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Thermochemistry Problems Worksheet Answers, Exercises of Chemistry

Laney CollegeChemistry

Problems are on first law and enthalpy with answers

Typology: Exercises

2020/2021

Uploaded on 06/11/2021

hambery
hambery 🇺🇸

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Thermochemistry,Problems,Page 1 of 7,
Thermochemistry,Problems,
The,First,Law,
1. The,complete,combustion,of,acetic,acid,,HC2H3O2(l),,to,form,water,,H2O(l),,and,
CO2(g),,at,constant,pressure,releases,871.7,kJ,of,heat,per,mol,of,acetic,acid.,,Write,
a,balanced,thermochemical,equation,for,this,reaction.,,Draw,an,enthalpy,
diagram,for,the,reaction.,,
,
2. Consider,the,following,reaction,,which,occurs,at,standard,state:,,,
2Cl(g)®,,Cl2(g), , ,∆H°,=,–P243.4,kJ,
a. Which,has,the,higher,,more,positve,,enthalpy,under,these,conditions,,2Cl,
(g),or,Cl2(g)?,
b. Consider,the,following,reaction:,,2,Mg(s),+,O2(g),®,,2,MgO(s),∆H,=,P1204,kJ.,,,
c. Is,the,reaction,exothermic,or,endothermic?,,,
d. Calculate,the,amount,of,heat,transferred,when,2.4,g,of,Mg(s),reacts,at,
constant,pressure.,,How,many,grams,of,MgO,are,produced,during,an,
enthalpy,change,of,96.0,kJ?,,
e. How,many,kilo,joules,of,heat,are,absorbed,when,7.50,g,of,MgO(s),are,
decomposed,into,Mg(s),and,O2(g),at,constant,pressure?,
Enthalpy,
Calorimetry,Problems,,
3. Determine,the,specific,heat,of,a,sample,of,Cu,from,the,fact,that,64.0J,are,needed,
to,raise,the,temperature,of,15.0,g,of,Cu,metal,from,22.0°C,to,33.0°C.,[0.388,J/g°C],
4. The,specific,heat,of,a,sample,of,copper,,is,0.385,J/gPK.,,How,many,joules,of,heat,
are,necessary,to,raise,the,temperature,of,a,1.42,kg,block,of,copper,from,25.0,°C,to,
88.5°C?,
5. A,50.0Pg,sample,of,water,at,100.00°C,was,placed,in,an,insulated,cup.,,Then,25.3Pg,
of,zinc,at,25.00°C,was,added,to,the,water.,,The,temperature,of,the,water,
dropped,to,96.68°C.,,What,is,the,specific,heat,of,the,zinc?[0.388,J/g°C],
6. When,a,6.50,g,sample,of,solid,NaOH,dissolves,in,100.0g,of,water,in,a,coffeePcup,
calorimeter,,the,temperature,rises,from,21.6°C,to,37.8°C.,,Calculate,the,∆H,(in,
kJ/mol,NaOH),for,the,solution,process.,(Assume,the,specific,heat,of,the,solution,
formed,is,4.18,J/gPK),
NaOH(s),®,,Na+(aq)+,OH–(aq),
7. A,house,is,designed,to,have,passive,solar,energy,features,,Brickwork,is,to,be,
incorporated,into,the,interior,of,the,house,to,act,as,a,heat,absorber.,,Each,brick,
pf3
pf4
pf5

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Thermochemistry Problems

The First Law

1. The complete combustion of acetic acid, HC 2 H 3 O2(l), to form water, H 2 O(l), and

CO2(g), at constant pressure releases 871.7 kJ of heat per mol of acetic acid. Write

a balanced thermochemical equation for this reaction. Draw an enthalpy

diagram for the reaction.

2. Consider the following reaction, which occurs at standard state:

2Cl(g)® Cl2(g) ∆H° = – -­‐‑243.4 kJ

a. Which has the higher, more positve, enthalpy under these conditions, 2Cl

(g)

or Cl

2 (g)

b. Consider the following reaction: 2 Mg

(s)

+ O

2 (g)

® 2 MgO

(s)

∆H = -­‐‑1204 kJ.

c. Is the reaction exothermic or endothermic?

d. Calculate the amount of heat transferred when 2.4 g of Mg

(s)

reacts at

constant pressure. How many grams of MgO are produced during an

enthalpy change of 96.0 kJ?

e. How many kilo joules of heat are absorbed when 7.50 g of MgO

(s)

are

decomposed into Mg

(s)

and O

2 (g)

at constant pressure?

Enthalpy

Calorimetry Problems

3. Determine the specific heat of a sample of Cu from the fact that 64.0J are needed

to raise the temperature of 15.0 g of Cu metal from 22.0°C to 33.0°C. [0.388 J/g°C]

4. The specific heat of a sample of copper is 0.385 J/g-­‐‑K. How many joules of heat

are necessary to raise the temperature of a 1.42 kg block of copper from 25.0 °C to

88.5°C?

5. A 50.0-­‐‑g sample of water at 100.00°C was placed in an insulated cup. Then 25.3-­‐‑g

of zinc at 25.00°C was added to the water. The temperature of the water

dropped to 96.68°C. What is the specific heat of the zinc?[0.388 J/g°C]

6. When a 6.50 g sample of solid NaOH dissolves in 100.0g of water in a coffee-­‐‑cup

calorimeter, the temperature rises from 21.6°C to 37.8°C. Calculate the ∆H (in

kJ/mol NaOH) for the solution process. (Assume the specific heat of the solution

formed is 4.18 J/g-­‐‑K)

NaOH

(s)

® Na

(aq)+ OH

(aq)

7. A house is designed to have passive solar energy features, Brickwork is to be

incorporated into the interior of the house to act as a heat absorber. Each brick

weighs approximately 1.8 kg. The specific heat of the brick is 0.85 j/g-­‐‑K. How

many bricks must be incorporated into the interior of the hose to provide the

same total heat capacity as 1.0 10

3

gal of water?

Bomb calorimetery

8. When 15.3 g of sodium nitrate was dissolved in water in a calorimeter, the

temperature fell from 25.00 °C to 21.56 °C. If the heat capacity of the solution

and the calorimeter is 1071 J/°C, what is the enthalpy change for the process

when 1 mol of sodium nitrate dissolves in water?

9. Camphor (C 10 H 16 O) has an energy of combustion of -­‐‑5903.6 kJ/mol. When a

sample of camphor with mass 0.1204 g is burned in a bomb calorimeter, the

temperature increases by 2.28°C. Calculate the heat capacity of the calorimeter.

Hesses Law

10. Consider the following hypothetical reactions

A ® B ∆H = 30 kJ

B ® C ∆H = 60 kJ

Use Hess’s law to calculate the enthalpy change for the reaction A ® C.

Construct an enthalpy diagram for substances A, B, and C and show how Hess’s

law applies.

11. Given the following enthalpies of reaction:

P4(s)+3O2(g)®P 4 O6(s) ∆H = – 1640.1 kJ

P4(s)+ 5O2(g) ®P 4 H10(s) ∆H = – 2940.1 kJ

Calculate the enthalpy change for the reaction: P 4 O6(s)+ 2 O2(g) ®P 4 H10(s)

12. From the following enthalpies of reaction

H2(g) + F2(g) ® 2HF(g) ∆H = – 537 kJ

C(s) + 2 F2(g) ® CF4(g) ∆H = – 680 kJ

2 C(s) + 2 H2(g)® C 2 H4(g) ∆H = – 52.3 kJ

Calculate the ∆H for the reaction of C 2 H4(g) with F2(g) to make CF4(g) and HF(g)

Heats of formation

13. Many cigarette lighters contain liquid butane, C 4 H10(l), which has a ∆H°ƒ = – 147.

kJ/mol. Using enthalpies of formation, calculate the quantity of heat produced

when 1.0 g of butane is completely combusted in air. (Write the equation for the

process first 1 )

14. Gasoline is composed primarily of hydrocarbons with eight carbon atoms. These

hydrocarbons are called octanes. One of the cleanest burning octanes is a

must be vaporized in the refrigeration cycle to convert all of the water at 22.0°C

to ice at -­‐‑5.0°C. The heat capacities for water and ice are 4.18 J/g-­‐‑°C and 2.08 J/g-­‐‑

°C respectively, and the enthalpy of fusion for water is 6.02 kJ/mol

ANSWERS

1. HC 2 H 3 O2(l) + 2O2(g) ® CO2(g + 2H 2 O(l),

HC 2 H 3 O2(l)+ 4 O2(g)

  • 871.7 kJ ¯

2 CO2(g + 2H 2 O(l)

  1. The enthalpy of reaction is ∆H° prod - ∆H° react. The reactant is at a higher potential than

the product. When the reaction occurs, heat is released, stabilizing the system and the

system moves to a lower energy potential.

  1. The reaction is exothermic, the enthalpy is negative

[1] HC

H

O

2(l)

+ 2O

2(g)

® 2 CO

2(g)

+ 2 H

O

(l)

HC

H

O

2(l)

+ 4 O

2(g)

  • 871.7 kJ ¯

2CO

2(g)

+ 2 H

O

(l)

[2] The enthalpy of reaction is ∆H° prod - ∆H° react. The reactant is at a higher potential

than the product. When the reaction occurs, heat is released, stabilizing the system and the

system moves to a lower energy potential.

[3] The reaction is exothermic. The ∆H is negative.

2.4 g MgX

1 mol Mg

24.0 g Mg

= 0.10 mol Mg X

  • 1204 kJ

2 mol Mg

= – 60.2 kJ

96 kJ X

2 mol MgO

  • 1204 kJ

X

40.0 g MgO

1 mol MgO

= 6.4 g MgO

7.50 g MgO X

1 mol MgO

40.0 g MgO

X

+1204 kJ

2 mol MgO

= 113 Kj

[4] q =

0.385 J

g–K

X 1.42 X 10

g Cu X (88.5°C – 25.0°C) X

1 K

1°C

= 3.47 X

J

[5] First, find the energy that the solution either absorbed or released. Then relate that

amount of energy to the moles of NaOH

q sol = (100.0 g water + 6.50 g NaOH) X (37.8°C - 21.6°C) X

4.18 J

g°C

= 7.211 X 10

J

∆H =

  • 7.211 kJ

6.50 g

X

39.90 g NaOH

1 mol NaOH

= 44.3 kJ/mol

[6]

A ® B A

B ® C 60 kJ B ­ 30 kJ ­ 90 kJ

Total 90 kJ C ­ 60 kJ

[7] Hess’s law is a law of summations. Partial paths add to the desired path

Given the following enthalpies of reaction:

P

O

6(s)

® P

4(s)

+ 3 O

2(g)

∆H = +1640.1 kJ

P

4(s)

+ 5 O

2(g)

® P

O

10(s)

∆H = - 2940.1 kJ

P

O

6(s)

+ 2 O

2(g)

® P

O

10(s)

  • 1640.1 kJ + – 2940.1 kJ = – 1300.0 kJ

[8] H

2(g)

+ F

2(g)

® 2HF

(g)

∆H = - 537 kJ

2 C

(s)

+ 4 F

2(g)

® 2 CF

4(g) ∆H = - 680 kJ 2

C

H

4(g)

® 2 C

(s)

+ 2 H

2(g)

∆H = – 52.3 kJ

C

H

4(g)

+ 5 F

2(g)

® 2 CF

4(g)

+ 2 HF

(g)

∆H = – 1949.3 kJ