Geology Reference
In-Depth Information
Chapter 3
−−
21
molm s
molm
=
m
2
s
−1
, since m
−2
/m
−4
) = m
2
, and the
−
4
3.1
N
0
= 18,032. Calculate ln (
N
0
/
N
) for each value of
t.
(As a check on arithmetic, the value for
t
= 25 is
0.09426.) ln (
N
0
/
N
) gives a linear plot against
time, indicating that the reaction is first-order.
After one half-life,
N
/
N
0
= 1/2, therefore ln
(
N
0
/
N
) = 0.6913. Reading from the graph, this value
is reached at
t
= 190 hours. If
n
= the number of
half-lives required for decay to 1/100, (1/2)
n
=
1/100, so
n
log(1/2) = log(1/100), therefore
n =
6.6
and t
1/100
= 1254 hours.
3.2
Room temperature = 25 °C = 298 K. The doubling of
reaction rate can be written:
k
mol terms cancel out.
Chapter 4
2
+
2
−
4.1
(a)
BaSO
→+
Ba
SO
(Equation 4.17)
4
4
solid
solution
m
m
m
m
2
−
2
+
SO
10
10
−
K
=⋅
a
a
=
Ba
⋅
=
4
BaSO
2
+
2
−
θ
θ
Ba
SO
4
4
When dissolved in pure water:
a
⋅
a
=
10
5
−
2
+
2
−
Ba
SO
4
Therefore if solution is ideal,
m
Ba
2
+
=
m
SO
=
2
k
−
=
10
−
5
mol kg
−
1
BaSO in saturated
308
298
2
4
4
solution.
The Arrhenius equation in log form gives simulta-
neous equations:
10
3
−
(b) In CaSO
4
solution,
a
=
a
=
−
+
2
2
SO
Ca
4
at 298 K: ln (
k
298
) = ln A −
E
a
/(8.314 × 298)
at 308 K: ln (2
k
298
) = ln A −
E
a
/(8.314 × 308)
Therefore ln A = ln
k
298
+
E
a
/2478 =
ln 2
k
298
+
E
a
/2561.
Rearranging, ln 2
k
298
− ln
k
298
= ln 2
=
E
a
(1/2478 − 1/2561).
E
a
= 0.6915/0.013 × 10
−3
= 52,900 J mol
−1
= 52.9 kJ mol
−1
3.3
Calculate ln (1/viscosity) for each temperature
and plot against 1/
T
(e.g. for
T
= 1325 °C = 1598 K,
1/
T
= 0.000626 K
−1
and ln(1/η) = −7.622). Slope of
graph = −34,030 K = −
E
a
/
R.
Thus
E
a
= 283 kJ mo1
−1
.
3.4
Half-life =
ln2
If
x.m
mol kg
-1
of BaSO
4
dissolves:
K
=⋅
a
a
=
10
−
10
BaSO
2
+
2
−
Ba
SO
4
4
(
)
≅
−
3
−
3
=
x
10
+
x
10
x
2
( ll)
Therefore
x.m
= 10
-7
mol kg
-1
.
sinceisverysma
x
2
4.2
CaF a
solid
→+
+
2
F
−
2
solution
=⋅
(
)
2
Ka
a
CaF
2
+
−
Ca
F
2
In pure water
a
=
2
a
−
2
+
F
Ca
⋅
(
)
=
(
)
=
2
3
Therefore
Ka
=
2
a
4
a
10
−
.
10 4
λ = 4.9 × 10
10
years.
Therefore ln(
N
0
/
N
) =
λt
= 1.42 × 10
−11
× 4.6 × 10
9
= 0.0653
therefore
N
0
/
N
= 1.068
∴
N
/
N
0
= 94%
Therefore percentage decayed = 6%.
3.5
Equation 3.11
⟮
Fick's First Law of Diffusion:
f D
c
x
/
CaF
2
+
2
+
2
+
Ca
Ca
Ca
2
87
Rb
≅
410
11
Therefore
a
Ca
2
+
=
.
0 00022
m
Ca
+
=
0 00022
.
molkg
−
1
2
Relative molecular mass of CaF
2
= 40 + (2 × 19) = 78
Therefore 0.00022 × 78 = 0.017 g CaF
2
will dissolve
in 1 kg water at 25 °C.
⟯
can be rearranged to express the diffu-
sion coefficient
D
i
in terms of
f
i
,
c
i
and
x
:
d
d
=−
i
i
i
4.3
CO HO HCO
2
+
2
2
3
f
c x
i
(
)
D
=
air
Ka P
=
/
=
0 031
.
i
(
)
dd
/
HCO
CO
2
3
2
i
a
HCO
=
0 031 0 00028
.
×
.
=
0 00000
.
868
=
10
506
−
.
The flux
f
i
has the units
mol m
−2
s
−1
(see text
above
Equation 3.11). The concentration gradient
d
d
2
3
c
x
According to equation 4.21: H CO
+
HHCO
+
−
i
has
2
3
3
aa
a
⋅
+
−
the units (mol m
−3
) m
−1
=
mol m
−
4
(see text
below
Equation 3.11). Therefore the units of
D
i
are
H CO
K
=
=
10
6.
−
3
HCO
2
3
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