7.16
(E A1 − E A2 ) = RT(ln k 2 − ln k 1 )
= 0.008314 kJ/(Kmol)(298 K)(ln 1 − ln 1000)
= −17.11 kJ mol
−1
7.17 It is a competitive inhibitor.
7.18 It is an uncompetitive or mixed inhibitor.
7.19 The initial velocity is reaching a value of 280 μM min
−1 at high substrate concentrations so this is assigned as the maximum velocity. The
half velocity would be 140 μM min
−1
, which occurs at 1 × 10
−5 M
and is assigned as the K m .
7.20 The temperature in the exponent will decrease, making the freeenergy dependence much sharper than at room temperature. There
is also a weak temperature dependence in the coefficient.
7.21 As the free-energy difference approaches the reorganization energy,
the activation energy decreases until the process becomes activationless
when the rate reaches a maximum. At larger free energies the difference between the free-energy difference and reorganization energy
increases again and the rate decreases.
CHAPTER 8
8.1 Probability = number of spades/number of cards = 13/52 = 1/4
8.2 Probability = number of cards with a four/number of cards
= 4/52 = 1/13
8.3 There are 9 possible outcomes. (a) The probability of three heads is
1 of 9. (b) The probability of two heads and one tail is 3 of 9.
8.4 Number = (52)(51)(50) = 132,600
8.5 (a) Probability = 1/4
3
= 1 out of 64; (b) probability = 1/4
3
= 1 out
of 64; (c) probability = 1/4
7
= 1 out of 16,384; (d) probability = 1/4
12
= 1 out of 16,777,216.
8.6 (a) Number of outcomes = 2
3
= 8; (b) number of energetically distinct
states = 4; (c) the three up spins have the highest energy, followed
in order by two up and one down, one up and two down, and three
down.
8.7 (a) e
−(E 2 −E 1 )/k B T
= e
−(9×10
−24 J/(1.38×10
−23 J K
−1
)(298 K)
= e
−0.021
≈ 0.979
(b) e
−(E 2 −E 1 )/k B T
= e
−(9×10
−24 J/(1.38×10
−23 J K
−1
)(4 K)
= e
−0.16
= 0.849
ln
ln
(
)
(
)
k
k
RT
E
E
E
E
A
A
A
A
2
1
1
2
1
2
1
−
= −
−
→
−
(ln
ln )
=
−
RT k
k
2
1
452
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 452
(E A1 − E A2 ) = RT(ln k 2 − ln k 1 )
= 0.008314 kJ/(Kmol)(298 K)(ln 1 − ln 1000)
= −17.11 kJ mol
−1
7.17 It is a competitive inhibitor.
7.18 It is an uncompetitive or mixed inhibitor.
7.19 The initial velocity is reaching a value of 280 μM min
−1 at high substrate concentrations so this is assigned as the maximum velocity. The
half velocity would be 140 μM min
−1
, which occurs at 1 × 10
−5 M
and is assigned as the K m .
7.20 The temperature in the exponent will decrease, making the freeenergy dependence much sharper than at room temperature. There
is also a weak temperature dependence in the coefficient.
7.21 As the free-energy difference approaches the reorganization energy,
the activation energy decreases until the process becomes activationless
when the rate reaches a maximum. At larger free energies the difference between the free-energy difference and reorganization energy
increases again and the rate decreases.
CHAPTER 8
8.1 Probability = number of spades/number of cards = 13/52 = 1/4
8.2 Probability = number of cards with a four/number of cards
= 4/52 = 1/13
8.3 There are 9 possible outcomes. (a) The probability of three heads is
1 of 9. (b) The probability of two heads and one tail is 3 of 9.
8.4 Number = (52)(51)(50) = 132,600
8.5 (a) Probability = 1/4
3
= 1 out of 64; (b) probability = 1/4
3
= 1 out
of 64; (c) probability = 1/4
7
= 1 out of 16,384; (d) probability = 1/4
12
= 1 out of 16,777,216.
8.6 (a) Number of outcomes = 2
3
= 8; (b) number of energetically distinct
states = 4; (c) the three up spins have the highest energy, followed
in order by two up and one down, one up and two down, and three
down.
8.7 (a) e
−(E 2 −E 1 )/k B T
= e
−(9×10
−24 J/(1.38×10
−23 J K
−1
)(298 K)
= e
−0.021
≈ 0.979
(b) e
−(E 2 −E 1 )/k B T
= e
−(9×10
−24 J/(1.38×10
−23 J K
−1
)(4 K)
= e
−0.16
= 0.849
ln
ln
(
)
(
)
k
k
RT
E
E
E
E
A
A
A
A
2
1
1
2
1
2
1
−
= −
−
→
−
(ln
ln )
=
−
RT k
k
2
1
452
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 452