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22=
. 2 kJ =
 =
; &n=
bsp; 1
kJ 0
kJ <=
u>AB  =
; ET
=3D 0 kJ 2 kJ 4 kJ 1
kJ 1 kJ 2 kJ 2
kJ 3 kJ 3 kJ &nbs=
p; The
most likely total energy is 2 kJ.
 =
; &n=
bsp;  =
; &n=
bsp;  =
; 52. 6 =
C(s) +
6 O2(g) <=
span
style=3D'font-size:12.0pt;line-height:130%'>→ 6 CO2(g)&nbs=
p; &=
nbsp; =
916;G°
=3D 6(!=
394 kJ) 3 H2(g) + 3/2 O2=
(g) → <=
/span>3 H2O(l)  =
; &n=
bsp; ΔG°
=3D 3(!=
237 kJ)
______________________________________________________________
56. a. ΔG° =3D 2(!=
270. kJ=
) ! 2(!502 kJ)=
=3D 464
kJ
=
&=
#916;G°
=3D 0 =3D ΔH° !=
TΔS°, T =3D =
=3D 2890 K This reaction will be spontaneous at standard
conditions (ΔG° < 0) when T > 2890 K. Here the favorable entropy term will
dominate.
=
ΔG°
=3D ΔH° − TΔS° =3D −184 × 103 J
− 298 K (20. J/K) =3D −1.90 × 105 J =3D W=
22;190.
kJ =
ΔG°
=3D −RT ln K, ln K =3D =
=3D 76.683
&n=
bsp; b. These are standard conditio=
ns so
ΔG =3D ΔG° =3D −190. kJ.
When ΔG is negative, the forward reaction is spontaneous so t=
he
reaction shifts right to reach equilibrium. 72. At
boiling point, ΔG =3D 0 so ΔS =3D =
For methane: ΔS =3D =
=
&nb=
sp; =
&nb=
sp; =
&nb=
sp; =
&nb=
sp; =
=3D
73.2 J/molC=
K
 =
; For
hexane: ΔS =3D=
=
=3D 84.5 J/molCK
Hexane has the larger molar volume at the boil=
ing
point so hexane should have the larger entropy. As the volume of a gas increases,
positional disorder increases. 76.  =
; Ba(NO3)2(s)
⇌ Ba2+(aq) + 2 NO3=
-(aq) K =3D
Ksp;
ΔG° =3D !=
561 + 2(!=
109) !=
(!=
797) =3D 18 kJ
|
662 =
&nb=
sp;
CHAPTER 16 SPONTANEITY, ENTROPY, AND FREE ENERGY