MEt04011MetThermodynamicsI(1)

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Ministry of Science and Technology

Department of Technical and Vocational Education

Promotion Examination for A.G.T.I. Year I

Date: 26.10.06 Time: 12:30-

15:30

Met-04011 Metallurgical Thermodynamics I

Answer any five.

1 (a) The combustion of acetylene fuel with nitrous oxide as oxidant at 25ºC, is

widely performed in flame emission spectrometry. Calculate the maximum

temperature attained if the best mixture corresponds to the reaction:

(C2H2) + 3 (N2O) = 2(CO) + (H2O) + 3 (N2)

Assume that water is undissociated.

ΔHº298 (C2H2) = 54.23 kcal/mole, ΔHº298 (N2O) = 19.70 kcal/mole,

ΔHº298 (CO) = -26.42 kcal/mole, ΔHº298 (H2O) = -57.80 kcal/mole,

Cp for (CO) = 6.8 + 1.0 x 10-3T – 0.11 x 105T-2 cal/deg/mole

Cp for (H2O) = 7.17 + 2.56 x 10-3T – 0.08 x 105T-2 cal/deg/mole

Cp for (N2) = 6.5 + 1.0 x 10-3T cal/deg/mole

Ans: (C2H2) + 3 (N2O) = 2(CO) + (H2O) + 3 (N2)

ΔHº298 (Reaction) = 2 ΔHºCO + ΔHºH2O + 3 ΔHº N2 – (ΔHºC2H2 + ΔHº N2O)

= (2 x -26.42) + (-57.80) + (0) – (54.23) – (3x19.70)

= -223.97 Kcal

Heat evolved = dTpC product

Tm

∫ ∑298

223970 = ( )∫ ++Tm

NOHCO dTCpCpCp298

2232

=

( )( ) ( )

dTTx..Tx.Tx..

Tx.Tx..Tm

∫

++++

+−+−−−

−−

2983253

253

10035191008010562177

102201002613

= ( )dTTx.Tx..Tm

∫ −− −+298

253 10140105672740

= [ ]Tm298Tx.Tx.T. 1523 10140107832740 −− ++



223970 = 40.27 (Tm – 298) + 3.78 x 10-3 (Tm2 – 2982) + 0.14 x 105

−

298

11

Tm

Neglecting Tm-1

223970 = 3.78 x 10-3 Tm2 + 40.27 Tm – 12383

3.78 x 10-3 Tm2 + 40.27 Tm – 236353 = 0

Tm = 4207 K

Maximum temperature = 4207 – 273 = 3934ºC

The enthalpy changes for the following reactions are as follows:

2B + 2 H2 + 3 O2 + aq = 2 H3BO3 (dil solution) ΔHº298 = - 512.8 kcal

B2O3 + 3 H2O (l) + aq = 2 H3BO3 (dil solution) ΔHº298 = - 4.12 kcalH2 + ½ O2 = H2O ΔHº298 = - 68.73 kcal

I (b) Calculate the standard heat of formation of B2O3 in terms of per mole of B2O3,

and per gram of B2O3. Atomic weight of B and O are 10.82 and 16.0 respectively.

Ans: 2B + 2 H2 + 3 O2 + aq = 2 H3BO3 (dil solution) ΔHº298 = - 512.8 kcal

B2O3 + 3 H2O (l) + aq = 2 H3BO3 (dil solution) ΔHº298 = - 4.12 kcal

2B + 3 H2 + 3 O2 + = 2BO3 ΔHº298 = - 508.68 kcal

3H2 + 3/2O2 = 3H2O ΔHº298 = - 206.19 kcal

2B + 3/2O2 = B2O3

ΔHº298 = - 302.49 kcal

Relative molecular mass of B2O3 = 2 x 10.82 + 16 x 3= 69.64 gm/mole

Standard ht of formation of B2O3 = - 302.49 Kcal/mole

= - gm/Kcal.

.

6469

49302

= - 4.3436 Kcal/gm

2 (a) Calculate the entropy change of the system and surroundings for the

isothermal freezing of one mole of super cooled liquid silver at 850ºC, when the

surroundings are also at the same temperature.



Heat of fusion at the melting point of 1234ºK is 2690 cal/mole.

Cp for solid silver is 5.09 + 2.04 x 10 -3T cal/deg/mole and liquid copper is 7.3

cal/deg/mole.

{Cu}850 + 273 = 1123 <Cu>1123

ΔSsys = ?, ΔSsyr = ?

ΔFfus = 2690 cal/mole , Tfus = 1234K

Cp(s) = 5.09 + 2.04x10-3T , Cp(l) = 7.3 cal/deg/mole

Ans: {Cu}1123 <Cu>1123 ΔS1 = ∫ 1124

1123T

dTCpl

{Cu}1124 <Cu>1124 ΔS2 =t

t

T

∆Η

<Cu>1124 <Cu>1123 ΔS3 = dTCps∫ 1123

1124

{Cu}1123 <Cu>1124

ΔSsys = ΔS1+ ΔS2 +ΔS3 =

( )dT

T

Tx..dT

T

.∫ ∫

−++

−+

1123

1234

31234

1123

10042095

1234

269037

= ( )1234112310021

1234

1123095

1234

2690

1123

123437 3 −++− −x.ln.ln.

= 0.688 – 2.1799 – 0.479769 – 0.11322= - 2.084789

∫ ∫ −++−=∆Η =∆Η −1234

1123

1123

1234

310042095269037 TdTx..dT.syssur

=7.3(1234-1123)-2690+5.09(1123-1234)+1.02x10-3(11232-12342)

= 810.3-2690-564.99-266.859

= -2711.549 cal

deg/cal..THSsys

sur 1973621234

5492711 +=+=∆−=∆



0847892197362 ..SSS sursystotal −+=∆+∆=∆

= 0.1125769 cal/deg

2 (b) Calculate the standard enthalpy and entropy change at 298 K for the

following reaction

2<Cu> + ½ O2 = <Cu2O> ΔGº = -400,500 – 3.92T log T + 29.5 T cal.

Ans: 2<Cu> + ½ O2 = <Cu2O>

ΔGº = -400,500 – 3.92T log T + 29.5 T cal

529923500400

.Tlog.T

,

T

G+−−=

∆

T.T

,

T

T/G 1

7021

5004002 −=∂

∆∂

T

.

T

,

T

702150040022

−=∆Η

−

T., 7021500400 +−=∆Η

= - 400,500 + 1.702x298

= - 400,500+507

= - 399993 cal

= - 399.993Kcal

52970217021 .Tln..T

G+−−=

∂∆∂

Tln..S 702179827 −=∆

298702179827298 ln..S −=∆

= - 18.10 cal/ deg

3.(a) The melting point of gallium is 303K at 1atm. The densities of solid and

liquid gallium are 5.885 g/cc and 6.08 g/cc. the heat of fusion of gallium is 18.5

cal/g. Calculate the change in melting point of gallium for an increase of pressure of

1 atm.

Ans: ( )sl

fus

V V TdT

dP

−∆Η

=

Let atomic wt. Of gallium=x



ρ

−ρ

×=

sl

xx

.x.

dT

dP

303

29341518

,cc/g.,cc/g. sl 8855086 =ρρ1cal=41.293cc atm

dP=1atm, dT=?

−

×=

8855086303

29341518

.

x

.

x

.x.

dT

dP

= -462.7atm/deg

atmdeg/x.dP

dT 310162 −−=

dT= - 2.16x10-3 deg at dP= 1 atm

∴melting point of gallium will be decreased by 2.16x10-3 deg

3.(b) In an investigation for the thermodynamic properties of a managanese, the

following heat contents were determined:

Hº700 -Hº298 = 2895cal/g-atom

Hº1000 -Hº298 = 2895cal/g-atom

Find applications suitable equation for

298HHT − and also for Cp as

function of temperature in the form (a+ bT). Assume that no structural

transformation takes place in the given temperature range.

Ans: atomg/calH −=−Η 2895298700

( )

700

298

2

700

298

298700

2981000

22895

5450

+=

+=−

+=−=−Η

∫

bTaT

bTaHH

bTaCp

atomg/calH



( ) ( )

( ) ( )

( ) ( )

( ) ( )

761753325451087831

761753108783132545

2982

107566329832545

2982

298

32545

107566321

276357649

4555987025450

29810002

2981000

127499

2005984022895

2987002

2987002895

23

23

22

3

22

298

3

22

2981000

22

.T.Tx.

.Tx.T.

Tx.

T.

T b

TaHH

.a

x. b)(x)(

)......(........... ba

ba

baHH

).......(........... ba

ba

ba

T

−+=

−+=

−+−=

−+−=−∴

=

=⇒

=+

+=

−+−=−

=+

+=

−+−=

−

−

−

−

4.(a) Calculate the vaccum needed for obtaining molybdenum metal according to

the reaction

MoS2 = Mo + S2 at 800ºC and 1000ºC.

MoS2 + 2H2 = Mo + 2H2S; K 1

2H2S = 2H2 + S2; K 2

At temperature 800ºC K 1 is 7.3x10-6 and K 2 is 2.2x10-4 and at temperature

1000ºC K 1 is 1.439x10-3 and K 2 is 5.6x 10-3.

Solution

MoS2 + 2H2 = Mo + 2H2S , K 1

2H2S = 2H2 + S2 , K 2

MoS2 = Mo + S2, K = K 1K 2



22

2s

MoS

sMoP

a

paK ==

at 800C K 1 = 7.3x10-6, K 2 = 2.2x10-4

K = K 1K 2 = 7.3x10-6

x 2.2x10-4

K = Ps2 = 1.606x10-9

at 1000C K 1 = 1.439x10-3, K 2 = 5.6x10-3

K = K 1K 2 = 1.439x10-3 x 5.6x10-3

K = Ps2 = 8.0584x10-6

4.(b) Calculate the mean of vaporization of palladium and vapor pressure at

1568C from the following vapor pressure data :7.516x10-7

mmHg at temperature1314C and 1.164x10-6mmHg at 1351C.

Ans: ΔHv = ?, T = 1568+273 = 1814K → P = ?

T1= 1341+ 273 = 1597K, p1= 7.516x10-7mmHg

T2= 1351+ 273 = 1624K, p2= 1.614x10-6mmHg

4

7

7

6

121

2

103046

731986

1597

1

1841

1

9871

145871

105167

145871

15971

16241

9871105167106141

11

−

−

−

−

=

=

−

−=

=∆

−∆−=

−

∆−=

x.P

.

.x.

Pln

calH

.H

x.x.ln

TTR

H

P

Pln

v

v

v

5.(a) Determine the lowest temperature at which copper oxide (Cu2O)can

dissociate in a vaccum of 10-5mmHg.

Ans: Cu2O(s) = 2Cu(s) + 1/2 O2 ΔG = 40500+3.92T log T – 29.5T

Solution

Cu2O(s) = 2Cu(s) +1/2 O2

ΔG = 40500 + 3.92T log T -29.5T

21

2

2

221

2 5754 oOCu

Cuo PlogT.a

a.PlnRTG −=−=∆



In a vaccum of 10-5mmHg, taking Po2 in air as 0.2 atm

( )

092314940500

61952992340500

6191063525754

106352760

11020

21

9

952

=+−=−+

=−=∆

==

−

−−

ddd

o

TlogT.T.

T..TlogT.

T.x.logT.G

atmx.xx.P

T 500 1000

Eqn 21240 3160

by graphically→

The lowest temperature = 1085 K

5.(b) Calculate the equilibrium constant an equilibrium partial pressure of oxygen

for the reaction Zr + O2 = Zr O2 ΔG = 259940 + 2.33T logT –

59.12T

at 2000K. Also predict the possibility of decomposing a pure zirconia crucible

under a vaccum of 10-5mmHg.

Zr + O2 = Zr O2 ΔG = 259940 + 2.33T logT – 59.12T

ΔG = 259940 + 2.33 x 2000 log 2000 – 59.12 x 20000

- RT ln K = 156885

lefttoRightactionReK Q

PQ

mmHgx.k

P

PP.a

aK

x.K

.x.

K ln

O

O

OOZr

ZrO

∴⟩

==

==

==

=

−=−

=

−

5

2

17

2

22

2

18

101

10385811

1

102177

473920009871

156885

Zirconia crucible will be decomposed.



6.(a) In separate experiments, pure nitrogen at atmospheric pressure is passed

over either mercury or a mercury-thallium amalgam(68.4 atom percent thallium).

At temperature less than 100C the vapor pressure of thallium is negligible, relative

to that of mercury. The flow rate of nitrogen is controlled such that the gas is just

saturated with mercury vapor as it passes over the pure mercury or the amalgam.

The vapor carried over in the gas stream is condensed in a cold trapped an

weighted. In experiments with the samples at 26 C, the weight of mercury trapped

from the pure mercury an the amalgam were respectively 2.20 and 0.75mg per

hundred liter of nitrogen, the nitrogen volume being measured at 20 C and 1

atmosphere pressure. Assuming that equilibrium is continuously maintained over

the samples, calculate the partial pressure of mercury over the amalgam and the

activity coefficient of mercury in the amalgam at 26 C. (atomic weight of Hg =

200.6 g / mole)

Ans: Hg(pure,l)=Hg(g,in N2)

K P=HgpureHg

Hg

a

ρ………….(1)

= PHg(pure Hg)Hg(l,amalgam)=Hg(g,in N2)

K P=amlgamaHg

Hg

a

P

…………….(2)

Assuming Hg vapor behave ideally, (1) = (2)

PHg(pureHg)=

amlgamaHg

Hg

a

p

AHg=)pureHg(Hg

)amlgama(Hg

P

P…………….(3)

Volume of (100liter at 20C) at 26 C = literx 102273

299100 =

.NTPatcc..x.

.occupiesmgHg.,mole/g.MHg 239022

6200

422226200 =

=



Catccx. 26273

2992390=

= 0.262 cc at 26 C

∴ partial pressure of Hg over pure Hg = 7601020002620 x.

mmHgx.PHg40 10519 −=

at NTP 0.75mg Hg occupies = cc..x.

.0820750

6200

422=

at 26 C 0.75mg Hg occupies =0.082x cc.0890273

299=

∴ partial pressure of Hg over amalgam = 760102000

0890 x.

PHg=6.65x10-4mmHg

(3)

0813160

340

34010519

106564

4

0

..

.

N

a

.x.

x.

P

Pa

Hg

HgHg

Hg

HgHg

===ρ

=== −

−

6.(b) Chromium plates are bright annealed at 1000K in a wet hydrogen

atmosphere. The pressure of wet hydrogen is 1atm. (i) Calculate the permissible

water content in the hydrogen if there is to be no oxidation at 1000K (ii) Will

annealed chromium plates be oxidized when cooled to 500K in the furnace

atmosphere, as calculated in (i)? neglect the possibility of dissolution of hydrogen

in chromium.

Given: 2<Cr>+3(H2O) = <Cr 2O3>+3(H2) ΔGº= - 91050+22.8Tcal.

Ans: 2<Cr>+3(H2O) = <Cr 2O3>+3(H2) ΔGº= - 91050+22.8Tcal

68250

100082291050

82291050

1000

−=−+−=∆

+−=∆

K lnRT

x.G

T.GT



RighttoLeft.QK

K Q

x.K ln

K lnRT

x.G

x.P

.P

P

.PP

x.P

PK

.K ln

OH

OH

OH

OH

H

OH

H

>==

=

−=−+−=∆

=

=−

=

=

=

=

−

16533

5009871

9650

9650

50082291050

10061

27938361

2793836

102658

34834

500

500

5

2

2

2

2

2

14

3

2

21000

Cr. Plate will be oxidized at 500 ºK

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MEt04011MetThermodynamicsI(1)