Met Phy Chem Lecture

Metallurgical Engineering Board Exam Refresher Course

METALLURGICAL PHYSICAL CHEMISTRY(Day 1)

Engr. Jonah D. Longaquit-GamutanJune 2, 2012 (Saturday)

COE Room 106

TOPIC

TOPIC OUTLINE

THERMODYNAMICS1. Behavior of Gases and Vapors: Ideal Gas Law2. First Law of Thermodynamics: Heat and Work3. Second Law of thermodynamics: Entropy4. The Gibbs-Helmholtz Equation5. Theory of Solutions: Activity6. Chemical Equilibrium

KINETICS1. Order of Reactions2. Arrhenius Law: Activation Energy3. Nernst Equation4. Electrochemistry: Faradays Law

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Isolated System does not permit the transfer of both matter & energyOpen System allows the transfer of both matter & energy Closed System does not permit the transfer of matter but allows the

transfer of energy

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State defined by the following variables: Temperature (T) = C, K, R, F Pressure (P) = Pa, psi, atm, mmHg, Torr, bar Volume = m3, ft3, L Composition = wt.%, x, M, m

Definition of Terms

System portion of the universe chosen for studySurroundings universe less the system

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Boyles Law: P1V1 = P2V2

Amontons Law: P1/T1 = P2/T2

Charles Law: V1/T1 = V2/T2

Ideal Gas Law: PV = nRT

Avogadros number: 1 mole = 6.022 x 1023 particlesGas constant: R = 0.08206 Latm/molK

= 1.987 cal/molK= 8.314 J/molK

STP conditions: 0C and 1atm(for gases) 1gmole = 22.4L

1lbmole = 359.05ft3

Behavior of Gases and Vapors

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Daltons Law of Partial Pressures total pressure exerted by a mixture of gases is equal to the sum of the partial pressures (pressure exerted by a gaseous component if it occupied the container alone)

Behavior of Gases and Vapors

Ptotal = P1 + P2 + P3 + + Pi

where: P1, P2, P3, , Pi = partial pressures

such that Pi = xiPtotal

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Sample Problems

1. In an industrial process, nitrogen is heated to 500K in a vessel of constant volume. If it enters the vessel at 100 atm and 300K, what pressure would it exert at the working temperature if it behaved as a perfect gas?

2. The mass percentage composition of dry air at sea level is approximately 75.5% N2, 23.2% O2 and 1.3% Ar. What is the partial pressure of each component when the total pressure is 1 atm?

3. In an experiment to measure the molar mass of a gas, 250cm3 of the gas was confined in a glass vessel. The pressure was 152 Torr at 298K and the mass of the gas was 33.5mg. What is the molar mass of the gas? R = 62.36 Torrdm3/molK

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First Law of Thermodynamics

Concepts:

1. Work motion against an opposing force2. Heat transfer of energy due to temperature difference3. Energy capacity to do work

Internal Energy, U Total sum of the kinetic and potential energies of a system Units: 1 J = 1 kg m2/s2 = beat of the human heart

1 cal = 4.184 J = energy required to raise the temperature of 1g

of water by 1 C

Statement of the First Law: The internal energy of an isolated system is constant.

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Sign conventions:When q > 0, heat is absorbed by the system

q < 0, heat is released to the surroundings

When w > 0, work is done by the systemw < 0, work is done by the surroundings

Reversible Processes upon completion of these type of processes, both system and surroundings can be restored to their respective initial states

Mathematical Statement: U = q - w

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Work Calculations

Expansion Work work arising from a change in volume

dw = force x displacementdw = PA dx = PdVw = PdV

Free Expansion expansion against zero opposing force

w = 0

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Isobaric Expansion

w = PdV = P dVw = P (V2 V1) = PV

Isothermal Expansion

w = PdV =

Work Calculations

2

1

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Calorimeter device used to measure energy transferred as heat at adiabatic conditions

At constant volume,qv = nCvdT

And since w = 0,U = qvU = nCv T

where: Cv = heat capacity

Heat Calculations

***Therefore, heat supplied or obtained in a constant-volume system is equal to the change in internal energy.***

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Enthalpy, H

H = U + pV = U + nRT

At constant pressure, H = nCp T

Relationship between Cp and Cv: Cp Cv = nR

Enthalpy Calculations:

When Cp is not constant, it can be expressed as: p 2

p 2

2 22 2

21 1

1

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Enthalpy, H

Standard Enthalpy Changes, H change in enthalpy for a process in which the reactants and products are in their standard states

Enthalpies of Physical ChangeH vap = EvaporationH fus = Fusion or MeltingH sub = Sublimation

Enthalpies of Chemical ChangeH r = ReactionH c = CombustionsH f = Formation

Example. Given a reaction 2A + B = 3C + D

H r = H products H reactantsH r = [3 H C + H D] [2 H A + H B]

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Enthalpy, H

Hesss Law standard enthalpy of an over-all reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided

where: Cp = difference of molar heat capacities of products and reactants

**For chemical reactions, since H = nCpdTthen H2 H1 = n CpdT

H2 = H1 + n CpdT

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Sample Problems

4. One mole of N2 gas at 1000K is allowed to expand reversibly and isothermally from 1L to 5L. Calculate the work done by the gas.

5. Consider an isobaric compression of 0.450mol of an ideal gas from 22.4L and 1.00atm to 10.5L. Given that Cp for the gas is 2.5R, calculate T, q, w, U and H.

6. What is the change in molar enthalpy of N2 when it is heated from 25 C to 100 C? Given a = 28.58, b = 3.77x10-3 and c = 0.50x10-5

values.

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Sample Problems

7. Given the following, determine Hf for diborane (B2H6) at 298K:B2H6 + 3O2 = B2O3 + 3H2O H1 = -1941 kJ/mol2B + 1.5O2 = B2O3 H2 = -2368 kJ/molH2 + 0.5O2 = H2O H3 = -241.8 kJ/mol

8. Find the enthalpy change at 800K for the reaction: CaO + CO2 = CaCO3, given:

H 298 (kJ) a b c (J/mol K)CaO -634.3 49.62 4.52x10-3 -6.95x105

CO2 -393.5 44.14 9.04x10-3 -8.54x105

CaCO3 -1206.7 104.52 21.92x10-3 -25.94x105

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Second Law of Thermodynamics

Kelvin-Planck Statement:No process is feasible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

Entropy Statement:The entropy of an isolated system increases in the course of a spontaneous change such that Stotal > 0.

Stotal = Ssystem + Ssurroundings

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Second Law of Thermodynamics

General Equation:

When Stotal > 0, spontaneous changeStotal < 0, non-spontaneous change

Entropy Calculations: Expansion Phase Transition Heating

Hesss Law is also applicable to entropy values

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Entropy Calculations

Entropy of Expansion of Gas

f

i

Given U = 0:

f

i

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At constant pressure, q = H such that: trans

*If transition is exothermic, S < 0 (non-spontaneous)endothermic, S > 0 (spontaneous)

Entropy due to Phase Transition

transtrans

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p

p12

1

2

p121

2

For one mole of substance,

Entropy due to Increase in Temperature

p121

2

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Third Law of Thermodynamics

At OK, the entropy of a system is zero. All energy of thermal motion has been quenched and the crystal structure is perfect.

The entropy of all perfect crystalline substances is zero at 0K.

S 0 as T 0

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The Fundamental Equation

Combining the 1st and 2nd Laws: dU = dq dw

**Applies to both reversible and irreversible changes.**

Since dw = pdV and dq = TdS,

dU = TdS - pdV

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Gibbs-Helmholtz Equation

If G < 0, forward reaction is feasibleG > 0, reverse reaction is feasibleG = 0, the system is in equilibrium

Hesss Law is also applicable to Gibbs free energy values

At standard state conditions,

G = H - TS

where: G = Gibbs free energy (energy available to do work)

Greaction = Gproducts - Greactants

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Sample Problems

9. Calculate the entropy change when argon at 25C and 1bar in a container of volume 0.5L is allowed to expand to 1L and is simultaneously heated to 100C. Assume Cv=1.5R.

10. Calculate the entropy of liquid iron at its melting point, 1808K, given that for iron, Hf = 15.4 kJ/mol, S298 = 27.2 J/molK and Cp = 25.2 J/molK.

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Sample Problems

11. Calculate G for the following reductions at 500K and predict the feasibility of the reaction.

a) CuO + H2 = Cu + H2O H500 = -87 kJ/mol S500 = 47 J/molK

b) ZnO + H2 = Zn + H2O H500 = 104kJ/mol S500 = 60 J/molK

12. Using Hesss Law, calculate G for the reaction: Fe + CO2 = CO + FeO. Given the following data at 1600C:

2CO + O2 = 2CO2 G1 = -242 kJ/mol2Fe + O2 = 2FeO G2 = -292 kJ/mol

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Chemical Equilibrium

Established after a certain time such that reactants and products coexist at equilibrium.

Rateforward = Ratereverse

Equilibrium Constant, K expressed in terms of concentration:Gases = Partial Pressure, piLiquid solutions = Molarity, MSolids = Activity, ai

If K >> 1, equilibrium lies to the rightK

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Chemical Equilibrium

Le Chateliers Principle dictates changes in equilibrium direction due to the following factors that affect the position of equilibrium:

Concentration Pressure Temperature

General Equation:

G = G + RTlnK (for non-standard conditions)

At equilibrium, G = 0. Such that: G = -RTlnK

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Sample Problems

13. If for the following reaction: Fe + H2O = Fe3O4 + H2 at 1173K, G = -14.76 kJ. Calculate:

a) K at 1173Kb) pH2 if pH2O = 0.0065atm

14. Calculate K of the following reaction at 727C: FeO(s) + CO(g) = Fe(s) + CO2(g)

Given the following data:FeO(s) + H2(g) = Fe(s) + H2O(g) G = 3150 1.85T calCO2(g) + H2(g) = CO(g) + H2O(g) G = 8600 7.8T cal

Will FeO form if an iron sheet is heated to 727C in an atmosphere containing 10%CO, 2%CO2 and 88%N2?

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Ellingham Diagrams

Developed by H.J.T. Ellingham in 1944 Also known as the standard free energy temperature

(G-T) diagrams Plots of G versus temperature, conveniently

allowing G to be read off directly at any temperature Depicts the dependence of G on temperature

Why study?Feasibility of metallurgical processes, specifically the

reduction of metal ores from oxide forms, can be easily gauged using these diagrams

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Ellingham Diagrams

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Construction of G-T Diagrams

The standard free energy change for this reaction can be

calculated from the equation: G = H TS

Consider the reaction: M(s) + O2(g) = MO2(s)

This equation is of the straight line form: y = mx + b

where: y = G m = -S (slope)x = T b = H (y-intercept)

G is then plotted as per mole of O2 for comparative purposes.

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Reaction Kinetics

Rate of Reaction Rate at which the concentration of a reactant decreases or the rate at which concentration of a product increases

eg. Given the reaction: A B,

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Reaction Kinetics

Experimental Rate Laws State that the reaction rate is proportional to simple powers of reactant concentration

In general, rate [concentration]n

where: n = order of reaction

In case of A + B products, rate [A]x[B]ywhere: p = over-all order of reaction = x + y

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Reaction Kinetics

Zero-order ReactionsGiven the reaction: A products

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Reaction Kinetics

First-order ReactionsGiven the reaction: A products

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Reaction Kinetics

Second-order ReactionsGiven the reaction: A products

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Reaction Kinetics

Half-life, t Time required to decrease the concentration of the reactants to half its initial value

Zero-order:

First-order:

Second-order:

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Reaction Kinetics

Arrhenius Equation Describes the relationship between rate of reaction and temperature using thermodynamic functions

where k = rate of reactionA = constant (frequency factor)Ea = activation energyR = gas constantT = temperature in K

or

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Sample Problems

15. Consider the first order reaction: A products, which has a rate constant, k = 2.95x10-3 sec-1. What percent of A remains after 150 seconds?

16. The reaction A products is second order in A. Initially [AO] = 1.00M and after 25 minutes, [A] = 0.25M. What is the rate constant for this reaction?

17. The rate constant for the reaction H2(g) + I2(g) = 2HI(g) is 5.4x10-4

M-1s-1 at 326C. At 410C the rate constant was found to be 2.8x10-2 M-1s-1. Calculate the activation energy.

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Thermodynamics of Solutions

SolutionMixture of two or more substances which constitutes into a single phase

Solvent major component of a solutionSolute minor components in a solution

Solution = Solvent + Solute

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Activity, ai Defined as the ratio of the vapor pressure, Pi, of the component in solution to the vapor pressure, Pi

o, of the pure component at a given temperature

Activity Coefficient, i Ratio of the activity, ai, to the mole fraction, Xi

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Colligative Properties depend on the number of solute particles and not on the nature of the solute or solvent

1. Vapor Pressure Lowering P = Pxiwhere: P = extent of vapor pressure lowering

P = vapor pressure of pure solventxi = mole fraction of the solvent in the solution

2. Boiling Point Elevation tb = kbmwhere: tb = tbsolution tbsolvent

kb = ebullioscopic constantm = molality

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Colligative Properties depend on the number of solute particles and not on the nature of the solute or solvent

3. Freezing Point Depression tf = kfmwhere: tf = tfsolvent tfsolution

kf = cryoscopic constantm = molality

4. Osmotic Pressure, V = nRT minimum pressure required to prevent osmosis

where: V = volume of solution n = moles of soluteT = absolute temperature

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Electrochemistry

Types of Electrochemical Cells:

Galvanic Cell produces electricity as a result of the spontaneous reactions occurring inside it

Electrolytic Cell non-spontaneous reaction is driven by an external source of current

ELECTROLYSIS a redox reaction brought about by the passage of a direct current through a solution of an electrolyte

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STANDARD REDUCTION POTENTIALS

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Electrochemistry

Determining Standard State Electrode Potentials

Ecell = Ered + Eox

Example. Find the standard cell potential for an electrochemical cell with the following cell reaction:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

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Electrochemistry

Nernst Equation

where: E = non-standard cell potential (V)E = standard cell potentialn = number of electrons involvedF = Faradays constant (96,500 C)K = equilibrium constant

Given that G = nFE (at non-standard and standard states)

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Electrochemistry

Faradays Law of ElectrolysisThe mass (m) of any substance discharged (deposited or dissolved) at an electrode is proportional to the quantity of electricity passed.

1 coulomb (C) = 1 Ampere-second (A s)

where: m = mass (g)MW = molecular weight (g/mol)I = amount of current (A)t = time (sec)n = number of electrons involvedF = Faradays constant (96500 C)

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18. Predict the cell potential for the following reaction when the pressure of the oxygen gas is 2.50 atm, the hydrogen ion concentration is 0.10 M, and the bromide ion concentration is 0.25 M: O2(g) + 4 H

+(aq) + 4 Br-(aq) 2 H2O(l) + 2 Br2(l). Given that:4 Br-(aq) 2 Br2(l) + 4 e

- Eored = + 1.077 VO2(g) + 4 H

+(aq) + 4 e- 2 H2O(l) Eo

red = + 1.229 V

19. Calculate the volume of H2 gas at 25C and 1.00atm that will collect at the cathode when an aqueous solution of Na2SO4 is electrolyzed for 2 hours with 10A current.

20. Determine the oxidation number of chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.5 hours with a 10A current deposits 9.71 grams of chromium metal at the cathode.

Sample Problems

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Met Phy Chem Lecture