The Nature of Energy

6.2 Conservation of Energy

6.3 Heat Capacity

6.4 Energy and Enthalpy

6.5 Thermochemical Equations

6.6 Enthalpy change for chemical Rections

6.7 Where does the Energy come from?

6.8 Measuring Enthalpy Changes: Calorimetry

6.9 Hess's Law

6.10 Standard Enthalpy of Formation

6.11 Chemical Fuels for Home and Industry

6.12 Food Fuels for Our Bodies


Chapter 18. Thermodynamics: Directionality of Chemical Reactions

18.1 Reactant-Favored and Product-Favored Processes

18.2 Probability and Chemical Reactions18.3 Measuring Dispersal or Disorder: Entropy18.4 Calculating Entropy Changes18.5 Entropy and the Second Law of

Thermodynamics18.6 Gibbs Free Energy18.7 Gibbs Free Energy Changes and Equilibrium Constants18.8 Gibbs Free Energy, Maximum Work, and

Energy Resources18.9 Gibbs Free Energy and Biological Systems18.10 Conservation of Gibbs Free Energy18.11 Thermodynamic and Kinetic Stability


What is Hess's Law of Summation of Heat?To heat of reaction for new reactions.

Two methods?

1st method: new DH is calculated by adding DHs of other reactions.

2nd method: Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.


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

kJ

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 -277.1 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 entropy 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:

DSo

= So

(products) - So

(reactants)

Based on Hess’s Law second method:

DSo

= So

(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


DGfo

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:

DGo

= S npDGfo

products – S nrDGfo

reactants

Standard free energy of formation, DGfo


Substance 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

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 negative DHo

rxn.

Reaction is “entropy driven”

NH4NO3(s) + heat NH4NO3(aq)


Calculating DGorxn

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

DGo

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

C

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

DHfo

, kJ/mol 0.00 0.00 -46.11

So

, 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

∆Go

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 =

S

H = T

rxn

rxn


Keq is related to reaction favorability and so to Gorxn.

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

of K.

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

by:

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

DGo

-RT

-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

o

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

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 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

DGo

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)

chemistry 102(001) fall 2012

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