The MOLECULES
of LIFE
Physical and Chemical Principles
Selected Solutions for Students
Prepared by James Fraser and Samuel Leachman
Chapter 9
Free Energy
Problems
True/False and Multiple Choice
2.
The work done by biological systems is most commonly
which type of work?
a.
b.
c.
d.
e.
4.
6.
pressure
chemical
home
volume
heat
The value of ∆f Go for elemental oxygen is:
a. –15 kJ•mol–1.
b. 0 kJ•mol–1.
c. Much larger than the free energy of elemental
hydrogen and much less than the free energy of
elemental uranium.
d. Equal to the free energy of elemental tungsten.
e. Both b and d.
At equilibrium:
a. There is no change in temperature over time.
b. The free energy of the system is minimized.
c. The combined entropy of the system and
surroundings is maximized.
d. Only b and c.
e. a, b, and c.
Fill in the Blank
8.
The standard state has a pressure of ___________.
Answer: 1 atm
10. In biochemistry, the standard state for water is ____ M.
Excepting H+, for all other solutions, the standard state
is ______ M.
Answer: 55, 1
Quantitative/Essay Problems
12. Consider the distribution of molecules in energy levels
in the diagram below, where each “X” represents one
molecule. A molecule in energy level 1 contributes one
unit of energy to the total internal energy, a molecule
in energy level 2 contributes two units of energy to the
total internal energy, etc. The system is at 273 K.
Energy level 4
X
Energy level 3
XX
Energy level 2
XXXXX X
Energy level 1
XXXXX XXXXX XXXXX
What are the values of the (a) internal energy (U), (b)
entropy (S), and (c) Helmholtz free energy (A) of the
system? Assume that kB = 1 energy unit•K–1.
Answer:
a. Utotal = ΣNiUi = 1 × 4 + 2 × 3 + 6 × 2 + 15 × 1 =
4 + 6 + 12 + 15 = 37 units
b. W = (Total)!/(Energy level 4)! × (Energy level 3)! ...
= 24!/(1!2!6!15!) = 329,491,008
S = kB ln W = 19.6 units•K–1
c. A = U – TS = 37 units – 273 K × 19.6 units•K–1 =
–5317 units
14. Assume that entropy and enthalpy changes are
independent of temperature. A system in state A has an
enthalpy of –22 kJ and an entropy of 7 J•K–1. In state B,
it has an enthalpy of –12 kJ and an entropy of 15 J•K–1.
At what temperatures will state B be favored?
Answer:
The state B will be favored when GB < GA
Therefore:
HB – TSB < HA – TSA
(HB – HA)/(SB – SA) < T (because SB – SA > 0)
(– 12,000 J + 22,000 J)/(15 J•K–1 – 7 J•K–1) = 1250 K
1250 K < T. The temperature must be greater than
1250 K.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
2
Chapter 9: Free Energy
16. What is the expression for the equilibrium constant for
the reaction A
B + C?
Answer:
[B][C]/[A]
18. Consider the following reactions in which A, B, C, and D
are elements in their most stable states:
A+B→Y
C+D→Z
Y+Z→J
∆Go
Using the following table of
values, what is the
value of ∆f Go for the formation for J?
Reaction
∆Go (kJ•mol–1)
A+B→Y
–70
C+D→Z
23
Y+Z→J
15
Answer:
∆f G(Y) – (∆f G(A) + ∆f G(B)) = –70 kJ•mol–1 → ∆f G(Y) =
–70 kJ•mol–1
∆f G(Z) – (∆f G(C) + ∆f G(D)) = 23 kJ•mol–1 → ∆f G(Z) =
23 kJ•mol–1
∆f G(J) – (∆f G(Y) + ∆f G(Z)) = 15 kJ•mol–1
∆f G(J) = 15 kJ•mol–1 – 70 kJ•mol–1 + 23 kJ•mol–1 =
–32 kJ•mol–1
20. Consider the same reaction as in Problem 19. Assume
that enthalpy and entropy do not vary with temperature.
At what temperature will the reaction begin to proceed
spontaneously in the opposite direction (that is, at what
temperature will ∆Goreactants < ∆Goproducts)?
22. A chemist wants to develop a fuel by converting water
back to elemental hydrogen and oxygen using coupled
ATP hydrolysis to drive the reaction. Given that the
value of ∆ f Go for water is –237 kJ•mol–1 and that one
mole of ATP hydrolyzed to ADP yields –30 kJ•mol–1, how
much ATP is needed to yield three moles of H2 gas?
Answer:
Reaction 1:
2H2+O2 → 2H2O
The ATP reaction:
2H2O + X ATP → 2H2 + O2 + ADP + phosphate
The molar ratio of water to hydrogen gas is 1:1,
therefore 3 moles of H2O will need to be converted.
∆Greaction1 = –711 kJ must be balanced by ATP
hydrolysis.
∆Greaction1 > ∆GreactionATP
–711 kJ > –30 kJ × (moles of ATP)
–711 kJ/–30 kJ < moles of ATP
23.7 < moles of ATP
Therefore at least 23.7 moles of ATP must be
hydrolyzed to yield 3 moles of hydrogen gas. This also
demonstrates how ATP hydrolysis can be used to drive
an otherwise unfavorable process.
24. Explain why the absolute value of the work done by a
process at constant pressure can never be greater than
the absolute value of the Gibbs free-energy change for
the process.
Answer:
The maximum amount of work that can be performed
by a process at constant temperature and pressure
is the Gibbs free energy. It incorporates the minimum
energy that must necessarily be lost as heat. When the
temperature isn’t held constant, additional energy is
lost as heat, which cannot be used as work.
Answer:
At 298 K, the reaction proceeds to products in Problem
19. To proceed towards reactants:
∆Greactants < ∆Gproducts
0 < ∆Gproducts – ∆Greactants
0 < ∆f Hproducts – T∆f Sproducts – (∆f Hreactants – T∆f Sreactants)
0 < –300,000 J – T × 60 J•K–1 + 240 J•K–1 × T + 30,000 J
+ 200 J•K–1 × T
0 < –270,000 J + 380 J•K–1 × T
710.5 K < T
Therefore the temperature must be at least 710.5 K.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science