10/8/2010
TUT 8th October 2010
Chem 251
y 7.9a
Exercises from the book
The equilibrium constant for the gas-phase
isomerization of borneol (C10H17OH) to
isoborneol at 503 K is 0.106. A mixture
consisting of 7.50 g of borneol and 14.0 g of
isoborneol in a container of volume 5.0 dm3 is
heated to 503 K and allowed to come to
equilibrium. Calculate the mole fractions of the
q
two substances at equilibrium
y 7.12b
Estimate the temperature at which
CuSO4·5H2O undergoes dehydration.
y 7.15b
Devise cells in which the following are the
reactions and calculate the standard emf in each
case:
a)
b)
c)
Atomic Element weight
C
12.011
H
1.0079
O
15.9994
ΔHf / ΔSf / Compound
(kJmol‐1) (J K‐1mol‐1)
CO2(g)
‐393.51 213.785
‐241.82
188.83
H2O(g)
‐2279.70
300.40
CuSO4•5H2O(s)
CuSO4(S)
‐771.36
109.00
0 205.152
O2(g)
Cu(g)
338.32
166.38
‐285.83
69.95
H2O(l)
P(white) + 3H+ + 3e‐ —>PH3(g) 2 H+ + 2 e‐ —>H2(g) AgBr(s) + e‐ —>Ag(s) + Br− 2 H2O + 2 e‐ —>H2(g) + 2OH− Bi(s) + 3 H+ + 3 e‐ —>BiH3 Na+ + e‐ —>Na(s) La3+ + 3 e‐ —>La(s) I3 − + 2 e‐ —>3 I− I2(s) + 2 e‐ —>2 I− 2 Na(s) + 2H2O(l) Æ 2 NaOH(aq) +
H2(g)
H2(g) + I2(g) Æ 2 HI(aq)
H3O+(aq) + OH–(aq) Æ 2H2O(l)
B ↔ IB
7.9(a)
Mass Cmpd
(g)
Mole
B/IB (MW)
Borneol (B)
7.5 0.048942 153.2437
Isoborneol (IB)
14 0.091358
⎛
⎞
0.04..
⎟⎟ × 1.15.. Q = ΡIB
ΡB = ⎜⎜
⎝ (0.04.. + 0.09.. ⎠
Ρ
⎛
⎞
0.09..
ΡIB = ⎜⎜
⎟⎟ × 1.15..
⎝ (0.04.. + 0.09.. ⎠
Parameter Value
Units
R = 0.08205746dm3 atm K‐1 mol‐1
T = 503K
V = 5dm3
Q =
1.86666667 ‐
P = nRT/V
1.15816885atm
B
B ↔ IB
0.404012391 0.754156
Initial
+X
‐X
change
0.4.. + X
0.75.. ‐ X
eqm
1.047168945
0.111 eqm (value)
Use K
not Q
0.106 =
−0.063 0.0000 +0.0713
−0.83 −0.8 −2.71 −2.379 +0.53 +0.54 0.75... + X
0.4... + X
χ IB = 0.11... ×1.15..
ΡB = χ B × ΡT
chk (k) = χB
χIB
nB
nIB
0.106
0.904159132
0.095840868
0.126852989
0.013446417
1
10/8/2010
Δr Sθ =
Ans… 7.12(b)
y Rxn of interest is
∑ vS θ − ∑ vS θ
m
Pr oducts
Δr H θ =
∑ vH θ − ∑ vH θ
m
Pr oducts
y CuSO4•5H2O(s) ↔ CuSO4(s) + 5H2O(g)
m
Re ac tan ts
m
Re ac tan ts
At some temperature, water vapor will be produced, & the value for K will be
the amount of water vapor produced (since the solids in any K expression are
always unity). We can assume this is occurring at 1 bar (std pressure), & thus the
value for K = 1, ∴lnK = 0, & ΔrG = O (ΔrG = -RTlnK).
Alternatively, for spontaneous reaction to occur, we know the free energy must
be minimized (-ve), thus we can assume that this reaction would proceed at ΔrG
=0
Either way: ΔrG ° = ΔrH ° - T ΔrS°
0 = ΔrH ° - T ΔrS°
T = = ΔrH ° / T ΔrS°
ΔrH ° = [((-771.36) + (5*(-241.82)) – (-2279.7)] = + 299.2
ΔrS°= [(109) + (5*(188.83)) – (300.4) = 752.8
T = 299200 J mol-1 / 752.8 J K-1 mol-1 = 397 K
Ans…7.15(b)
(a). Right left electrode is determined by the direction the rxn is written, thus combine ½
rxns
2Na+ (aq) + 2e– Æ2 Na(s)
2H2O(l) + 2e– Æ 2OH–(aq) + H2(g)
Pt|Na(s) |Na+(aq)|| OH –(aq) | H2(g)|Pt
E°= –2.71V
E°= –0.83V
E°= +1.88 V
(b) H2(g) + I2(g) Æ 2 HI(aq)
(b).
I2(s) + 2e– Æ 2I– (aq)
E°= +0.54V
2H+ (aq) + 2e– Æ H2(g)
E°= +0.00V
Pt| H2(g)|H+(aq)|| I– (aq)| I2(s) |Pt
E°= +0.54V
(c). H3O+(aq) + OH–(aq) Æ 2H2O(l)
2H+ (aq) + 2e– Æ H2(g)
E°= +0.00V
2H2O(l) + 2e– Æ 2OH–(aq) + H2(g)
E°= –0.83V
Pt| H2(g)|H+(aq)|| OH –(aq) | H2(g)|Pt E°= +0.83V
2