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 =

0.71 J/g•°C450 g

150 °C

∆H = (450g) (150 °C)

Calculator 450 x 0.71 x 150 = 47,925 J

∆H = m • C • ∆T

(0.71 J )g•°C

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?

C = ∆H

m x ∆T

∆H = m = C =∆T =

_____ J/g°C50 g

110 °C

= 770 J

50 g x 110 °C

C = 0.14 J/g•°C

∆H = m • C • ∆T 770 J

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?

C =

m x ∆Tm x ∆T∆H

m x ∆T

Calculator770 ÷ 50 ÷ 110 = 0.14

or770 ÷ (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 =

0.385 J/g•°C175.0 g

22.0°C – 125.0°C

∆H = (175g) (-103 °C)

Calculator 175 x 0.385 x -103 = -6939.625 J

∆H = m • C • ∆T

(0.385 J )g•°C

But water Absorbed the heat so… ∆H = + 6939.625 J

∆T = – 103.0°C

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.

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?

∆T = ∆H

m x C

∆H = m = C =∆T =

4.184 J/g°C5.00 g

_____ °C

= 372

(5 x 4.184)

Tf = 23.0 + 17.8 = 40.8 °C

∆H = m • C • ∆T 372 J

∆T =

m x Cm x C∆H

m x C

Calculator372 ÷ 5 ÷ 4.184 = 17.78

Ti =Tf = _____ °C

23.0 °C

= 17.8 °C

Heat was added so temp.increases by 17.8.

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?

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 =

0.878 J/g•°C2.0 x 102 g = 200 g

30°C

∆H = (200g) (30 °C)

Calculator (2.0 x 102) x 0.878 x 30 = 5268 J

∆H = m • C • ∆T

(0.878 J )g•°C

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 with an initial temperature of 22.0 °C . The final temperature of the water and the lead is 25.0 °C.

C = ∆H

m x ∆T

Water∆H = m = C =∆T =

______ J/g•°C85.0 g

25-99=-74.0°C

= - 1248.924 J

(85.0) (-74.0 °C)

C = 0.199 = 0.20 J/g•°C

∆H = m • C • ∆T

Lead∆H = m = C =∆T =

25-22=3.0°C

∆T = Tf - Ti

99.5 g4.184 J/g•°C

_________ J

_________ J∆H = (99.5g) (3.0 °C)(4.184 J )

g•°C∆H = 1,248.924 J

Energy water gainedfrom the hot metal so…

-

1,248.924

Water so…C is known

7. Find the standard heat of formation for the following reaction.

4 NH3(g) + 5 O2 (g) 4 NO(g) + 6 H2O (l)

∆H° = ∆Hf°(products) - ∆Hf°(reactants)

-46.19x 4-184.76 kJ

- 1168.56 kJ

∆H° = - 1168.56 kJ

∆H° = -1,353.32 kJ – (-184.76 kJ) =

Substance

∆Hf°

(kJ/mol)

NH3(g) -46.19

O2(g) 0.0

NO(g) 90.37

H2O(g) -285.8

0.0x 5 0 kJ+

-184.76 kJ(reactants)

90.37x 4361.48 kJ

-285.8x 6-1714.8 kJ+

-1,353.32 kJ(products)

∆H° = ∆Hf°(products) - ∆Hf°(reactants)

8. Heat of Reaction:

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

xg H2

mol H2

1.01x 22.02

1

HHydrogen

1.01

2.02

1x =

-572 kJ

mol H2

- 707.9

-708 kJ

2

Calculator: 5.00 ÷ 2.02 x -572 ÷ 2 =

9. Heat of Solution:

∆Hsoln = -445.1 kJ/mol

Determine the heat of solution when 40.00 g of NaOH is dissolved in water.

40.00 g NaOH

xg NaOH

mol NaOH

1

HHydrogen

1.01

40.00

1x =

-445.1 kJ/mol

mol NaOH

- 445.1

-445.1 kJ1

Calculator: 40.00 ÷ 40.00 x -445.1 = 11

Nasodium22.99

8

OOxygen16.00

10. Heat of Solution:

∆Hsoln = 25.7 kJ/mol

Determine the heat of solution when 25.58 g of NH4NO3 is dissolved in water.

25.58 g NH4NO3

xg NH4NO3

mol NH4NO3

1

HHydrogen

1.01

80.06

1x =

25.7 kJ

mol NH4NO3

8.211416438

8.211 kJ1

Calculator: 25.58 ÷ 80.06 x 25.7 = 7

NNitrogen

14.01

8

OOxygen16.00

11. Heat of Combustion:

2 C2H2 + 5 O2 4 CO2 + 2 H2O ∆H = -2600 kJ

How much heat is produced when 35.00 g of C2H2 is reacted with excess O2?

35.00 g C2H2

xg C2H2

mol C2H2

1

HHydrogen

1.01

26.04

1x =

-2600 kJ

mol C2H2

- 1747.311828

-1747 kJ

2

Calculator: 35.00 ÷ 26.04 x -2600 ÷ 2 =6

CCarbon12.01

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)

Use the enthalpy changes for the combustion of aluminum and iron.

∆H = ______ kJ

2 Al(s) + 1.5 O2 (g) Al2O3 (s) ∆H1 = -1,669.8 kJ1)

2 Fe(s) + 1.5 O2 (g) Fe2O3 (s) ∆H2 = -824.2 kJ2)

Aluminum is supposed to be a reactant so leave 1st reaction alone.

2 Al(s) + 1.5 O2 (g) Al2O3 (s) ∆H1 = -1,669.8 kJ1)

Iron is supposed to be a product so reverse 2nd reaction and change sign for ∆H.

2 Fe(s) + 1.5 O2 (g) ∆H2 = +824.2 kJ

2)

Fe2O3 (s)

+2 Al(s) + Fe2O3 (s) 2 Fe(s) + Al2O3 (s) ∆H = -845.6 kJ-845.6

It is called Hess’s Law of Heat SUMMATION

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Solid phase: Temp. from -20.0 °C to 0 °C.

∆T = 0°C – (-20.0°C) = 20.0°C

m = C(ice)=∆H =

150.0 g

2.1 J/g•°C

∆H = m • C • ∆T∆H = (150.0g) (20 °C)(2.1 J )

g•°C

∆H = 150 x 2.1 x 20 = 6,300 J ∆H = 6.3 kJ

______ J

∆T = Tf - Ti

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Melting: Temperature stays at 0 °C.

∆T = 0°C – 0°C = 0°CCan’t use ∆H = m • C • ∆T

150.0 g H2O

Set up unit conversions to solve:

= 50.0 kJ

Use ∆H(fus.) = 6.01 kJ/mol

x

1

HHydrogen

1.01

8

OOxygen16.00

1.01x 2

2.02

H2O

16.00x 1

+16.00 = 18.02 g

g H2O

mol H2O

18.02

1x

6.01 kJ1 mol H2O

Calculator: 150 ÷ 18.02 x 6.01 = 50.02774695 kJ

It takes6.01 kJ to

melt 1 moleof water.

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Liquid phase: Temp. from 0 °C to 100 °C.

∆T =100°C – 0°C = 100°C

m = C(liquid)=∆H =

150.0 g

4.184 J/g•°C

∆H = m • C(liquid) • ∆T∆H = (150.0g) (100 °C)(4.184 J )

g•°C

∆H = 150 x 4.184 x 100 = 62,760 J

∆H = 62.8 kJ

______ J

∆T = Tf - Ti

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Boiling: Temperature stays at 100 °C.

∆T =100°C – 100°C = 0°CCan’t use ∆H = m • C • ∆T

150.0 g H2O

Set up unit conversions to solve:

= 338.8 kJ

Use ∆H(fus.) = 40.7 kJ/mol

x

1

HHydrogen

1.01

8

OOxygen16.00

1.01x 2

2.02

H2O

16.00x 1

+16.00 = 18.02 g

g H2O

mol H2O

18.02

1x

40.7 kJ1 mol H2O

Calculator: 150 ÷ 18.02 x 40.7 = 338.790231 kJ

It takes40.7 kJ to boil 1 moleof water.

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Gas phase: Temp. from 100 °C to 120.0 °C.

∆T =120.0°C – 100 °C = 20.0°C

m = C(ice)=∆H =

150.0 g

1.7 J/g•°C

∆H = m • C(steam) • ∆T∆H = (150.0g) (20 °C)(1.7 J )

g•°C

∆H = 150 x 1.7 x 20 = 5,100 J ∆H = 5.1 kJ

______ J

∆T = Tf - Ti

13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C?

Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C

∆H(fus) = 6.01 kJ/mol ∆H(vap) = 40.7 kJ/mol

Solid phase

Melting

Liquid phase

Gas phase

Boiling

6.3 kJ

50.0 kJ

62.8 kJ

338.8 kJ

5.1 kJ +

Total energy = 463.0 kJ