1. Chapter 11 (Practice Test)
Thermochemistry

2. = ∆H m • C • ∆T ∆H = m = C = ∆T = ∆H = 450 g 0.71 J/g•°C 150 °C (450g)
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 =
450 g
0.71 J/g•°C ∆H =
m • C • ∆T
150 °C
∆H =
(450g)
( J )
g•°C
(150 °C)
Calculator 450 x 0.71 x 150 = 47,925 J
Answer w/ Sig. Figs.
∆H = 48,000 J or kJ

3. = ∆H m • C • ∆T C = ∆H = m = C = ∆T = C = 0.14 J/g•°C 770 J ∆H m x ∆T
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?
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 =
m =
C =
∆T = ∆H =
m • C • ∆T
770 J ∆H
m x ∆T
50 g
m x ∆T
m x ∆T
_____ J/g°C
110 °C
C =
Calculator
770 ÷ 50 ÷ 110 = 0.14 or
770 ÷ (50 x 110) = 0.14
C = ∆H
m x ∆T
770 J =
50 g x 110 °C
C = 0.14 J/g•°C

4. = ∆H m • C • ∆T ∆H = m = C = ∆T = ∆H = 175.0 g 0.385 J/g•°C
3. Calculate the heat absorbed by the water in a calorimeter when grams of copper cools from °C to 22.0 °C The specific heat capacity of copper is J/g•°C.
3. Calculate the heat absorbed by the water in a calorimeter when grams of copper cools from °C to 22.0 °C The specific heat capacity of copper is J/g•°C.
∆H =
m =
C =
∆T =
175.0 g
0.385 J/g•°C ∆H =
m • C • ∆T
22.0°C – 125.0°C
∆T = – 103.0°C
∆H =
(175g)
( J )
g•°C
(-103 °C)
Calculator 175 x x -103 = J
But water Absorbed the heat so…
∆H = J

5. = ∆H m • C • ∆T ∆T = ∆H = m = C = ∆T = Ti = Tf = _____ °C
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 = J/g•°C?
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 = J/g•°C?
∆H =
m =
C =
∆T = ∆H =
m • C • ∆T
372 J ∆H
m x C
5.00 g
m x C
m x C
4.184 J/g°C
_____ °C
∆T =
Ti =
Tf = _____ °C
23.0 °C
∆T = ∆H
m x C
372 = =
17.8 °C
(5 x 4.184)
Calculator
372 ÷ 5 ÷ = 17.78
Heat was added so temp.
increases by 17.8.
Tf = = 40.8 °C

6. = ∆H m • C • ∆T ∆H = m = C = ∆T = ∆H = 2.0 x 102 g = 200 g
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 = J/g•°C)
∆H =
m =
C =
∆T =
2.0 x 102 g = 200 g
0.878 J/g•°C ∆H =
m • C • ∆T
30°C
∆H =
(200g)
( J )
g•°C
(30 °C)
Calculator (2.0 x 102) x x 30 = 5268 J
Answer to correct sig.figs.
∆H = 5 kJ

7. - = ∆H m • C • ∆T = ∆T Tf - Ti Water ∆H = m = C = ∆T = Lead ∆H = m =
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 with an initial temperature of 22.0 °C . The final temperature of the water and the lead is 25.0 °C.
Water so…C is known
Water
∆H =
m =
C =
∆T = ∆H =
m • C • ∆T
_________ J
99.5 g
∆H =
(99.5g)
( J )
g•°C
(3.0 °C)
4.184 J/g•°C
∆H = 1, J 1,
25-22=3.0°C
Energy water gained
from the hot metal so… ∆T =
Tf - Ti
Lead
∆H =
m =
C =
∆T = -
_________ J
C = ∆H
m x ∆T J =
85.0 g
(85.0) (-74.0 °C)
______ J/g•°C
C = = 0.20 J/g•°C
25-99=-74.0°C

8. ∆H° = ∆Hf°(products) - ∆Hf°(reactants)
7. Find the standard heat of formation for the following reaction.
∆H° = ∆Hf°(products) - ∆Hf°(reactants)
Substance
∆Hf°
(kJ/mol)
NH3(g)
-46.19
O2(g)
0.0
NO(g)
90.37
H2O(g)
-285.8
4 NH3(g) + 5 O2 (g)  4 NO(g) H2O (l)
-46.19 x kJ
0.0
x 5
0 kJ
90.37 x kJ
-285.8 x kJ + + kJ
-1, kJ
(reactants)
(products)
∆H° = ∆Hf°(products) - ∆Hf°(reactants)
∆H° = -1, kJ – ( kJ) = kJ
∆H° = kJ

9. H 8. Heat of Reaction: 2 2 H2(g) + O2 (g)  2 H2O (l) ∆H = -572 kJ
How much heat is produced when 5.00 g of H2 (at STP) is reacted with excess O2?
5.00 g H2 1
mol H2 x x =
-708 kJ
mol H2
2.02
g H2
1.01
x 2
2.02 1 H
Hydrogen
1.01
Calculator: 5.00 ÷ 2.02 x -572 ÷ 2 =

10. Na O H 9. Heat of Solution: ∆Hsoln = -445.1 kJ/mol -445.1 kJ/mol 1 x x
Determine the heat of solution when g of NaOH is dissolved in water.
40.00 g NaOH 1
mol NaOH x x = kJ
40.00 1
mol NaOH
g NaOH 11 Na
sodium
22.99 8 O
Oxygen
16.00 1 H
Hydrogen
1.01
Calculator: ÷ x =

11. N O H 10. Heat of Solution: ∆Hsoln = 25.7 kJ/mol 25.7 kJ 1 x x =
Determine the heat of solution when g of NH4NO3 is dissolved in water.
25.58 g NH4NO3 1
mol NH4NO3 x x =
8.211 kJ 1
80.06
mol NH4NO3
g NH4NO3 7 N
Nitrogen
14.01 8 O
Oxygen
16.00 1 H
Hydrogen
1.01
Calculator: ÷ x 25.7 =

12. 11. Heat of Combustion:
2 C2H O2  4 CO H2O ∆H = kJ 2
-2600 kJ
How much heat is produced when g of C2H2 is reacted with excess O2?
35.00 g C2H2 1
mol C2H2 x x =
-1747 kJ
mol C2H2
26.04
g C2H2 6 C
Carbon
12.01 1 H
Hydrogen
1.01
Calculator: ÷ x ÷ 2 =

13. + 12. (Hess’s Law) It is called Hess’s Law of Heat SUMMATION -845.6
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) O2 (g)  Al2O3 (s)
∆H1 = -1,669.8 kJ 2)
2 Fe(s) O2 (g)
2 Fe(s) O2 (g)  Fe2O3 (s)
Fe2O3 (s)
∆H2 = kJ
∆H2 = kJ
Iron is supposed to be a product so reverse 2nd reaction and change sign for ∆H.
It is called Hess’s Law of Heat SUMMATION
Aluminum is supposed to be a reactant so leave 1st reaction alone. 1)
2 Al(s) O2 (g)  Al2O3 (s)
∆H1 = -1,669.8 kJ +  2)
-845.6
2 Al(s) + Fe2O3 (s)  2 Fe(s) + Al2O3 (s)
∆H = kJ

14. = ∆T Tf - Ti = ∆H m • C • ∆T ∆T = m = C(ice)= ∆H = ∆H =
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= J/g°C
Csteam= 1.7 J/g°C
Solid phase: Temp. from °C to 0 °C.
∆T =
0°C – (-20.0°C) = 20.0°C ∆T =
Tf - Ti
m =
C(ice)=
∆H =
150.0 g
2.1 J/g•°C ∆H =
m • C • ∆T
______ J
∆H =
(150.0g)
(2.1 J )
g•°C
(20 °C)
∆H = 6.3 kJ
∆H = 150 x 2.1 x 20 = 6,300 J

15. H O ∆T = Can’t use ∆H = m • C • ∆T Use ∆H(fus.) = 6.01 kJ/mol
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= 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
It takes
6.01 kJ to
melt 1 mole
of water. 1 H
Hydrogen
1.01 8 O
Oxygen
16.00
Use ∆H(fus.) = 6.01 kJ/mol
6.01 kJ
1 mol H2O
Set up unit conversions to solve:
1.01
x 2
2.02
H2O
16.00
x 1
+16.00
= g 1
mol H2O
150.0 g H2O x x
= 50.0 kJ
18.02
g H2O
Calculator: 150 ÷ x 6.01 = kJ

16. = ∆T Tf - Ti = ∆H m • C(liquid) • ∆T ∆T = m = C(liquid)= ∆H = ∆H =
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= J/g°C
Csteam= 1.7 J/g°C
Liquid phase: Temp. from 0 °C to 100 °C.
∆T =
100°C – 0°C = 100°C ∆T =
Tf - Ti
m =
C(liquid)=
∆H =
150.0 g
4.184 J/g•°C ∆H =
m • C(liquid) • ∆T
______ J
∆H =
(150.0g)
( J )
g•°C
(100 °C)
∆H = 150 x x 100 = 62,760 J
∆H = 62.8 kJ

17. H O ∆T = Can’t use ∆H = m • C • ∆T Use ∆H(fus.) = 40.7 kJ/mol
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= 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
It takes
40.7 kJ to
boil 1 mole
of water. 1 H
Hydrogen
1.01 8 O
Oxygen
16.00
Use ∆H(fus.) = 40.7 kJ/mol
40.7 kJ
1 mol H2O
Set up unit conversions to solve:
1.01
x 2
2.02
H2O
16.00
x 1
+16.00
= g 1
mol H2O
150.0 g H2O x x
= kJ
18.02
g H2O
Calculator: 150 ÷ x 40.7 = kJ

18. = ∆T Tf - Ti = ∆H m • C(steam) • ∆T ∆T = m = C(ice)= ∆H = ∆H =
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= J/g°C
Csteam= 1.7 J/g°C
Gas phase: Temp. from 100 °C to °C.
∆T =
120.0°C – 100 °C = 20.0°C ∆T =
Tf - Ti
m =
C(ice)=
∆H =
150.0 g
1.7 J/g•°C ∆H =
m • C(steam) • ∆T
______ J
∆H =
(150.0g)
(1.7 J )
g•°C
(20 °C)
∆H = 5.1 kJ
∆H = 150 x 1.7 x 20 = 5,100 J

19. + 50.0 kJ 62.8 kJ 338.8 kJ 463.0 kJ 6.3 kJ Solid phase Melting
13. How much heat is absorbed by g of ice at °C to steam at 120 °C?
∆H(fus) = 6.01 kJ/mol
∆H(vap) = 40.7 kJ/mol
Cice= 2.1 J/g°C
Cliquid= J/g°C
Csteam= 1.7 J/g°C
6.3 kJ
Solid phase
50.0 kJ
Melting
62.8 kJ
Liquid phase
338.8 kJ
Boiling +
5.1 kJ
Gas phase
Total energy =
463.0 kJ