Answers to Chemistry 254 final exam, April 2009 Question 1. a) 10 pnts

1

1 1

378.7 ( 393.5 0) 14.8

248.7 151.7 130.7 33.7r

r

H kJ mol

S J mol K

−

− −

Δ = − − − + =

Δ = − − = −

b) 10 pnts

1

5

14800 298.15*( 33.7) 24842.6

ln( ) 10.0269 4.42*10

r r r

rp p

G H T S JmolGK K

RT

−

−

Δ = Δ − Δ = − − =Δ

= − = − → =

c) 10 pnts

4

1 1ln( (600)) ln( (298)) ( )600 298

14800 1 110.026 ( ) 10.026 3.004 7.0228.314 600 298

(600) 8.94*10

rp p

p

HK KR

K −

Δ= − −

= − − − = − + = −

=

Assumption is made that rHΔ is constant (the value at 298). d) 10 pnts

2

1

1

3 1

ln( )

(600) (298) (600 298.15)

14600 (45.2 37.1 28.8)*301.8514800 20.7*301.85 8549

ln( )2.86*10

p r

r r v

p

d K HdT RT

H H C

J molJ mol

d KK

dT

−

−

− −

Δ=

Δ = Δ + Δ −

= + − −

= − =

=

e) 10 pnts.

1

0

2

( ) (100)

100 8.94*10

p x x

x p

PK K KP

K K

νΔ −

−

= =

= =

Question 2: a) 10 pnts molecule n(t=0) ( )n ξ ( )x ξ H2 1 1-ξ (1-ξ )/(10-ξ ) CO2 9 9-ξ (9-ξ )/(10-ξ ) HCOOH 0 ξ ξ /(10-ξ ) total 10 10-ξ 1 b) 5 pnts. The minimum value of ξ is 0, the maximum value of ξ is 1, 0 1ξ≤ ≤ otherwise (at least) one of the mole numbers is negative. (5 pnts) c) 10 pnts.

(10 )(10 )(1 ) (9 ) (1 )(9 )

(10 ) (10 )

xQ

ξξ ξξ

ξ ξ ξ ξξ ξ

−−= =− − − −− −

d) 10 pnts

2

2 2

2

2

(10 )0.1

(1 )(9 )

0.1(1 )(9 ) 10

0.1(9 10 ) 10

1.1 11 0.9 0

11 11 4*0.9*1.1 (15.82,0.082)2*1.1

eq eqx

eq eq

eq eq eq eq

eq eq eq eq

eq eq

eq

Kξ ξξ ξ

ξ ξ ξ ξ

ξ ξ ξ ξ

ξ ξ

ξ

−= =

− −

− − = −

− + = −

− + =

± −= =

The value of 15.82 is not allowed (negative number of moles), hence 0.082eqξ = e) 5pnts:

The number of moles of HCOOH at equilibrium is simply 0.082eqξ = f) 5 pnts:

If only 0.01 moles of HCOOH is formed, equilibrium has not been reached in half an hour. Using a catalyst would speed up the time to reach equilibrium and hence the yield of the reaction for a given reaction time. g) 5 pnts:

The maximum amount of HCOOH that can be obtained is the equilibrium value. Hence finding 0.15 moles of HCOOH is not possible, if we trust the thermodynamic data. Note: we used ideal behaviour. It might be that fugacity coefficients alter the theoretical expectation.

Question 3: Quintessential Thermo Part I: a) 12 pnts: 0U HΔ = Δ = as the process is isothermal 10 *(10 20) *100 / 10000 10extw P V bar L J Lbar J kJ= − Δ = − − = = 10q w kJ= − = −

12

1

10ln ln 5.7620gas

VS nR R JKV

−⎛ ⎞Δ = = = −⎜ ⎟

⎝ ⎠

1862.3gasG H T S JΔ = Δ − Δ = b) 8 pnts 10bath systemH q kJΔ = − =

110000 30.95323.15

systembath

qS JK

T−−

Δ = = =

0bath bath bathG H T SΔ = Δ − Δ = c) 5 pnts 130.95 5.76 25.19 0bath gasS S JK −Δ + Δ = − = >

The process is spontaneous, or physically possible as 0SΔ > 1862.3gas bathG G JΔ + Δ = Part II:

a) 12 pnts: Now the process is reversible isothermal:

0U HΔ = Δ = as the process is isothermal

10 10

20 20

10ln 1862.320

L L

L L

dVw PdV nRT nRT JV

= − = − = − =∫ ∫

1862.3q w J= − = −

12

1

10ln ln 5.7620gas

VS nR R JKV

−⎛ ⎞Δ = = = −⎜ ⎟

⎝ ⎠

1862.3gasG H T S JΔ = Δ − Δ = b) 8 pnts: 1862.3bath systemH q JΔ = − =

11863.3 5.76323.15

systembath

qS JK

T−−

Δ = = =

0bath bath bathG H T SΔ = Δ − Δ = c) 5 pnts: 15.76 5.76 0bath gasS S JK −Δ + Δ = − =

The process is reversible as 0totalSΔ = 1862.3gas bathG G JΔ + Δ =

Further comments: This was a difficult question, as I learned while going through your answers. Parts a) were straightforward as they discuss the process for a gas. Here you can use that , , ,H S U GΔ Δ Δ Δ are state functions, and yield the same answer in parts I and II. Only the work and heat is different. Part b) is confusing. The gas in the vessel transfers heat to the bath, and hence the temperature of the bath will change, but only a tiny bit, because it is very large. So you should not use 0HΔ = because the bath is isothermal (many of you did). I mentioned that the bath is isothermal in the question, such that you can calculate

bathSΔ (but I should really have said: approximately isothermal). Finally, to calculate GΔ you can use again that the temperature stays constant:

G H T SΔ = Δ − Δ . In reality this is not quite true, and there would be a complicated additional contribution bathS TΔ . Knowing part b) it is straightforward to do part c), and the different processes make sense: I can compress a gas isothermally using a constant external pressure (it is a physically allowed process, 0totalSΔ > ). I would lower a mass, if that is what I use to apply the pressure. In part II, the reversible process, we would lower ‘grains of sand’, as much as possible from any given height. This is the reversible process and 0totalSΔ = . Many of you used that , ,S H GΔ Δ Δ are state functions and therefore obviously the answers of part b) and c) are the same. You see above that for the heat bath they are not. Why? How can this be? Again the answer is subtle. In the analysis we forget about the mass that is lowered in the process (in order to apply the pressure). In part I, we lowered a large mass (to apply 10 bar of pressure). In part II we lowered grains of sand from different heights to apply at each moment the reversible pressure. Obviously the change in potential energy of the mass is not the same, and this also affects the state of the heat bath, as both are part of the environment. The final states are really not the same. It would be most appropriate to take a finite large heat bath, with a finite large heat capacity and do the problem exactly for all pieces. In the limit that the heat capacity is infinite (no temperature change) we get the result of the question (except for bathGΔ , which I don’t think is 0). In summary: ‘quintessential thermodynamics’ is hard. I wanted to remind you of that.

4. Maintaining suitable ice conditions at the Vancouver winter games. a) 10 pnts:

3 15 3 1

3

3 15 3 1

3

18.02*10 1.9587*10920

18.02*10 1.8074*10997

icem

waterm

kg molV m molkg m

kg molV m molkg m

− −− −

−

− −− −

−

= =

= =

b) 10 pnts

1

5 3 1

5 1 3

5 1

1

( )

6010273.15 *( 0.1513*10 )145.4*10145.4*10145.4

fusions lwater ice

s l s l fusion m m

HdP SdT V T V V

J molK m mol

JK mPaK

barK

−

− −

−

− −

− −

−

−

ΔΔ⎛ ⎞ = =⎜ ⎟ Δ −⎝ ⎠

=−

= −

= −

= −

c) 10 pnts

27 1 2

4

5

60 *9.8 1.96*100.3 *1*10

196*10 196

Mg kg msP kgm sA m m

Pa bar

−− −

−= = =

= =

d) 10 pnts

196 1 195P bar bar barΔ = − =

1195 195145.4 1.35145.4s l

P dP bar barK T KT dT T K

−

−

Δ ⎛ ⎞≈ → = − → Δ = = −⎜ ⎟Δ Δ −⎝ ⎠

e) 10 pnts.

To create ice suitable for sprinting the ideal temperature would be a bit lower than -1.35 ˚C, while for long distance skating the temperature should be a bit higher (following our thermodynamic analysis).