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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 1 of 3: using the Hof data

B

A

C

CH4(g) + 2O2(g)

2H2O(l) + CO2(g) C

=

-

A

B

1 For the following thermchemical data, (1) (2) (3) (4)

1/2 H2(g) + 1/2 Cl2(g)  HCl(g) 2C(s) + 3H2(g) + 1/2 O2(g)  CH3CH2OH(l) 2C(s) + 1.5 H2(g) + 1/2 O2(g) + 1/2 Cl2(g)  CH3COCl(l) 4C(s) + 3.5 H2(g) + O2(g)  CH3COOC2H5(l)

Hof [HCl(g)] = -92.3 kJmol-1 Hof [CH3CH2OH(g)] = -278.0 kJmol-1 Hof [CH3COCl(l)] = -275.0 kJmol-1 Hof [CH3COOC2H5(l)] = -481.0 kJmol-1

Calculate the enthalpy change for the reaction: (5) CH3COCl(l) + CH3CH2OH(l)  CH3COOC2H5 + HCl(g) -20.3kJmol-1

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 1 of 3: using the Hof data

2

Given the following thermochemical data Molecule Hof / kJmol-1

Ethanoic acid -487

Ethanol -278

Ethyl ethanoate -481

Water -286

Calculate the standard enthalpy of the following reaction. Ethanoic acid + ethanol  ethyl ethanoate + water

-2 kJmol-1

3 Given the following data (1) (2) (3)

C(s) + O2(g)  CO2(g) H2(g) + 1/2O2(g)  H2O(l) 3C(s) + 4H2(g)  C3H8(g)

Ho = -393 kJmol-1 Ho = -286 kJmol-1 Ho = -104 kJmol-1

Calculate the standard enthalpy of combustion of propane. (4)

C3H8(g) + 5O2(g)

 3CO2(g) + 4H2O(l)

Hoc [C3H8(g)] = ??? kJmol-1

-2219 kJmol-1

4 Given the following thermochemical data (1) (2) (3)

1/2 H2(g) + 1/2 Cl2(g)  HCl(g) 2C(s) + 3H2(g)  C2H6(g) 2C(s) + 2H2(g) + Cl2(g) C2H4Cl2(l)

Hof [HCl(g)] = -92.3 kJmol-1 Ho f [C2H6(g)] = -84.7 kJmol-1 Ho f [C2H4Cl2(g)] = -166.0 kJmol-1

Calculate the standard enthalpy for the reaction (4) (4)

2Cl2(g) + C2H6(g)  C2H4Cl2(l) + 2HCl(g)

Ho= ??? kJmol-1 -265.9 kJmol-1

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 1 of 3: using the Hof data

[12DSE/paperI/8(d)/2marks] (d)

According to the literature, the standard enthalpy changes of formation of K2CO3(s), KHCO3(s), CO2(g) and H2O(l) are as follows: Compound K2CO3(s) KHCO3(s) CO2(g) H2O(l)

Hof, 298 / kJmol-1 -1146 -959 -394 -286

(i)

Using the given information, calculate the standard enthalpy change of decomposition of KHCO3(s).

(ii)

Suggest why the answers obtained from (c)(ii) and d(i) are different. From calorimeter: standard enthalpy change of decomposition of KHCO3(s) = +44.15 kJmol-1

[08AL/paperII/1(c)(ii)(I)(II)/4 marks]

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

[05AL/paperII/5(a)(i)(ii)/8 marks] Most of the petroleum stock located on Earth is likely to be used up in 50 to 100 years if petroleum consumption is maintained at the current rate. With a view to cutting down petroleum consumption, some countries have adopted an alternative fuel for motor vehicles — gasoline which contains ethanol. (i)

Based on the standard enthalpy changes of formation given below, calculate the standard enthalpy changes for the complete combustion of octane and ethanol respectively. Compound

△HӨf, 298 / kJ mol−1

C8H18(l) C2H5OH(l) CO2(g) H2O(l)

−250 −278 −394 −286

(ii) Assuming that gasoline contains only octane, compare the enthalpy change of combustion values, in kJ g−1, of gasoline and an alternative fuel containing gasoline and 10% ethanol by mass.

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

[04AL/paperII/2(a)(iii)/3 marks] In a similar experiment, the molar enthalpy change for the reaction of magnesium carbonate with hydrochloric acid was found to be –90 kJ. In addition, given that the standard enthalpy change of formation of H2O(l) is –285 kJ mol-1 and that of CO2(g) is –393 kJ mol-1, estimate the enthalpy change of formation of MgCO3(s) under the conditions of the experiment.

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 1 of 3: using the Hof data

[03AL/paperI/3(a)/3 marks] 3.

(a)

Chlorine dioxide, ClO2, can be prepared in the laboratory according to the following equation: 2AgClO3(s) + Cl2(g)  2AgCl(s) + 2ClO2(g) + O2(g)

H 298 = +10 kJ mol-1

Given that the standard enthalpy changes of formation at 298 K of AgClO3(s) and AgCl(s) are – 30 kJ mol-1 and –127 kJ mol-1 respectively, calculate Hf,298 [ClO2(g)]. Hence, comment on the stability of ClO2(g) under standard conditions.

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 1 of 3: using the Hof data

[02AL/paperII/2(a)(i)/3 marks] 2.

(a)

The table below lists the standard enthalpy changes of combustion ( H c,298 ) of three substances. Substance C (graphite) H2(g) C3H6(g) cyclopropane (i)

H c,298 / kJ mol-1 -394 -286 -2090

Calculate the standard enthalpy change of formation of cyclopropane.

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) [00AL/paperII/4(a)/6 marks]

4.

(a)

Interesting!

To find H - common type 1 of 3: using the Hof data

Standard enthalpy change of combustion , HC, 298 , can be calculated using standard enthalpy change of formation data or bond dissociation enthalpy data. (i)

(ii)

(iii)

Given the following standard enthalpy changes of formation, calculate HC, 298 [CH4(g)] Compound

Hf, 298 / kJ mol-1

CO2(g)

-393

H2O(l)

-285

CH4(g)

-75

Given the following bond dissociation enthalpies (E), calculate HC, 298 [CH4(g)] . Bond

E / kJ mol-1

C-H

+433

O=O

+497

C=O

+806

H-O

+465

Comment on the values calculated in (i) and (ii) .

HC, 298 by Method A

HC, 298 by Method B

COMPARE!

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Page 10


Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) [10AL/paperII/2(a)(i)(ii)/6 marks]

Challenging

To find H - common type 1 of 3: using the Hof data

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Page 11


Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 2 of 3: using enthalpy change cycle

5 5

6

7

8

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 2 of 3: using enthalpy change cycle

5

6 Enthalpy change cycle:

Enthalpy change cycle:

By Hess’s Law:

By Hess’s Law:

Hof [CH4(g)]+ Hoc [CH4(g)] =Hof [CO2(g)] + 2 x Hof [H2O(l)] Hof [CH4(g)]

= -393+(2)(-286)-(-890) = - 75kJmol-1

-125 kJmol-1

7

8 Enthalpy change cycle:

Enthalpy change cycle:

By Hess’s Law:

By Hess’s Law:

-275 kJmol-1

+50 kJmol-1

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) To find H - common type 3 of 3: using the algebraic-equation method

9

A student wanted to look up the standard enthalpy change of neutralization between copper (II) oxide and hydrochloric acid, Honeut, from a data book. However he could not find such value in the data book. (a) Write a balanced equation for the reaction between copper (II) oxide and hydrochloric acid. State clearly the state symbols of the reactants and products. (b) Instead, the student found the following data from the data book: Standard enthalpy change / kJmol-1

Reaction (1) Cu(s) + Cl2(g)  CuCl2(s)

H o1 = -206.0

(2) CuCl2(s) + aq  CuCl2(aq)

H o2 = -51.5

(3) H2(g) + 1/2O2(g)  H2O(l)

H o3 = -285.9

(4) 1/2H2(g) + 1/2Cl2(g)  HCl(g)

H o4 = -92.3

(5) Cu(s) + 1/2O2(g)  CuO(s)

H o5 = -155.0

(6) HCl(g) + aq  HCl(aq)

H o6 = -73.9

Express Honeut in terms of Ho1 to Ho6 and hence calculate the value of Honeut.

(a) (b)

CuO(s) + 2HCl(aq)  CuCl2(aq) + H2O(l) Honeut = Ho1 + Ho2 + Ho3 - 2Ho4 - Ho5 - 2Ho6 Honeut = - 56.0 kJmol-1

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

10

To find H - common type 3 of 3: using the algebraic-equation method

-431.2 kJmol-1

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) Energy changes in chemical reactions

(from Curriculum and Assessment Guide, S4-6 Chemistry)

TWO possible cases

2

1

e.g. cracking

(1) increase surrounding temperature

e.g. combustion

energy released can…

temp

(2) work done to surrounding

‘system’

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

For an exothermic (releasing energy to surrounding) reaction:

no work done!

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

1

2

3

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

4

5

6

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

Find H by bond enthalpies

H= ???

and

and

BOND BREAKING

BOND FORMATION

requires energy

releases energy endothermic

exothermic

7

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

H = H

(in bond breaking)

+ H

(in bond formation)

8 exothermic

endothermic

9

10

-184 kJ -103 kJ

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

11

-242 kJ mol-1 -38 kJmol-1 CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) [03AL/paperII/1(c)/7 marks] (c)

Rocket propellants consist of two classed of substances, fuels and oxidizers. When hydrazine is used as a fuel, fluorine or oxygen can be used as the oxidizer. The reaction involved can be represented by the following equations:

(i)

H2NNH2(g) + 2F2(g)  N2(g) + 4HF(g)

(1)

H2NNH2(g) + O2(g)  N2(g) + 2H2O(l)

(2)

You are provided with the following bond enthalpy data: Bond N-N NN N-H F-F F-H O=O O-H

Bond enthalpy / kJ mol-1 163 945 390 158 565 498 464

*Calculate the enthalpy changes, in kJ mol-1, for reactions (1) and (2). (ii)

With reference to the chemistry involved in these two reactions, discuss the relative merits of using fluorine and oxygen as an oxidizer in powering rockets.

CH36 Standard enthalpy changes / CH37 Hess’s Law / CH35 Energy changes in chemical reactions

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy)

12

13

14 15

16

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Topic VIII - Chemical reactions and energy (Part B – Hess’s Law + Bond enthalpy) [05CE/MC] 12.

Which of the following reactions is endothermic ? Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) C9H20(l) → C2H6(g) + C3H6(g) + C4H8(g)

A. B. C. D.

[02CE/MC] 33.

Which of the following reactions is/are endothermic? (1) the fermentation of glucose (2) the cracking of naphtha (3) the reaction of lime with dilute hydrochloric acid A. B. C. D.

(1) only (2) only (1) and (3) only (2) and (3) only

[01CE/MC] 2.

Which of the following processes is exothermic? A. B. C. D.

melting of ice evaporation of ethanol sublimation of iodine dissolving sodium hydroxide pellets in water

[00CE/MC] 25.

Which of the following processes is endothermic? A. B. C. D.

cracking of petroleum fractions fermentation of glucose solution manufacture of ammonia by Haber process oxidation of sulphur dioxide to sulphur trioxide in the contact process

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Topic 8 chemical reaction and energy (part b)