CMT458TEST3apr2009Ans.doc

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CMT 458/ APR09/TEST 3 TIME = 1hr Total marks = 35

Answer ALL questions

All the symbols used represent their usual meanings.

1. Two of the four differential energy expressions are:

(i) dH = TdS + VdP(ii) dG = -SdT + VdP

a) Derive the two relations above starting from the definitions of H and G .

b) From these two relations obtain the following energy function derivatives:

=

∂∂

P T

G=

∂∂

S P

H

c) What is the Maxwell relation for dG = -SdT + VdP

(10marks)

H = U + PV

dH = dU + PdV + VdP

= (TdS – PdV) + PdV + VdP= TdS + VdP

G = H - TSdG = dH - TdS - SdT

= (TdS + VdP) – TdS - SdT

= VdP – SdT

S T

G

P

−=

∂∂

V P

H

S

=

∂∂

Maxwell relation:

T P P

S

T

V

∂∂−=

∂∂

2. Give a statement of the third law of thermodynamics and explain the significance

of this law to chemists

(3 marks).It is impossible to attain absolute zero of temperatureThe entropy of a pure crystalline substance at absolute zero is zero:S(0 K) = 0

Allows us to determine the absolute entropies of substances i.e provide

basis for calc. of ∆S for chemical rxn.

3. A reaction is not spontaneous at a certain temperature e.g 300K but it is

spontaneous at a higher temperature e.g 500 K. Do you agree with this statement?

Explain your answer ( 5 marks)

Using ∆G = ∆H -T∆S as our guide,Cmt458/test3/apr09



Role of temp can be seen if both ∆H and ∆S have the same signs

If both ∆H and ∆S are negative,

Then ∆G is negative if Temp is low…

Not possible because here not spont at lower (300 K) but spont at higher temp (500K)

If both ∆H and ∆S are positive,

The both ∆G is negative if Temp is high…

This is possible since at lower temp (300K) it is not spont but becomes spont at higher

temp

4. a) What is fugacity?

b) The virial equation for hydrogen is given as )1( bP RT PV += where

b = 6.4 x 10-4 atm-1

Calculate the fugacity of hydrogen at 600 atm and 300 K.

(7 marks)a) Idealized pressure to account for non-ideal bahaviour of gas

b) ∫ −

= P

dP P

Z

P

f

0

1ln

)1( bP RT PV +=

bP RT

PV Z +== 1

b P

Z =

−1

bP bdP dP P

Z

P

f P P

==−= ∫ ∫ 00

1ln = 6.4 x 10-4 x 600 atm

)exp(bP P f = = 881 atm

5. A gas mixture containing 1 mol of CO2, 1 mol of H2 and 2 mol of CO is fed into a

furnace at 900oC. The mixture reacts according to the following reaction.

CO2g) + H2(g) CO(g) + H2O(g)

The dependence of ∆Go

on temperature is given by∆Go = 36400 – 32.01T (Joules)

i) Calculate ∆Go and K p at 900oC

ii) Show that 2)1(

)2(

x

x x K p −

+= where x is the amount of CO2 that has reacted.

Cmt458/test3/apr09



iii) If x = 0.267, calculate the amount (in atm) of each species at equilibrium if the

total pressure in the furnace is 1 atm.

(10 marks)

i) ∆Go = 36400 – 32.01 (900 + 273)

= -1148 J

K RT G ln0 −=∆

∆−= RT

G K

o

exp

= exp (0.1177)

= 1.125

ii)CO2g) + H2(g) CO(g) + H2O(g)

Init 1 1 2 0

React -x -x +x +xEquil 1-x 1-x 2+x x

Total mol = 1-x + 1-x + 2+x + x = 4

Pi

P x

4

1−

P x

4

1− P

x

4

2+

P x

4

2

2

41

44

2

P x

P x

P x

K p

−

+

= = 2

)1(

)2(

x

x x

−

+

iii)

PCO2 = PH2 = P

x

4

1−

==

−

)1(4

267.01atm

0.183 atm

PCO= P

x

4

2+

= 0.5668 atm

PH2O=

P x

4 = 0.0668 atm

END OF TEST

Cmt458/test3/apr09



____________________________________________________________________

These relations/constants MAY be useful

R = 8.314 Jmol-1K

= 0.0821 Latm.mol-1

.K -1

1 L = 10-3 m3

1 L.atm = 101.3 J

+

=∆

1

2

1

2 lnlnV

V R

T

T C S v

−∆=

21

12

1

2lnT T

T T

R

H

K

K

∫ −

= P

dP P

Z

P

f

0

1ln

Cmt458/test3/apr09

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CMT458TEST3apr2009Ans.doc