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
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