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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