BioPHYSICAL chemistry

7.16

(EA 1 −EA 2 ) =RT(lnk 2 −lnk 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 280μM min−^1 at high sub-

strate concentrations so this is assigned as the maximum velocity. The

half velocity would be 140μM min−^1 , which occurs at 1 × 10 −^5 M

and is assigned as the Km.

7.20 The temperature in the exponent will decrease, making the free-

energy 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 differ-

ence 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 = 23 =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 )/kBT=e−(9×^10

− (^24) J/(1.38× 10 − (^23) JK− (^1) )(298 K)
=e−0.021≈0.979
(b)e−(E^2 −E^1 )/kBT=e−(9×^10
− (^24) J/(1.38× 10 − (^23) JK− (^1) )(4 K)
=e−0.16=0.849
lnkkln ( ) ( )
RT

2 1 EEAA 12 EEAA 12

1

−=− −→−=−RT(lnk 21 ln )k

452 ANSWERS TO PROBLEMS

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