BioPHYSICAL chemistry

3.3 For a spontaneous process, the change in the Gibbs energy will be

negative.

3.4 The process is endothermic.

3.5

3.6

3.7 (a)

(b)

3.8 The entropy change is 30 J K−^1 and the enthalpy change is −85 J.

3.9 The entropy change of the cold block is opposite in sign and larger

in magnitude than the entropy change of the hot block.

3.10 The change in the Gibbs energy is negative.

3.11 The change in entropy will always be greater than the ratio of the

heat divided by the temperature.

3.12 The total entropy change is zero.

3.13 The change in internal energy is zero since the process is adiabatic

and no work is performed. The process of burning is spontaneous

so the change in entropy is positive. Since the system is adiabatic

the change in entropy for the surroundings is zero.

3.14 C 12 H 12 O 11 (solid) +12O 2 (gas) →12CO 2 (gas) +11H 2 O (liquid)

ΔH=12(−393.5 kJ mol−^1 ) +11(−285.8 kJ mol−^1 ) −1(−2222 kJ mol−^1 )

=−5644 kJ mol−^1

3.15

3.16 Hybrid cars make use of regenerative braking to charge batteries,

use electric motors when the gasoline engine is inefficient, and turn

off of the gasoline engine when idle.

3.17 Drugs that are selected based upon enthalpic optimization are usu-

ally more selective with a higher binding affinity than that obtained

using entropic optimization.

3.18 Although the reaction is favorable, it is a complex multi-electron

process that involves a large number of intermediatesteps.

Δ

Δ

S

H

T (. )

==.

×

+

=

510 −

273 15 80

1

(^51) Jmol
K
4 42 J/(Kmol)
Ef ciency

K

K

1 =− 11 =− =

333

373

T

T

cold

hot

0 011.

Ef ciency

K

K

1 =− 11 =− =

293

373

T

T

cold

hot

0 021.

ΔSnR

V

V

nal

initial

==ln^30

ΔSnR

V

V

nal nR nR

initial

===−ln^3 ln. 05 069.

444 ANSWERS TO PROBLEMS

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