Chapter 11 (Practice Test)
Thermochemistry
1. The specific heat capacity of graphite is 0.71 J/g•°C.
Calculate the energy required to raise the temperature of
450 g of graphite by 150 °C.
∆H
= m • C • ∆T
∆H = (450g) (0.71
∆H
m
C
∆T
=
= 450 g
= 0.71 J/g•°C
= 150 °C
J ) (150 °C)
g•°C
Calculator 450 x 0.71 x 150 = 47,925 J
Answer w/ Sig. Figs.
∆H = 48,000 J
or
48 kJ
2. It takes 770 joules of energy to raise the temperature
of 50 g of mercury by 110 °C. What is the specific heat
capacity of mercury?
∆H
∆H
m
mx
x ∆T
∆T
= m • C • ∆T
m x ∆T
C=
C=
∆H
m x ∆T
=
770 J
50 g x 110 °C
C = 0.14 J/g•°C
∆H
m
C
∆T
=
=
=
=
770 J
50 g
_____ J/g°C
110 °C
Calculator
770 ÷ 50 ÷ 110 = 0.14
or
770 ÷ (50 x 110) = 0.14
3. Calculate the heat absorbed by the water in a calorimeter
when 175.0 grams of copper cools from 125.0 °C to 22.0 °C
The specific heat capacity of copper is 0.385 J/g•°C.
∆H
= m • C • ∆T
∆H
m
C
∆T
=
= 175.0 g
= 0.385 J/g•°C
= 22.0°C – 125.0°C
∆H = (175g) (0.385 J ) (-103 °C)
∆T = – 103.0°C
g•°C
Calculator 175 x 0.385 x -103 = -6939.625 J
But water Absorbed the heat so… ∆H = + 6939.625 J
4. Assume 372 Joules of heat are added to 5.00 g of water
originally at 23.0 °C. What would be the final temperature of
the water? (Note: Find ∆T first then find the final temperature)
The specific heat capacity of water = 4.184 J/g•°C?
∆H
∆H
m
mx
xC
C
= m • C • ∆T
mxC
∆T =
∆H
∆T =
mxC
=
372
(5 x 4.184)
Heat was added so temp.
increases by 17.8.
∆H
m
C
∆T
Ti
Tf
= 17.8 °C
=
=
=
=
=
=
372 J
5.00 g
4.184 J/g°C
_____ °C
23.0 °C
_____ °C
Calculator
372 ÷ 5 ÷ 4.184 = 17.78
Tf = 23.0 + 17.8 = 40.8 °C
5. How much heat is required to raise the temperature of
2.0 x 102 g of aluminum by 30 °C.
(specific heat of Al = 0.878 J/g•°C)
∆H
= m • C • ∆T
∆H
m
C
∆T
=
= 2.0 x 102 g = 200 g
= 0.878 J/g•°C
= 30°C
∆H = (200g) (0.878 J ) (30 °C)
g•°C
Calculator (2.0
x
102) x 0.878 x 30 = 5268 J
Answer to correct sig.figs.
∆H = 5 kJ
6. Find the specific heat capacity of Lead if an 85.0 g
sample of lead with an initial temperature of 99.0 °C is
placed into 99.5 g of
water so…C
with an
Water
is initial
knowntemperature of
22.0 °C . The final temperature of the water and the
lead is 25.0 °C.
Water
= m • C • ∆T
∆H
∆H = (99.5g) (4.184
∆H = 1,248.924
J
1,248.924
∆T
J ) (3.0 °C)
g•°C
Energy water gained
from the hot metal so…
= Tf - Ti
∆H
m
C
∆T
∆H
C=
=
m
m x ∆T
(85.0) (-74.0 °C)
C
C = 0.199 = 0.20 J/g•°C
∆T
∆H
- 1248.924 J
= _________ J
= 99.5 g
= 4.184 J/g•°C
= 25-22=3.0°C
Lead
= -_________ J
= 85.0 g
= ______ J/g•°C
= 25-99=-74.0°C
7. Find the standard heat of formation for the following reaction.
∆H° = ∆Hf°(products) - ∆Hf°(reactants)
4 NH3(g) +
-46.19
x
4
-184.76 kJ
+
5 O2 (g)  4 NO(g) + 6 H2O (l)
0.0
x 5
0 kJ
-184.76 kJ
(reactants)
90.37
x
4
361.48 kJ
+
-285.8
x
6
-1714.8 kJ
Substance
∆Hf°
(kJ/mol)
NH3(g)
-46.19
O2(g)
0.0
NO(g)
90.37
H2O(g)
-285.8
-1,353.32 kJ
(products)
∆H° = ∆Hf°(products) - ∆Hf°(reactants)
∆H° = -1,353.32 kJ – (-184.76 kJ) = - 1168.56 kJ
∆H° = - 1168.56 kJ
8. Heat of Reaction:
2 H2(g) +
O2 (g) 
2 H2O (l)
-572 kJ
∆H = -572
How much heat is produced when 5.00 g H
of2 H2 (at STP) is
reacted with excess O2?
x
1.01
x 2
2.02
1 mol H2
2.02 g H2
x
mol H2
= -708 kJ
1
H
Calculator: 5.00 ÷ 2.02 x -572 ÷ 2 = - 707.9
Hydrogen
1.01
9. Heat of Solution:
∆Hsoln = -445.1 kJ/mo
kJ
Determine the heat of solution when 40.00 g of
NaOH
NaOH is
dissolved in water.
1 mol NaOH
x
x
40.00 g NaOH
11
8
Na
O
22.99
16.00
sodium
= -445.1 kJ
1 mol NaOH
1
H
Calculator: 40.00 ÷ 40.00 x -445.1 = - 445.1
Oxygen
Hydrogen
1.01
10. Heat of Solution:
∆Hsoln = 25.7 kJ/mol
kJ
Determine the heat of solution when 25.58 g of
NHNH
4NO
4NO
3 3 is
dissolved in water.
1 mol NH4NO3
x
x
=
8.211 kJ
80.06 g NH4NO3 1 mol NH4NO3
7
8
N
O
14.01
16.00
Nitrogen
1
H
Calculator: 25.58 ÷ 80.06 x 25.7 = 8.211416438
Oxygen
Hydrogen
1.01
11. Heat of Combustion:
2 C 2H 2 +
5 O2  4 CO2 + 2 H2O
-2600 kJ
kJ
∆H = -2600
How much heat is produced when 35.00 g C
of2HC22H2 is reacted
with excess O2?
1 mol C2H2
x
x
26.04 g C2H2
6
C
= -1747 kJ
mol C2H2
1
H
Calculator: 35.00 ÷ 26.04 x -2600 ÷ 2 = - 1747.311828
Carbon
12.01
Hydrogen
1.01
12. (Hess’s Law)
Calculate the enthalpy change (∆H) in kJ for the following reaction.
2 Al(s) + Fe2O3 (s)  2 Fe(s) + Al2O3 (s)
∆H = ______ kJ
Use the enthalpy changes for the combustion of aluminum and iron.
1)
2 Al(s) + 1.5 O2 (g)  Al2O3 (s)
∆H1 = -1,669.8 kJ
2) 2 Fe(s) + 1.5 O2 (g)  Fe2O3 (s)
∆H2 = +824.2
-824.2 kJ
nd reaction and
IronIt
is supposed
to
be
a
product
so
reverse
2
is called
Hess’s
Law
of HeatsoSUMMATION
Aluminum
is supposed
to be
a reactant
leave 1st reaction alone.
change sign for ∆H.
1) 2 Al(s) + 1.5 O2 (g)  Al2O3 (s)
∆H1 = -1,669.8 kJ
2)
+

-845.6 kJ
2 Al(s) + Fe2O3 (s)  2 Fe(s) + Al2O3 (s) ∆H =-845.6
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
Csteam= 1.7 J/g°C
Solid phase: Temp. from -20.0 °C to 0 °C.
∆T
∆T
∆H
= 0°C – (-20.0°C) = 20.0°C
= Tf - Ti
= m • C • ∆T
m = 150.0 g
C(ice)= 2.1 J/g•°C
∆H = ______ J
∆H = (150.0g) (2.1 J ) (20 °C)
g•°C
∆H = 150 x 2.1 x 20 = 6,300 J
∆H = 6.3 kJ
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
Csteam= 1.7 J/g°C
Melting: Temperature stays at 0 °C.
∆T
= 0°C – 0°C = 0°C
Can’t use ∆H = m • C • ∆T
Use ∆H(fus.) = 6.01 kJ/mol
kJ 1 mol H2O
Set up unit conversions to solve:
150.0 g H2Ox
1
mol H2O
18.02 g H2O
x
1
It takes
6.01 kJ to
melt 1 mole
of water.
8
H
O
1.01
16.00
Hydrogen
Oxygen
H2O
=1.0150.0
16.00 kJ
x
2 x
1
2.02 +16.00= 18.02 g
Calculator: 150 ÷ 18.02 x 6.01 = 50.02774695 kJ
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
Csteam= 1.7 J/g°C
Liquid phase: Temp. from 0 °C to 100 °C.
∆T
∆T
∆H
= 100°C – 0°C = 100°C
= Tf - Ti
= m • C(liquid)
m = 150.0 g
C(liquid)= 4.184 J/g•°C
∆H = ______ J
• ∆T
∆H = (150.0g) (4.184 J ) (100 °C)
g•°C
∆H = 150 x 4.184 x 100 = 62,760 J
∆H = 62.8 kJ
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
Csteam= 1.7 J/g°C
Boiling: Temperature stays at 100 °C.
∆T
= 100°C – 100°C = 0°C
Can’t use ∆H = m • C • ∆T
Use ∆H(vap.) = 40.7 kJ/mol
kJ 1 mol H2O
Set up unit conversions to solve:
150.0 g H2Ox
1
mol H2O
18.02 g H2O
x
1
It takes
40.7 kJ to
boil 1 mole
of water.
8
H
O
1.01
16.00
Hydrogen
Oxygen
H2O
=1.01338.8
kJ
16.00
x
2 x
1
2.02 +16.00= 18.02 g
Calculator: 150 ÷ 18.02 x 40.7 = 338.790231 kJ
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
Csteam= 1.7 J/g°C
Gas phase: Temp. from 100 °C to 120.0 °C.
∆T
∆T
∆H
= 120.0°C – 100 °C = 20.0°C
= Tf - Ti
= m • C(steam)
m = 150.0 g
C(steam) = 1.7 J/g•°C
• ∆T ∆H = ______ J
∆H = (150.0g) (1.7 J ) (20 °C)
g•°C
∆H = 150 x 1.7 x 20 = 5,100 J
∆H = 5.1 kJ
13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to
steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
Cice= 2.1 J/g°C
∆H(vap) = 40.7 kJ/mol
Cliquid= 4.184 J/g°C
6.3 kJ
Solid phase
Melting
50.0 kJ
Liquid phase
62.8 kJ
Boiling
Gas phase
+
Total energy =
338.8 kJ
5.1 kJ
463.0 kJ
Csteam= 1.7 J/g°C