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Calculate H for the reaction below, which requires catalytic acid to proceed C…
Related Topics
Wize High School Grade 12 Chemistry Textbook > Energy Changes
Ways to Calculate Enthalpy of Reactions
10 Activities
Calculate
Δ
H
°
\Delta H\degree
Δ
H
°
for the reaction below, which requires catalytic acid to proceed
C
6
H
12
(
l
)
+
H
2
O
(
l
)
→
C
6
H
13
O
H
(
l
)
C_6H_{12}\left(l\right)+H_2O\left(l\right)\to C_6H_{13}OH\left(l\right)
C
6
H
12
(
l
)
+
H
2
O
(
l
)
→
C
6
H
13
O
H
(
l
)
Given:
C
6
H
12
(
l
)
+
H
2
(
g
)
→
C
6
H
14
(
l
)
Δ
H
°
=
−
120.4
k
J
m
o
l
C_6H_{12}\left(l\right)+H_2 (g)\to C_6H_{14}\left(l\right)\ \ \ \ \ \ \ \ \Delta H\degree=-120.4\ \frac{kJ}{mol}
C
6
H
12
(
l
)
+
H
2
(
g
)
→
C
6
H
14
(
l
)
Δ
H
°
=
−
120.4
m
o
l
k
J
C
6
H
14
(
l
)
+
1
/
2
O
2
(
g
)
→
C
6
H
13
O
H
(
l
)
Δ
H
°
=
−
142.6
k
J
m
o
l
C_6H_{14}\left(l\right)+1/2\ O_2 (g)\to C_6H_{13}OH\left(l\right)\ \ \ \ \ \ \ \ \Delta H\degree=-142.6\ \frac{kJ}{mol}
C
6
H
14
(
l
)
+
1/2
O
2
(
g
)
→
C
6
H
13
O
H
(
l
)
Δ
H
°
=
−
142.6
m
o
l
k
J
Δ
f
H
°
o
f
H
2
O
(
l
)
=
−
285.8
k
J
m
o
l
\Delta_fH\degree\ of\ H_2O\left(l\right)=-285.8\ \frac{kJ}{mol}
Δ
f
H
°
o
f
H
2
O
(
l
)
=
−
285.8
m
o
l
k
J
-548.8 kJ/mol
+283.5 kJ/mol
22.8 kJ/mol
-312.2 kJ/mol
548.8 kJ/mol
I don't know
Check Submission
More Ways to Calculate Enthalpy of Reactions Questions:
Enthalpy of Formation
What is the enthalpy of the following reaction when 0.35 g of Mg
2+
is produced? Use the following information:
ΔH°
f
[Mg(OH)
2
(s)] = −924.5 kJ
ΔH°
f
[NH
4+
(aq)] = −132.5 kJ
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: Enthalpy of reactions
The following reaction enthalpies are provided:
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
,
Δ
H
1
=
−
285.8
k
J
/
m
o
l
(
1
)
H_{2(g)}+\frac{1}{2}O_{2(g)}\to H_2O_{(l)},\Delta H_1=-285.8kJ/mol\hspace{70pt} (1)
H
2
(
g
)
+
2
1
O
2
(
g
)
→
H
2
O
(
l
)
,
Δ
H
1
=
−
285.8
k
J
/
m
o
l
(
1
)
C
(
s
)
+
O
2
(
g
)
→
C
O
2
(
g
)
,
Δ
H
2
=
−
293.5
k
J
/
m
o
l
(
2
)
C_{(s)}+O_{2(g)}\to CO_{2(g)},\Delta H_2=-293.5kJ/mol \hspace{80pt} (2)
C
(
s
)
+
O
2
(
g
)
→
C
O
2
(
g
)
,
Δ
H
2
=
−
293.5
k
J
/
m
o
l
(
2
)
Calorimetry: Temperature change
1. A 56.0 g block of iron heated to 90 °C is placed in a 4.8 kg water bath at 20 °C. What will be the final temperature of the iron? heat capacity of iron c = 0.450 J /gK, heat capacity of water is 4.185 J/gK
Enthalpy of Reactions
Consider the following reaction:
N
2
H
4
(
I
)
+
O
2
→
N
2
(
g
)
+
2
H
2
O
(
I
)
Δ
H
=
−
622.2
k
J
N_2H_4(I)+O_2\rightarrow N_2(g)+2\ H_2O(I)\quad \Delta H=-622.2\ kJ
N
2
H
4
(
I
)
+
O
2
→
N
2
(
g
)
+
2
H
2
O
(
I
)
Δ
H
=
−
622.2
k
J
Given the following data, calculate the heat of reaction for the same reaction where water is a gaseous product instead of a liquid:
The standard enthalpy of formation for
N
O
2
(
g
)
NO_2\left(g\right)
N
O
2
(
g
)
is 33.18 kJ/mol, and the standard enthalpy of partial combustion of nitrogen to NO(g) is 90.25 kJ/mol. Calculate the standard enthalpy of combustion of NO(g) to NO
2
(g).
Hess’s Law: Indirect Determination of ΔH
The standard enthalpy of formation of ethanol
(
C
2
H
5
OH
)
{\small \left(\text{C}_2\text{H}_5\text{OH}\right)}
(
C
2
H
5
OH
)
is -277.7 kJ/mol. Write the chemical equation to which this value applies.
Solution:
Calculate the ΔH for the reaction below IF
5
(g) → IF
3
(g) + F
2
(g)
IF (g) + F
2
(g) → IF
3
(g) ΔH = -390 kJ IF (g) + 2 F
2
(g) → IF
5
(g) ΔH = -745 kJ
Liquification of fuels is an important process that allows fuel to be transported more safely and more easily around the world. Oxidizing methane transforms it from a gas to a liquid, methanol, which is much easier to transport. Using the data provided calculate
Δ
H
\Delta H
Δ
H
for the oxidation of methane (CH
4
) shown below
C
H
4
(
g
)
+
1
/
2
O
2
(
g
)
→
C
H
3
O
H
(
l
)
Δ
H
f
o
(
C
H
4
(
g
)
)
=
−
74.9
k
J
m
o
l
−
1
Δ
H
f
o
(
C
H
3
O
H
(
l
)
=
−
238.6
k
J
m
o
l
−
1
\begin{array}{c} CH_{4(g)}\ +\ 1/2\ O_{2(g)} \rightarrow CH_3OH_{(l)} \\ \\ \Delta H^o_f(CH_{4(g)})=-74.9 kJ\ mol^{-1} \\ \Delta H_f^o(CH_3OH_{(l)}=-238.6\ kJ\ mol^{-1} \end{array}
C
H
4
(
g
)
+
1/2
O
2
(
g
)
→
C
H
3
O
H
(
l
)
Δ
H
f
o
(
C
H
4
(
g
)
)
=
−
74.9
k
J
m
o
l
−
1
Δ
H
f
o
(
C
H
3
O
H
(
l
)
=
−
238.6
k
J
m
o
l
−
1
Practice: Total Bond Energy of IF$_7$
The total bond enthalpy for F
2
is 156.9 kJ/mol and the total bond enthalpy of I
2
is 152.5 kJ/mol. Using the thermochemical equation below, what is the total bond enthalpy of IF
7
?
I
2
(
s
)
+
7
F
2
(
g
)
→
2
I
F
7
(
s
)
Δ
H
=
2669.2
k
J
.
m
o
l
I_{2(s)}+7\ F_{2(g)}\rightarrow 2\ IF_{7(s)}\ \ \Delta H=2669.2\ kJ.mol
I
2
(
s
)
+
7
F
2
(
g
)
→
2
I
F
7
(
s
)
Δ
H
=
2669.2
k
J
.
m
o
l
Enthalpy of Reactions
Given the following at 25°C and 1.00 atm:
R
e
a
c
t
i
o
n
Δ
H
°
1
2
N
2
(
g
)
+
O
2
(
g
)
→
N
O
2
(
g
)
33.2
k
J
N
2
(
g
)
+
2
O
2
(
g
)
→
N
2
O
4
(
g
)
11.1
k
J
\def\arraystretch{1.5}\begin{array}{cc}\rm Reaction&\Delta H\degree\\\cfrac{1}{2} N_2 (g)+O_2 (g)→NO_2 (g)&33.2 kJ \\N_2 (g)+2O_2 (g)→N_2 O_4 (g)&11.1 kJ\end{array}
Reaction
2
1
N
2
(
g
)
+
O
2
(
g
)
→
N
O
2
(
g
)
N
2
(
g
)
+
2
O
2
(
g
)
→
N
2
O
4
(
g
)
Δ
H
°
33.2
k
J
11.1
k
J
Enthalpy of Reactions
Estimate the heat of reaction at 298 K for the reaction shown, given the average bond energies below.
B
r
2
(
g
)
+
3
F
2
(
g
)
→
2
B
r
F
3
(
g
)
Br_2 (g)+3F_2 (g)→2BrF_3 (g)
B
r
2
(
g
)
+
3
F
2
(
g
)
→
2
B
r
F
3
(
g
)
B
o
n
d
B
o
n
d
E
n
e
r
g
y
B
r
−
B
r
193
k
J
/
m
o
l
F
−
F
155
k
J
/
m
o
l
B
r
−
F
249
k
J
/
m
o
l
\def\arraystretch{1.5} \begin{array}{c} \rm Bond &&&& \rm Bond \ Energy \\ \hline Br-Br &&&& 193 kJ/mol\\ F-F &&&& 155 kJ/mol\\ Br-F &&&& 249 kJ/mol\\ \end{array}
Bond
B
r
−
B
r
F
−
F
B
r
−
F
Bond
Energy
193
k
J
/
m
o
l
155
k
J
/
m
o
l
249
k
J
/
m
o
l
Enthalpy: Bond Dissociation Energies
Consider the following balanced chemical reaction and the bond dissociation energies for the partial combustion of hydrogen sulfide:
What is the enthalpy change Δ
r
H for this reaction?
Thermochemistry: Calorimetry
Consider the following balanced chemical equation:
2
C
ℓ
F
3
(
g
)
+
2
O
2
(
g
)
→
C
ℓ
2
O
(
g
)
+
3
F
2
O
(
g
)
∆
r
H
°
=
?
?
?
2 CℓF_3(g) + 2 O_2(g) → Cℓ_2O (g) + 3 F_2O (g) \qquad ∆_rH° = ???
2
C
ℓ
F
3
(
g
)
+
2
O
2
(
g
)
→
C
ℓ
2
O
(
g
)
+
3
F
2
O
(
g
)
∆
r
H
°
=
???
S
u
b
s
t
a
n
c
e
△
f
H
°
@
25
°
C
(
k
J
/
m
o
l
)
C
ℓ
F
3
(
g
)
?
?
C
ℓ
2
O
(
g
)
80.3
F
2
O
(
g
)
–
21.7
F
2
(
g
)
0
\def\arraystretch{1.5} \begin{array}{|c|c|} \hline \rm Substance & △_fH° \ @ \ 25°C \ (kJ/mol) \\ \hline CℓF_3 (g) & ?? \\ \hline Cℓ_2O (g) & 80.3 \\ \hline F_2O (g) & – 21.7 \\ \hline F_2 (g) & 0 \\ \hline \end{array}
Substance
C
ℓ
F
3
(
g
)
C
ℓ
2
O
(
g
)
F
2
O
(
g
)
F
2
(
g
)
△
f
H
°
@
25°
C
(
k
J
/
m
o
l
)
??
80.3
–21.7
0
Thermochemistry: Calorimetry
Ethene gas (C
2
H
4
) reacts with fluorine gas according to the following balanced equation:
C
2
H
4
(
g
)
+
6
F
2
(
g
)
→
2
C
F
4
(
g
)
+
4
H
F
(
g
)
Δ
r
H
°
=
−
3012
k
J
/
m
o
l
⋅
r
x
n
@
25
°
C
C_2H_4 (g) + 6 F_2 (g) → 2 CF_4 (g) + 4 HF (g) \qquadΔ_rH° = −3012\ kJ/mol·rxn\ @ \ 25°C
C
2
H
4
(
g
)
+
6
F
2
(
g
)
→
2
C
F
4
(
g
)
+
4
H
F
(
g
)
Δ
r
H
°
=
−
3012
k
J
/
m
o
l
⋅
r
x
n
@
25°
C
S
u
b
s
t
a
n
c
e
△
f
H
°
@
25
°
C
(
k
J
/
m
o
l
)
C
2
H
4
(
g
)
?
?
F
2
(
g
)
0
C
F
4
(
g
)
–
933
H
F
(
g
)
–
273.3
\def\arraystretch{1.5} \begin{array}{|c|c|} \hline \rm{Substance} & △_fH° \ @ \ 25°C(kJ/mol) \\ \hline C_2H_4 (g) & ?? \\ \hline F_2 (g) & 0 \\ \hline CF_4 (g) & – 933 \\ \hline HF (g) & – 273.3 \\ \hline \end{array}
Substance
C
2
H
4
(
g
)
F
2
(
g
)
C
F
4
(
g
)
H
F
(
g
)
△
f
H
°
@
25°
C
(
k
J
/
m
o
l
)
??
0
–933
–273.3
Thermodynamics: Enthalpy of reaction
Find ΔH
o
for the following reaction if 12.0 g of N
2
H
4
is consumed in excess N
2
0
4
.
2
N
2
H
4
(
l
)
+
N
2
O
4
(
g
)
→
3
N
2
(
g
)
+
4
H
2
O
(
l
)
2N_2H_{4(l)}+N_2O_{4(g)}\to 3N_{2(g)}+4H_2O_{(l)}
2
N
2
H
4
(
l
)
+
N
2
O
4
(
g
)
→
3
N
2
(
g
)
+
4
H
2
O
(
l
)
The following enthalpies of formation are provided:
Calculate ΔH
o
f
of SO
2(g)
given the following: (answer in kJ)
S
(
s
)
+
1.5
O
2
(
g
)
→
S
O
3
(
g
)
,
Δ
H
=
−
395.8
k
J
S_{(s)}+1.5O_{2(g)}\to SO_{3(g)}, \Delta H=-395.8kJ
S
(
s
)
+
1.5
O
2
(
g
)
→
S
O
3
(
g
)
,
Δ
H
=
−
395.8
k
J
2
S
O
2
(
g
)
+
O
2
(
g
)
→
2
S
O
3
(
g
)
,
Δ
H
=
−
198.2
k
J
2SO_{2(g)}+O_{2(g)}\to 2SO_{3(g)}, \Delta H=-198.2kJ
2
S
O
2
(
g
)
+
O
2
(
g
)
→
2
S
O
3
(
g
)
,
Δ
H
=
−
198.2
k
J
Intramolecular bonding: Strength of Covalent Bonds
Calculate the enthalpy for the following reaction.
2 C
2
H
6
(g) + 7 O=O (g) → 4 O=C=O (g) + 6 H-O-H (g)
The following are bond dissociation energies (kJ/mol):
Practice: Hydrogenation of Acetylene
Using the table below, answer the following questions about the hydrogenation reactions of acetylene:
a) Calculate the enthalpy of formation of ethylene, C
2
H
4
, considering ΔH
rxn
= -174.4kJ/mol
C
2
H
2
+
H
2
→
C
2
H
4
C_2H_2+H_2\to C_2H_4
C
2
H
2
+
H
2
→
C
2
H
4
Find the standard molar enthalpy of combustion of C
2
H
5
OH using the standard enthalpy of formations below.
Δ
H
f
(
C
2
H
5
O
H
(
l
)
)
=
−
277.7
k
J
/
m
o
l
\Delta H_f(C_2H_5OH_{(l)}) = -277.7\ kJ/mol
Δ
H
f
(
C
2
H
5
O
H
(
l
)
)
=
−
277.7
k
J
/
m
o
l
Δ
H
f
(
C
O
2
(
g
)
)
=
−
393.5
k
J
/
m
o
l
\Delta H_f(CO_{2(g)}) = -393.5\ kJ/mol
Δ
H
f
(
C
O
2
(
g
)
)
=
−
393.5
k
J
/
m
o
l
Thermochemistry Calculations
The standard enthalpy of fusion for water, ΔH
o
fusion
= 6.01 kJ/mol. Calculate the energy required (in kJ) to convert 26.0 g of ice at 0
o
C to liquid water at 20
o
C.
Heat of Formation
Determine the standard molar enthalpy of formation of solid Al
2
O
3
from the following equations.
Al (s) + 3 H
2
O (l) → Al(OH)
3
(s) + 3/2 H
2
(g)
Δ
H
=
+
100
k
J
m
o
l
\Delta H\ =+100\ \frac{kJ}{mol}
Δ
H
=
+
100
m
o
l
k
J
H
2
(g) + 1/2 O
2
(g) → H
2
O (l)
Δ
H
=
−
300
k
J
m
o
l
\Delta H=-300\ \frac{kJ}{mol}
Δ
H
=
−
300
m
o
l
k
J