2.5 The ITC profile reflects the binding of the drug to the target.
2.6 A favorable drug candidate would have an ITC profile that changes
at low concentrations of the drug, reflecting a tight binding to the
target of interest.
2.7 Since the expansion is adiabatic, q = 0. Since the external pressure
is constant, w = −PΔV = −(1 atm)(6 L) = −6 atmL = −606 J. The change
in the internal energy is the sum of the heat and work, or −606 J.
2.8
= −873.6 J
2.9 (a) The number of moles of carbon dioxide is 5 g/(44 g mol
−1
) =
0.114 mol. The maximum pressure that can develop is:
(b) The final volume of the expanded gas is:
w = −PΔV = −(2.75 atm)(2.78 L − 0.03 L) = −278 J
2.10
q = nC V ΔT = (0.0409 mol)(25 J/(Kmol))(596 K − 298 K) = 305 kJ
The change in the internal energy is equal to the heat transferred
since no expansion work was performed.
2.11 The change in enthalpy will be 1.0 J.
2.12 The internal energy will not change.
2.13
For an ideal gas, the internal energy depends only upon the temperature so the change in internal energy is zero. The change in internal
energy is the sum of the heat and work so the work is +5.76 kJ.
w
nRT
V
V
nal
initial
ln
(
)( .
= −
= −
3
1
8 314
mol
J/(Km mol)
K
L
L
kJ
)(
) ln
.
298
4
2
5 76
= −
T
T
P
P
nal
initial
nal
initial
3
3
=
=298
2
1
K
atm
atm m
K
= 596
Number of moles,
atm L
n
PV
RT
(
)( )
.
=
=
1
1
0 08206
L Latm
Kmol
K
mol
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
=
(
)
.
298
0 0409
V
V
P
P
nal
initial
initial
nal
3
3
( .
)
.
=
= 0 03
92 6
L
3 3
1
2 78
atm
atm
L
.
=
P
nRT
V
( .
) .
=
=
⎛
⎝
⎜ ⎜
⎞
⎠
⎟
0 114
0 08206
mol
Latm
Kmol ⎟ ⎟
=
(
)
.
.
298
0 03
92 63
K
L
atm
w
nRT
V
V
nal
initial
ln
( .
)( .
= −
= −
3
0 5
8 314
mol
J/( (Kmol)
K
L
L
)(
) ln
303
20
10
442
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 442