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Thermochemistry: Equilibrium (Keq)
Related Topics
Wize High School Grade 12 Chemistry Textbook > Energy Changes
Introduction to Thermochemistry
2 Activities
The Haber Bosch Process is an industrial process for “fixing” nitrogen, shown below.This process produces around 500 million tons of ammonia every year. Calculate K
eq
3
H
2
(
g
)
+
N
2
(
g
)
→
2
NH
3
(
g
)
3\ H_{2(g)}+N_{2(g)}\rightarrow\ 2\ \text{NH}_{3(g)}
3
H
2
(
g
)
+
N
2
(
g
)
→
2
NH
3
(
g
)
Δ
f
G
o
(
N
H
3
)
=
−
16.4
k
J
/
m
o
l
\Delta _fG^o(NH_3)=-16.4\ kJ/mol
Δ
f
G
o
(
N
H
3
)
=
−
16.4
k
J
/
m
o
l
749.6
749.4
749.5
749.7
I don't know
Check Submission
More Introduction to Thermochemistry Questions:
Which of the following is an Intensive Property?
Gibbs Free Energy
The reaction of
H
2
(
g
)
+
C
O
2
(
g
)
↔
H
2
O
(
g
)
+
C
O
(
g
)
a
t
2000
K
K
P
=
4.40
H_{2(g)}+CO_{2(g)}\leftrightarrow H_2O_{(g)}+CO_{(g)}\ at\ 2000K\ K_P=4.40
H
2
(
g
)
+
C
O
2
(
g
)
↔
H
2
O
(
g
)
+
C
O
(
g
)
a
t
2000
K
K
P
=
4.40
Calculate ΔG for this reaction
Thermochemistry: Temperature
The oxidation of NO is shown below. Using the provided thermodynamic data answer the following questions
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
2\ NO_{(g)}+O_{2(g)}\rightarrow2\ NO_{2)g)}
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
Δ
H
o
=
−
57.1
k
J
m
o
l
and
Δ
S
o
=
−
73
J
m
o
l
a
t
298
K
\Delta H^o=-57.1\frac{kJ}{mol}\ \text{and}\ \Delta S^o=-73\frac{J}{mol}\ at\ 298K
Δ
H
o
=
−
57.1
m
o
l
k
J
and
Δ
S
o
=
−
73
m
o
l
J
a
t
298
K
Thermochemistry: Gibbs Free Energy
The oxidation of NO is shown below. Using the provided thermodynamic data answer the following questions
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
2\ NO_{(g)}+O_{2(g)}\rightarrow2\ NO_{2)g)}
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
Δ
H
o
=
−
57.1
k
J
m
o
l
and
Δ
S
o
=
−
73
J
m
o
l
a
t
298
K
\Delta H^o=-57.1\frac{kJ}{mol}\ \text{and}\ \Delta S^o=-73\frac{J}{mol}\ at\ 298K
Δ
H
o
=
−
57.1
m
o
l
k
J
and
Δ
S
o
=
−
73
m
o
l
J
a
t
298
K
Using the following data, calculate the standard entropy change for the reaction:
2
A
l
(
s
)
+
3
Z
n
O
(
s
)
→
A
l
2
O
3
(
s
)
+
3
Z
n
(
s
)
2Al_{(s)}+3ZnO_{(s)}\to Al_2O_{3(s)} +3Zn_{(s)}
2
A
l
(
s
)
+
3
Z
n
O
(
s
)
→
A
l
2
O
3
(
s
)
+
3
Z
n
(
s
)
Report your answer in J/mol K to four significant figures. Do not include units in the answer field.
A spontaneous process:
Which of the following statements regarding spontaneous changes is
false
?
A freshly baked pie is placed near an open window to cool. Which of the following statements best describes this situation?
Thermodynamics: Heat and Work
Consider the following balanced chemical equation for the combustion of propane, C
3
H
8
:
C
3
H
8
(
g
)
+
5
O
2
(
g
)
→
4
H
2
O
(
ℓ
)
+
3
C
O
2
(
g
)
q
<
0
C_3H_8 (g) + 5 O_2 (g) → 4 H_2O (ℓ) + 3 CO_2 (g) \qquad q < 0
C
3
H
8
(
g
)
+
5
O
2
(
g
)
→
4
H
2
O
(
ℓ
)
+
3
C
O
2
(
g
)
q
<
0
Use this information to answer the next two questions. Choose one if the options inside the brackets
If a reaction is nonspontaneous at 298 K with a negative ΔH
o
, then the reaction is:
Select all that apply
The oxidation of NO is shown below. Using the provided thermodynamic data answer the following questions
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
2\ NO_{(g)}+O_{2(g)}\rightarrow2\ NO_{2)g)}
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
Δ
H
o
=
−
57.1
k
J
m
o
l
and
Δ
S
o
=
−
73
J
m
o
l
a
t
298
K
\Delta H^o=-57.1\frac{kJ}{mol}\ \text{and}\ \Delta S^o=-73\frac{J}{mol}\ at\ 298K
Δ
H
o
=
−
57.1
m
o
l
k
J
and
Δ
S
o
=
−
73
m
o
l
J
a
t
298
K
The oxidation of NO is shown below. Using the provided thermodynamic data answer the following questions
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
2\ NO_{(g)}+O_{2(g)}\rightarrow2\ NO_{2)g)}
2
N
O
(
g
)
+
O
2
(
g
)
→
2
N
O
2
)
g
)
Δ
H
o
=
−
57.1
k
J
m
o
l
and
Δ
S
o
=
−
73
J
m
o
l
a
t
298
K
\Delta H^o=-57.1\frac{kJ}{mol}\ \text{and}\ \Delta S^o=-73\frac{J}{mol}\ at\ 298K
Δ
H
o
=
−
57.1
m
o
l
k
J
and
Δ
S
o
=
−
73
m
o
l
J
a
t
298
K
Find ΔG
0
for the following reaction.
CH
4
(
g
)
+
2
O
2
(
g
)
→
2
H
2
O
(
g
)
+
CO
2
(
g
)
\text{CH}_{4(g)}+2\ O_{2(g)}\rightarrow 2\ H_2O_{(g)}+\text{CO}_{2(g)}
CH
4
(
g
)
+
2
O
2
(
g
)
→
2
H
2
O
(
g
)
+
CO
2
(
g
)
Consider the following vaporization:
C
H
3
C
H
2
O
H
(
l
)
→
C
H
3
C
H
2
O
H
(
g
)
CH_3CH_2OH_{(l)}\to CH_3CH_2OH_{(g)}
C
H
3
C
H
2
O
H
(
l
)
→
C
H
3
C
H
2
O
H
(
g
)
If ∆H
vap
= 39.3 kJ/mol and the boiling point of ethanol is 78.3°C, what is ΔS for the vaporization of 1.73 mols of ethanol at its boiling point?
Predict the sign of the ΔH, ΔS,(+ or -) and identify when the following reactions are spontaneous at low temperature, high temperature, or all temperatures (low, high, all). Use commas to write your answers ie. +,+,low
For freezing of a popsicle: ΔH is ___________, ΔS is ___________. Reaction is spontaneous at ____________________ temperatures
For evaporation of liquid water: ΔH is ___________, ΔS is ___________. Reaction is spontaneous at ____________________ temperatures
Arrange the following compounds in order of increasing entropy, assuming 1 mol of each compound:
A
r
(
g
)
F
e
(
s
)
C
H
3
C
H
2
C
H
3
(
g
)
H
O
C
H
2
C
H
2
C
H
2
O
H
(
l
)
Ar(g) \hspace{15pt}Fe(s)\hspace{15pt} CH_3CH_2CH_{3(g)} \hspace{15pt} HOCH_2CH_2CH_2OH_{(l)}
A
r
(
g
)
F
e
(
s
)
C
H
3
C
H
2
C
H
3
(
g
)
H
O
C
H
2
C
H
2
C
H
2
O
H
(
l
)
A
r
(
g
)
Ar(g)
A
r
(
g
)
Calculate ΔS
sys
ΔS
surr
and ΔS
Univ
for the combustion of hydrogen shown below (unbalanced). Is the process spontaneous?
H
2
(
g
)
+
O
2
(
g
)
→
H
2
O
(
g
)
H_{2(g)}+O_{2(g)}\rightarrow H_2O_{(g)}
H
2
(
g
)
+
O
2
(
g
)
→
H
2
O
(
g
)
ΔH
f
0
(H
2
O
(g)
) = -285.83 kJ/mol
Thermochemistry: Entropy Calculations
Using the ΔH
0
of fusion for water 6.03 kJ/mol and the ΔS
0
of fusion for water 22.1 J K
-1
mol
-1
, calculate the ΔS
univ
for ice melting at -10
o
C, 0
o
C and 10
o
C. Remember: the universe transfers heat in a reversible way.
Thermodynamics: Entropy Calculations
Using the following data, calculate the standard entropy of the reaction shown below. Enter your answer in units of J / (mol*K)
2
A
l
(
s
)
+
3
Z
n
O
(
s
)
→
A
l
2
O
3
(
s
)
+
3
Z
n
(
s
)
2\ Al_{(s)}+3\ ZnO_{(s)}\rightarrow Al_2O_{3(s)}+3Zn_{(s)}
2
A
l
(
s
)
+
3
Z
n
O
(
s
)
→
A
l
2
O
3
(
s
)
+
3
Z
n
(
s
)
S
o
(
A
l
(
s
)
)
=
28.3
J
m
o
l
−
1
K
−
1
S^o(Al_{(s)})=28.3\ J\ mol^{-1}K^{-1}
S
o
(
A
l
(
s
)
)
=
28.3
J
m
o
l
−
1
K
−
1