# Some applications of Thermodynamics - ChemGod.com

Mechanics

Oct 27, 2013 (4 years and 6 months ago)

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Some applications
of
Thermodynamics

At what temperatures is the following reaction
spontaneous?

2 H
2

(g) + O
2

(g)

2 H
2
O (g)

If you calculate
Δ
H
0

(Appendix II) =
-
483.6 kJ

If you calculate
Δ
S
0

(Appendix II)
=
-
89 J/K

2 H
2

(g) + O
2

(g)

2 H
2
O (g)

If you calculate
Δ
H
0

(Appendix II) =
-
483.6 kJ

If you calculate
Δ
S
0

(Appendix II)
=
-
89 J/K

Enthalpy change good!

The balance is…

Δ
G

Δ
G=
Δ
H
-
T
Δ
S

We assume that
Δ
H and
Δ
S aren’t changing
significantly with temperature:

Δ
G=
Δ
H
0

T
Δ
S0

Δ
G
=(
-
483.6 kJ)

T(
-
0.089 kJ/K) [UNITS! UNITS!]

Δ
G=(
-
483.6 kJ)

T(
-
0.089 kJ/K
)

Spontaneous means
Δ
G<0

Δ
G
=(
-
#)

T
(
-
#)

Δ
G
=(
-
#) +T(#)

If T is big enough,
Δ
G will become positive.

Δ
G=(
-
483.6 kJ)

T(
-
0.089 kJ/K
)

0=
-
483.6 kJ + T(0.089 kJ/K)

483.6 kJ = T(0.089 kJ/K)

5434 K = T

So for any T below 5434 K, the reaction is
spontaneous. Above that,
Δ
G becomes (+) and the
reaction is NOT spontaneous anymore.

Caveat

5434 K is a pretty extreme temperature
especially relative to STP (298 K). I have to
wonder how good my assumption of
Δ
H and
Δ
S
being constant actually is. We could try and
establish
Δ
H and
Δ
S
at 5434 K but that is much
harder to do.

Opposite Case

Δ
G=(
-
483.6 kJ)

T(
-
0.089 kJ/K)

Spontaneous means
Δ
G<0

Δ
G=(
-
#)

T(
-
#)

Suppose the signs were reversed:

Δ
H=(+#)

Δ
S=(+#)

Opposite Case

Spontaneous
means
Δ
G<0

Δ
G
=(+#)

T
(+#)

Now the bigger T is, the better! At low
temperatures the reaction is NOT spontaneous
but at higher temperatures
it is.