Problem Set #7, Chem 340, Fall 2013 –Due Friday, Oct 18, 2013

To hand in : 1. Atkins 3.11(b)

2. Atkins 3.18 (b)

This seems to have a sign error in

formula for V but answer is right,

high pressure should reduce

volume. Problem is definition of

missed – sign: =-(1/V)(dV/dP)

3. Atkins 3.19 (b)

4. Atkins 3.16

Chap. 3 did not

define i but it

equals Gm,i

Was in lecture

This is formula from book

and lecture

5. Atkins 3.26

6. Atkins 3.38

7. Atkins 3.40

8. Engel P5.36

P5.36)

Plotting the data for crystalline glycine as 0

mp,C versus T gives:

To obtain the molar entropy of crystalline glycine at 300 K we calculate the area under the curve for each

temperature increment, divide the area by the upper temperature of the increment, and adding that number for all

temperature increments. The area for each temperature increment can be obtained by calculating the areas under

the red and blue step functions, adding the two, and dividing by two. For example the molar entropy for the

increment (lined area):

K 1002

K mol J 35.280K K 100K mol J 43.2K 80K 100 K) 80K (100S

-1-1-1-1

m

-1-1

m K mol J 84.7K) 80K (100S

Doing this calculation for all increments and adding yields: -1-1

m K mol J 4.69 K) 300 (glycine,S

50 100 150 200 250 300T K

20

40

60

80

100

120

Cp

50 100 150 200 250 300T K

20

40

60

80

100

120

Cp

9. Engel P5.45

P5.45 The entropy of denaturation at 340 K can be calculated as:

1-1-

-1den

den mol K J 6.1882 K 340

mol J 640100

T

ΔHK 340ΔS

The entropy of denaturation at 310 K is:

K 310

K 340

pden

K 340

K 310

p

denden dTT

1ΔCK 340ΔSdT

T

ΔCK 340ΔSK 310ΔS

1-1-1-1-1-1-

den mol K J 4.1109 K 340

K 310lnmol K kJ .378mol K J 6.1882K 310ΔS

The enthalpy of denaturation at 310 K is:

-1-1-1-1

denden mol kJ K 43.93 mol J K 1109.4 K310ΔS K310ΔH T

10. Engel P6.13

1-1-1-

0

H

0

NH

0

i

0

i

0

reaction

mol kJ 75.352 mol kJ 1.2372

3mol kJ 3.801

ΔG2

3ΔG1ΔGΔG

23

Oaq

i

The equilibrium constant is then:

49-

1-1-

-1-10

reaction

p 10.455 K 982mol K J .3144728

mol kJ mol kJ 75.352Exp

T R

ΔGExpK

Equilibrium constant goes past

the Wednes. lecture, can skip

11. Engel P6.20

P6.20) Calculate the Gibbs energy change for the protein denaturation described in Problem

5.45 at T = 310.K and T = 340.K.

From P5.34 we have:

-1-1

den mol K J 4.1109 K 310ΔS

-1-1

den mol K kJ 43.93 K 310ΔH

K 310ΔGden is then:

1-1-1-1-

dendenden

mol kJ 3.31 mol K J 4.1109K 982mol kJ 43.93

K 310ΔS TK 310ΔHK 310ΔG

At 340 K we obtain:

1-

1--1

reaction

mol kJ 87.47

K 310

1

K 340

1mol kJ 343.9

K 310

mol J 13300K 340K 340 ΔG

Extra problems 1. Atkins 3.12 (b)

2. Atkins 3.16 (b)

3. Atkins 3.17 (b)

4. Atkins 3.21 (b)

5. Atkins 3.24

s

6. Atkins 3.25

The double derivative method used

in class is much clearer than this. It

seems steps are missing

7. Atkins 3.28

8. Atkins 3.37

9. Engel P5.27

The standard entropy for this reaction is:

1-1-11-11-11-

11-11-

i

if,ireaction

mol K J 4.24 mol K J 13.011 mol K J 01mol K J 05.222

mol K J 13.822mol K J 0.072

S ν)K 298(ΔS

10. Engel P5.20

The standard entropy and enthalpy and heat capacity for this reaction are:

1-1-11-11-

11-11-

i

if,ireaction

mol K J 4.262 mol K J 13.826mol K J 0.076

mol K J 05.226mol K J 09.221

S ν)K 298(ΔS

1-

11

11

i

if,ireaction

mol kJ 7.2802

mol kJ 5.3936mol kJ 8.2856

mol kJ 06mol kJ 1.12731

H ν)K 298(ΔH

1-1-

11-11-

1-11-1

reaction p,

mol K J 8.278

mol K J 7.136mol K J 5.376

mol K J 9.426mol K J 19.221)K 298(C

At 310 K the entropy and enthalpy for this reaction are then:

1-1-1-1-1-1-

i

freactionp,reactionreaction

mol K J 7.290 K 15.298

K 330lnmol K J 278.8mol K J 4.262

T

TlnC)K 298(ΔS)K 310(ΔS

1-1-1-1-

ifreactionreactionreaction

mol kJ 8.2793 K 15.298K 330mol K J 278.8mol kJ 802.72

TTC)K 298(ΔH)K 310(ΔH

And the entropies for the surroundings and universe:

-1

3 -1 1reactionsurroundings

2793.8 kJ mol-dq HΔS 8.47 10 J mol K

T T 330 K

-1 -1 3 -1 1

universe reaction surroundings

3 -1 1

ΔS ΔS ΔS 290.47 J K mol 8.47 10 J mol K

8.76 10 J mol K

11. Engel P5.41

denS corresponds to the area under the trs

pC curve (see figure P4.31). Dividing the area into

small fragments, adding them up, and multiplying by the molecular weight of the protein yields:

-1-1

den mol K J 708 ΔS

Calculating the enthalpy of denaturation in the same way the entropy of denaturation was

calculated in P5.41 yields:

-1-1

den mol kJ mol kJ 216.63 ΔH

The entropy of denaturation is then:

1-1--1

den

denden mol K J 712.6

K 304

mol kJ 216.63

T

ΔHΔS

Engel P 4.31

12. Engel P6.4

a) for the isothermal reversible path

1 1 3

ln ln

10.0 L2.50 mol 8.314 J mol K 298 K ln 9.97 10 J

50.0 L

f

i

P

f i

i fP

P VG VdP nRT nRT

P V

1 1 3

ln

50.0 L2.50 mol 8.314 J mol K 298 K ln 9.97 10 J

10.0 L

f

i

V

f

iV

VA PdV nRT

V

b) Because A and G are state functions, the answers are the same as to part a) because the systems go

between the same initial and final states, T,Vi → T,Vf.

G – A = H – U = PV) = nRT). Therefore, G = A for an ideal gas if T is constant.

13. Engel P6.9

We first need to know how much energy is generated in the metabolism of glucose:

1-

1-1-1-1-

0

OHC

0

O

0

CO

0

OH

0

i

0

i

0

reaction

mol kJ 2878.4

mol kJ 6.1091mol kJ .006mol kJ 4.9436mol kJ 237.16

ΔG1ΔG6ΔG6ΔG6ΔGΔG6126222

i

Now we need to convert the work performed by the horse:

1

1-1-2-1-1-

horse

h kJ 2685.5

hmin 60 min s m 9.81ft m 0.3048ft 100lb kg 0.4536lb 330

h g mw

That means that the horse needs the following amount of glucose:

g 168.1 mol g 180.18mol kJ 2878.4

h kJ 2685.5M

w

wm 1

1

1

glucose

glucose

reaction

glucose