THERMOCHEMISTRY CHA4 FORM5

7/27/2019 THERMOCHEMISTRY CHA4 FORM5

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Thermochemistry

1.0 Energy Changes in Chemical Reaction

Definition:

Poem;

P.K.

EXOTHERMIC

ENDOTHERMIC

Very important

Bond breaking REQUIRES energy

Bond formation RELEASES energy

Thermochemistry is the study of changes in heat

energy which take place during chemical reaction

Heat energy, we shalllater studyIs involved in reactions ofthermochemistry,

A chemical reaction that GIVES OUT/

RELEASES heat to the surroundings

A chemical reaction that ABSORBS heat

from the surroundings


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EXOTHERMIC REACTION

Energy Profile Diagram

Study the diagram carefully.

So that you will be able to draw the Energy Level Diagram.

What is activation energy?

Activation energy is the energy barrierthat must beovercome by the colliding particles of the reactants in order

for reaction to occur / tobecome the products

H = The change in the amount of heat in a chemical

reaction is called the heat of reaction.

Energy

Reaction path

Energy

releases

during bondformation

Energy

requires

during

bondbreaking

Reactants

Products

Energy Requires < Energy Releases

X

Y

Z

Activation Energy H = ve


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What we can say about exothermic reaction?

The conclusion is;

- Bond breaking requires less energy than bond formation,

- So, the energy released to form the bond is higher /

greater than the energy absorbed to break the bonds.- Then, excess heat energy is released to the surroundings.

- Thus, during the reaction temperature of the mixture

inreases,

- The total energy of product is lower than the total energy

of reactant.

Energy Level Diagram for Exothermic Reaction

Simply mean like this;Project = rm100 000 [energy requires]Complete = rm150 000 [energy releases]

Loss = rm50 000 [ H, heat changes]Saving = less rm50 000 [negative]

Energy

Reactants

Products

H = (negative)

H = Hproducts

Hreactants

H = ve


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ENDOTHERMIC REACTION

Energy Profile Diagram

Study the diagram carefully.

What we can say about endothermic reaction?

The conclusion is

- Bond breaking requires more energy than bond formation,

- So, the energy released to form the bond is lower / less

than the energy absorbed to break the bonds.

- Then, heat energy is absorbed from the surroundings.

- Thus, during the reaction temperature of the mixture

decreases,

- The total energy of product is higher than the total energy

of reactant.

Energy

Reaction path

Energy

releases

during bond

formation

Energy

requires

during

bondbreaking

Reactants

Products

Energy Required > Energy Released

Q

P

R

Activation Energy

H = +ve


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Energy Level Diagram for Endothermic Reaction

Simply mean like this;Project = rm100 000 [energy requires]

Complete = rm50 000 [energy releases]

Profit = rm50 000 [energy heat changes]

Saving = up to rm50 000 [positive]

Example of exothermic reaction

Most of the chemical reaction is exothermic such as

- neutralization

- combustion- acid and metal

[Tip: better to memorize endothermic reaction, because not

many reaction is endothermic]

Energy

Reactants

Products

H = + (positive)

H = Hproduct

Hreactants

H = +ve


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Example of endothermic reaction

1. Salt dissolves in water

NH4Cl (s) NH4+ (aq) + Cl- (aq)

KNO3 (s) K+ (aq) + NO3

- (aq)

2. Salt crystallization

CuSO4.5H2O (s) Cu2+ (aq) + SO4

2- (aq) + 5H2O (l)

3. Thermal decomposition

ZnCO3 (s) ZnO(s) + CO2 (g)

2Mg(NO3)2 (s) 2MgO

(s) + 4NO2 (g) + O2 (g)

4. Salt dissociation

NH4Cl (s) NH4+ (aq) + Cl- (aq)

CaCO3 (s) CaO(s) + CO2 (g)

5. Reaction between acid with sodium hydrogen carbonate

and potassium hydrogen carbonate;HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

HCl(aq) + KHCO3(aq) KCl(aq) + H2O(l) + CO2(g)

6. photosynthesis

6CO2 + 6H2O C6H12O6 + 6O2

7. process of melting, evaporation and boiling.


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Example 1:

Mg + H2SO4 MgSO4 + H2

H = 467 kJ

Energy level diagram

Exercises

Based from the following equations, construct and explain

energy level diagram for the reaction.

(1) CaCO3 CaO + CO2 H = + 178 kJ

(2) 2H2 + O2 2H2O H = 572 kJ

(3) Zn + CuSO4 ZnSO4 + Cu H = 190 kJ

(4) H2 + I2 2HI H = + 53 kJ

Energy

Mg + H2SO

4

MgSO4 + H2

H = - 467 kJ

Explanation:

The reaction is an exothermic reaction

Temperature of mixture is increases

Total energy of 1 mole Mg and 1 mole H2SO

4is

higher than 1 mole of MgSO4 and 1 mole H2 by 467kJ

When 1 mole Mg reacts with 1 mole H2SO

4to form 1

mole of MgSO4

and 1 mole H2

, 467 kJ of heat is

released/produced.


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Effective Practice pg 148 no. 1, 2, 3 & 4

Example 2:

2HgO 2Hg + O2

H = +182 kJEnergy level diagram;

Kamal Ariffin Bin Saaim

SMKDBL

http://www.kemhawk.webs.com/

Energy

2HgO

2Hg + O2

H = + 182

Explanation:

The reaction is an endothermic reaction

Temperature of mixture is decreases

Total energy of 2 mole HgO is lower than 2 mole ofHg and 1 mole O

2by 182 kJ

When 2 mole HgO decompose to form 2 mole of Hg

and 1 mole O2, 182 kJ of heat is absorbed
http://www.kemhawk.webs.com/http://www.kemhawk.webs.com/


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exampleThe diagram shows an energy profile diagram.

Based on the above energy profile diagram, the amount of activation energy is..

A (Y X) kJ mol-1

B X kJ mol-1C (X Y) kJ mol-1

D Y kJ mol-1

25 The reaction between nitrogen and oxygen can be represented by the following equation:

N2 (g) + O2(g) 2NO(g) H = +181 kJ

Which of the following energy level diagrams represent the above reaction?

A B

C D

Energy

Reactants

X kJ mol-1

Y kJ mol-1

Products

2NO(g)

Energy

Energy

N2(g) + O

2(g)

2NO(g)

H = +181 kJ

N2(g) + O

2(g)

2NO(g)

H = +181 kJ

Energy

2NO(g)

N2(g) + O

2(g)

H = +181 kJ

Energy

N2(g) + O

2(g)

H = +181 kJ


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6 A pupil carried out an experiment to determine the value of heat of neutralization.

Diagram 6 shows the set up of the apparatus used in the experiment.

The following data was obtained;

a) Why was a polystyrene cup used in this experiment?

..[1 mark

(b) Given that the specific heat capacity of the solution is 4.2 Jg-1oC-1 and the

density of the solution is 1.0 gcm-3.

(i) Calculate the change of heat in the experiment.

(ii) Calculate the heat of displacement in the experiment.

[3 marks

1

Initial temperature of hydrochloric acid = 28oCInitial temperature of sodium hydroxide solution = 28oC

Highest temperature of the mixture of product = 41oC

100 cm3of 2.0 mol dm-3

Sodium hydroxide solution

100 cm3of 2.0 moldm-3

hydrochloric acid

Thermometer

DIAGRAM 6

Polystyrene cup


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(c) Draw the energy level diagram for the reaction.

[2 marks

(d) Based on the experiment, what is meant by theheat of neutralisation?

.......

[1 mark

(e) The pupil repeats the experiment by replacing hydrochloric acid with

ethanoic acid. All the other conditions remain unchanged.

(i) Predict the value of the heat of neutralisation?

.........

[1 mark]

(ii) Explain why?

........

....

........

....................................................................................................................

[2 marks

1


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Thermochemistry2.0 Heat of precipitation

- precipitate is unsoluble salt

- precipitate must be prepared through double bond

decomposition orprecipitation method

Do you still remember what is meant by double bond

decomposition?

[please refer to salts notes]

General equation double bond decomposition/precipitation;

Ionic equation for precipitation reaction.

1

The heat of precipitation is the heat changewhen one mole of a precipitate is formed from

their ions in aqueous solution

- two aqueous solution of a substances/soluble salt was mix

together.

- one of the solution contain cation, while another onecontain anion forinsoluble salt/precipitate that need

to be prepared.

- both of the aqueous solution exchanging theirions to

produce 2 substances, which is unsoluble salt (precipitate)

and one soluble salt

MX (aq) + NY (aq) MY (s) + NX (aq)

M+ (aq) + Y- (aq) MY (s)


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Salt Solubility in waterLi+, Na+, K+, NH4

+ All salt dissolve in water

Nitrate, NO3- All nitrate salt dissolve in water

ChlorideAll chloride salt dissolve dissolve in water except;PbCl2 - lead(II) chloride (dissolve in hot water)

AgCl - argentums/silver chloride

HgCl - hydroargentum chloride, mercury chloride

SulphateAll sulphate salt dissolve in water except;

PbSO4 , BaSO4 , CaSO4

CarbonateAll carbonate salt not dissolvein water except;

Li2CO3 Na2CO3 , K2CO3 , (NH4)2CO3

Oxide All oxide not dissolve in water except;Na2O , K2O , CaO

HydroxideAll hydroxide not dissolve in water except;

NaOH, KOH, Ca(OH)2 , Ba(OH)2

Formula to determine the heat change;

Heat released/absorbed, H = mc [unit = J or kJ]

Symbo

lDescription Unit

m mass of solution 1cm3 = 1 g

c specific heat capacity of solution 4.2 J g-1oC-1

temperature change oC

1


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To determine precipitation heat of silver chloride, AgCl

1

Method to determine the heat of precipitation


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In this experiment you must have the following data;

Data tabulation

Thermometer

Polystyrene

cup

Procedure

25 cm3 sodium chloride solution 0.5 mol dm-3 is measured with

measuring cylinder50ml, and poured intopolystyrene cup, recordthe temperature with termometer (0-110)oC.

25 cm3 silver nitrate solution 0.5 mol dm-3 is measured with

measuring cylinder 50ml, and poured into another polystyrene

cup, record the temperature with termometer (0-110)oC.

Sodium chloride solution is added to silver nitrate solution

quickly.

The reacting mixture is stirred using thermometer.Highest temperature obtained is recorded.

Repeat all the step by using different substance.

Precaution steps;

Use polystyrene cup. (polystyrene cup is insulator, to avoid loss of

heat)

Stir the mixture.

25 cm3 sodium

chloride solution

0.5 mol dm-3

Thermometer

Polystyrene

cup

25 cm3 silver

nitrate solution

0.5 mol dm-3

1


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Initial temperature of sodium chloride

NaCl /oCx oC

Initial temperature of silver nitrate,

AgNO3 /oC

y oC

Average initial temperature for bothsolution

Highest temperature for the solution z oC

Temperature changez (x + y) oC = oC

2

Chemical equation for the reaction;

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Ionic equation for the reaction;

Ag+ (aq) + Cl- (aq) AgCl (s)Calculation of heat of precipitation for AgCl;

1. Calculate the number of mole of precipitate formed

No. of mol NaCl = = = 0.0125 mol

No. of mol AgNO3 = = = 0.0125 mol

FBCE;

No. of mol AgCl = 0.0125 mol

1

(x + y) oC

2

MV

1000

0.5 X 25

1000

MV

1000

0.5 X 25

1000

Formula of mole


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2. Calculate the heat released/given out

[From the experiment]

Total volume of the mixture = 25 cm3 AgNO3 + 25 cm3 NaCl

= 50 cm3

Mass of solution = 50 g

Temperature change = oC

Heat given out, H =

= 50 4.2 J

= kJ

Therefore, heat given out during the experiment is kJ

3. Calculate the heat of precipitation

0.0125 mol of AgCl produces kJ

Therefore;

1 mol of AgCl produces = kJ mol-1

= kJ mol-1

H = kJ mol-1

Thus;

The heat of precipitation of silver chloride, AgCl;

1

50 4.2 1000

1 .

0.0125

m = mass of solution( 1cm3 = 1 g)

c = specific heat

capacity of solution

(4.2 J g-1oC-1)

= temperature change

oC

Heat given out

No. of mole

50 4.2

1000

50 4.2 12.5

50 4.2

1000

mc

1st

formula

2nd formula

50 4.2

1000


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H = kJ mol-1

Example 1: Precipitation forlead(II) sulphate

Chemical equation;

Pb(NO3)2 + K2SO4 PbSO4 + 2KNO3

H = -50 kJmol-1

Ionic equation;

Pb2+ + SO42- PbSO4 H = -50 kJmol

-1

50 kJ heat released when 1 mol of lead(II) ions react with 1 mol of

sulphate ions to form 1 mol precipitate of lead(II) sulphate.

The heat of precipitation for PbSO4 = 50 kJmol-1

Example 2: Precipitation forsilver chloride

Energy

Pb2+ + SO4

2-

PbSO4

H = - 50 kJmol-1


1

50 4.2

12.5


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Chemical equation ;

AgNO3 + KCl AgCl + KNO3

H = 65.5 kJmol-1

Ionic equation;

Ag+ + Cl- AgCl H = 65.5 kJmol-1

65.5 kJ heat released when 1 mol of silver ions react with 1 mol of

chloride ions to form 1 mol precipitate of silver chloride.

The heat of precipitation for AgCl = 65.5 kJmol-1

Example 3: Precipitation forcopper(II) hydroxide

Energy

Ag+ + Cl-

AgCl

H = 65.5 kJmol-1


1


20/75

Chemical equation;

CuSO4 + 2NaOH Cu(OH)2 + Na2SO4

H = -60 kJmol-1Ionic equation;

Cu2+ + 2OH- Cu(OH)2 H = -60 kJmol-1

60 kJ heat released when 1 mol of copper(II) ion react with 2 mol of

hydroxide ion to form 1 mol precipitate of copper(II) hydroxide.

The heat of precipitation for Cu(OH)2 = 60 kJmol-1

Calculation for heat of precipitate

Example 1

Energy

Cu2+

+ 2OH-

Cu(OH)2

H = 60 kJmol-1


2


21/75

When 50 cm3 calcium nitrate solution, Ca(NO3)2 2 mol dm-3

is added to 50 cm3 sodium carbonate solution, Na2CO3

2.0 mol dm-3,precipitate of calcium carbonate, CaCO3 is

produce.Temperature of the mixture solution decrease 3.0 oC.

Calculate the heat of precipitation of calcium carbonate.

[Specific heat capacity of solution: 4.2 J g-1oC-1.

Density of solution: 1 g cm-3]

Chemical equation;

Ca(NO3)2 + Na2CO3 CaCO3 + 2NaNO3

Ionic equation:

Ca2+ + CO3 CaCO3

Step 1: Calculate the number of mole of precipitate formed

No. of mol Ca(NO3)2 = = =

No. of mol Na2CO3 = = =

FBCE;

1 mol Ca2+ react with 1 mol CO32- , produce 1 mol calcium carbonat

Therefore;

0.1 mol Ca

2+

react with 0.1 mol CO32-

, produce 0.1 mol CaCO3Thus;

No. of mol CaCO3 = 0.1 mol

Step 2 : Calculate the total heat absorb in exp.

[From the experiment]2

MV

1000

2 X 50

10000.1 mol

MV

1000

2 X 50

10000.1 mol


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Total volume of the mixture = 50 cm3Ca(NO3)2 + 50 cm3Na2CO3

= 100 cm3


Temperature change = 3o

C

Heat absorbs, H =

= 100 4.2 3 J

= 1260 J

Therefore, heat given out during the experiment is 1.26 kJ

Step 3 : Calculate the heat of precipitstion

0.1 mol CaCO3absorb at 1260 J of heat.

Therefore;

1 mol CaCO3absorb heat = J mol-1

= 12600 J mol-1

= 12.6 kJ mol-1Thus

The heat of precipitation CaCO3, H = + 12.6 kJ mol-1

Step 4 : Draw energy level diagram

Example 2:

Energy

Ca2+ + CO3

2-

CaCO3

H = +12.6 kJ mol-1

2

1 kJ = 1000 J

1260

0.1

si n (+) must write

m = mass of solution

( 1cm3 = 1 g)

c = specific heat

capacity of solution(4.2 J g-1oC-1)

= temperature change

oC

mc

1st formula


23/75

Thermochemical equation for precipitate of magnesium carbonate,

Mg(NO3)2.

Mg(NO3)2 (ak) + Na2CO3 MgCO3 (p) + 2NaNO3 (ak)

H = +25 kJmol-1

Calculate the changes of temperature when 50 cm3 magnesium nitrate

solution , Mg(NO3)2 2.0 mol dm-3 is added to 50 cm3 sodium carbonate

solution, Na2CO3 2.0 mol dm-3/

Chemical equation has been given the question.

Ionic equation: Mg2+ + CO32- MgCO3

Step 1: Calculate the number of mole of precipitate formed

No. of mol Mg(NO3)2 = = =

No. of molNa2CO3 = = =

FBCE;

1 mol Mg2+ react with 1 mol CO32-

, to produce 0.1 mol MgCO3

Therefore;

0.1 mol Mg2+

react with 0.1 mol CO32-

, to produce 0.1 mol MgCO3Thus;

No of mole MgCO3 = 0.1 mol

Step 2 : Calculate the total heat absorb in exp.2

MV

1000

2.0 X 50

10000.1 mol

MV

1000

2.0 X 50

10000.1 mol


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Heat absorb = mc

(endothermic) = (50 + 50) x 4.2 x J

= 100 x 4.2 x J

= 420 J

[from chemical equation, (H = +25 kJ)]

The heat of precipitation for magnesium carbonate is +25 kJ,

Therefore;

1 mol precipitate of MgCO3 absorbs heat 25 kJ,

Thus;

0.1 mol precipitate MgCO3 absorb heat,

= 0.1 x 25 kJ,

= 2.5 kJ

= 2.5 x 1000 J

= 2500 J

Thus;

The heat absorbs in experiment is 2500 J

Step 3 : determine the value of , temperature change in exp.

In experiment,

Total heat absorb = mc

420 J = 2500 J

=

= 5.95 oC

The reaction is endothermic, temperature decrease 5.95 oC

2

m = solution mass

( 1cm3 = 1 g)

c = specific heat

capacity

(4.2 J g-1

o

C-1

) = tem erature

1 kJ = 1000 J

Change to J, we want

substitute it into formula

2500 J420 J


25/75

Energy level diagram of precipitate ofMgCO3

3. thermochemical equation for precipitate of copper(II)hydroxide,Cu(OH)2 given below,

Cu2+ (ak) + 2OH- (ak) Cu(OH)2 (p) H = -60 kJ

How many its volume solution for copper(II)sulphate, CuSO4 1.0

mol dm-3 that need to mixture with 50 cm-3 sodium hydroxide

solution, NaOH 2.0 mol dm-3 to increase temperature of the solution

mixture to 6.3oC?

Step 1: write the chemical equation for this reaction

Chemical equation has been state at the question.

Step 2 : calculate no. of mol for the substance

No. of mol CuSO4/Cu2+

= = =

No. of molNaOH /OH- = = =

Energy

Mg2+ + CO3

2-

MgCO3

H = +25 kJ mol-1

2

MV

1000

1.0 X V

1000

0.001 V mol

MV

1000

2.0 X 50

10000.1 mol


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Step 3 : ratio for no. of mol

FBCE;

2 mol NaOH /OH- react with 1 mol CuSO4/Cu2+ , to produce1 mol

Cu(OH)2, copper(II) hydroxide,

so ;

0.1 molNaOH /OH- react with mol CuSO4/Cu2+, to produce

molCu(OH)2.

So ;

0.1 molNaOH /OH- react with 0.05 mol CuSO4/Cu2+, to produce 0.05

mol Cu(OH)2.

* carefull to calculate value of V, we must include the value of

temperature changes, (6.3 oC) from the question that give above.

Step 4 : calculate the total heat release/absorb in this exp.

Total heat released = mc(t/b exothermic) = (50 + V) x 4.2 x 6.3 J

= (50 + V) x 26.46 J

= (50 + V) 26.46 J

(V not yet known)

[from chemical equation], (H = -60 kJmol

-1

)]

Heat of precipitate for copper(II)hydroxide is -60 kJ, (from question)

1 mol precipitate of Cu(OH)2 releaseing heat 60 kJ,

So ;

0.05 mol precipitate of Cu(OH)2 releasing heat,2

= 6.3 oC

(substitute in

equation)

0.1

20.1

2

Dont calculate the volume of

CuSO4 from this value


27/75

= 0.05 x 60 kJ,

= 3 kJ

= 3 x 1000 J

= 3000 J

So ;

Total heat released by 0.05 mol precipitate of Cu(OH)2 in experiment

is 3000 J

Step 5 : determine the value ofV, solution volume of CuSO4, in exp

In experiment,

Total heat released = mc J

(50 + V) 26.46 J = 3000 J

(50 + V) =

(50 + V) = 113.38 cm3

V = (113.38 50) cm3

V = 63.38 cm3

Energy level diagram for precipitate ofCu(OH)2

energy

Cu

2+

(ak) + 2OH

-

(ak)

Cu(OH)2

(p)

H = -60 kJ mol-1

2

3000 J

26.46 J

1 kJ = 1000 J

Change to J, we want

substitute it into equation


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4. when solution of 500 cm3 M2+ 2.0 mol dm-3 is mixture with

500 cm3 solution of ion Cl- , heat of solution increase to oC.

Calculate the heat of precipitate for the reacted ion M2+ and ion Cl-

to precipitate the formula substance MCl2.

Step 1: write the chemical equation

Ionic equation

M2+ (ak) + 2Cl- (ak) MCl2 (p)


No. of mol M2+ = = =

No. of mol Cl- = = =


DPKYS;1 mol M2+ react with 2 mol Cl- to produce

1 mol precipitate MCl2

so;

1 mol M2+ reacted with 2 mol Cl- to produce

1 mol precipitate MCl2


Total heat released = mc

(t/b exothermic) = (500 + 500) x 4.2 x J

= 1000 x 4.2 x J

= 4.2 x kJ

2

MV1000

2.0 X 5001000 1 mol

MV

1000

M X 500

10000.5 M mol

1 kJ = 1000 J

Dont use this no. of mol

because the actual no. of

mol we dont know yet


29/75

So ;

Total heat released by 1 mol precipitate of MCl2 in experiment is 4.2 x

kJ,

Step 5 determine the heat of precipitate of MCl2

Because the total heat released by 1 mol precipitate of

MCl2 in this experiment is 2 x kJ ,

So ;

Heat of precipitate for MCl2 is -4.2 x kJ mol-1

H= -4.2 x kJ mol-1

energy level diagram for heat of precipitate ofMCl2

5. In one experiment to determine the heat of precipitate between ion

M2+ and ion SO42- , found that when 250 cm3

of M2+ 2 mol dm3 solution is mix with 250 cm3 solution of

ion SO42- , temperature of the solution increase at 30 oC.

Calculate heat of precipitation for the reaction between

ion M2+ and ion SO42-

Energy

M2+ (ak) + Cl- (ak)

MCl2

(p)

H = -4.2 x kJ mol-1

2


30/75


Ionic equation

M2+

(ak) + SO42-

(ak) MSO4 (p)


No. of mol M2+ = = =

No. of mol SO42-= = =


DPKYS;

1 mol ofM2+ react with 1 mol SO42- to produce

1 mol precipitate ofMSO4

So ;

0.5 mol M2+ react with 0.5 mol SO42- to produce

0.5 mol precipitate ofMSO4


Total heat release = mc

(t/b exothermic) = (250 + 250) x 4.2 x 30 J= 63000 J

= 63 kJ

So ;

Total heat release by 0.5 mol precipitate of MSO4 in experiment is

63 kJ3

MV

1000

2.0 X 250

10000.5 mol

MV

1000

M X 250

1000

0.25 M mol

1 kJ = 1000 J

Dont use this no. of mol

because the actual no. of

mol we dont know yet


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Step 5 determine the heat of precipitate of MSO4In experiment;

0.5 mol precipitate of MSO4 release heat 63 kJ

So ;

1 mol precipitate of MSO4 releasing heat ;

=

= 126 kJ mol-1

So ;

Heat of precipitation of MSO4 is -126 kJ mol-1

H = -126 kJ mol-1

Energy level diagram for heat of precipitation ofMSO4

Energy

M2+ (ak) + SO4

2- (ak)

MSO4

(p)

H = -126 kJ mol-1

3

63 kJ

0.5


32/75

6. mixture between 75 cm3 hydrochloric acid solution 0.15 mol dm-3

and 75 cm3 silver nitrate solution 0.15 mol dm-3 has increase the

temperature at 1.9 oC.

i. how much the quantity of heat of energy that release in this

experiment?ii. heat of precipitate for this reaction

iii. draw energy level diagram for this reaction.


Chemical equation;

AgNO3 (ak) + HCl (ak) AgCl (p) + HNO3 (ak)

Ionic equation;

Ag+ (ak) + Cl- (ak) AgCl (p)


No. of mol HCl = = =

No. of mol AgNO3 = = =


DPKYS;

1 mol HCl react with 1 mol AgNO3 to produce

1 mol precipitate of AgCl

So ;

0.01125 mol HCl react with 0.01125 mol AgNO3 to produce

0.01125 mol precipitate ofAgCl

3

MV

1000

0.15 X 75

10000.01125 mol

MV

1000

0.15 X 75

1000

0.01125 mol


33/75



(t/b exothermic) = (75 + 75) x 4.2 x 1.9 J

= 1197 J

= 1.197 kJSo ;

Total heat release by 0.01125 mol precipitate of AgCl in experiment is

1.197 kJ

Step 5 determine the heat of precipitate of MSO4In experiment;

0.01125 mol precipitate of releasing heat at1.197 kJ

So ;

1 mol precipitate of AgCl releasing heat;

=

= 106.4 kJ mol-1

So ;Heat of precipitate of AgCl is -106.4 kJ mol-1

H = -106.4 kJ mol-1

Energy level diagram for heat of precipitation ofAgCl

Kamal Ariffin B Saaim

Energy

Ag+ (ak) + Cl- (ak)

AgCl (p)

H = -106.4 kJ mol-1

3

1 kJ = 1000 J

1.197 kJ mol-1

0.01125


34/75

Thermochemistry

3.0 Heat of Displacement

What is meant by displacement reaction?

Example 1;

Zn + Cu2+ Zn2+ + Cu H= 210 kJmol-1

Energy

Zn + Cu2+

Zn2+

+ Cu

When 1 mole Cu is displaced by Zn, 210 kJ heat energy

is released/given out.

H= 210 kJmol-1

3

The Heat of Diplacement is the heat change

when one mole of metal is diplaced from its solution

by a more electropositive metal.

Metal that more electropositive will displace

metal that is less electropositive from its salt solution.


35/75

Example 2;

Zn + Pb2+ Zn2+ + Pb H= 112 kJmol-1

Example 3;

Mg + Fe2+

Mg2+

+ Fe H= 80 kJmol-1

Energy

Zn + Pb2+

Zn2+ + Pb

When 1 mole Pb is displaced by Zn, 112 kJ heat energy


H= 112 kJmol-1

Energy

Mg + Fe2+

Mg2+ + Fe

When 1 mole Fe is displaced by Mg, 80 kJ heat energy


H= 80 kJmol-1

3


36/75

To determine the heat of diplacement of copper by zinc

Procedure;

50 cm3 copper(II) sulphate solution of 0.1 mol dm-3 is

measured with measuring cylinder 50ml and poured into

polystyrene cup, record the temperature withtermometer (0-110)oC.

1.0 g metal powder is weighed by using electronic balance and

quickly added into the polystyrene cup that contain copper(II)

sulphate solution.

The mixture is stirred using the thermometer.

The highest/maximum temperature of heat is recorded.

Repeat the step by using different substance. [if necessary]

Thermometer

Polystyrene

cup

50.0 cm3 copper(II)

sulphate solution

0.1 mol dm-3

Beaker that

contain

1 g zinc powder

(excess)

3

Method to determine heat o dis lacement


37/75

Data tabulation

Initial temperature of copper(II) sulphate ,

CuSO4 /oC

y

Highest/maximum temperature for the

solution /o

C

z

Temperature change /oC (z y) =

[Note : mass of zinc is used in excess]

Chemical equation for the reaction

Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4 (aq)

Ionic EquationZn + Cu2+ Cu + Zn2+

Calculation the heat of displacement of copper by zinc;

1. Calculate the number of mole of Cu formed

No of moles of Zn = . mass . = 1.0 = 0.015 mol

molar mass 65

No. of moles CuSO4 = = = 0.005 mol

(No. of moles of CuSO4) 0.005 < 0.015 (No of moles of Zinc)

[Important notes: calculation MUST based on CuSO4 solution

because quantity of zinc used in excess]

FBCE;1 mole of CuSO4 produces 1 mole of Cu

Therefore;

0.005 mole of CuSO4 produces 0.005 mole of Cu

Thus;

No. of mole of Cu formed = 0.005 mol

3

M V

1000

0.1 X 50

1000


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2. Calculate the heat released/given out

Heat release/given out = mc

(in exp.)= 50 x 4.2 x J

= kJ

3. Calculate the heat of diplacement for 1 mole copper by zinc

0.005 mole of copper diplaced produced kJ

Therefore;

When 1 mole of copper is diplaced by zinc, the heat released is

= kJ mol-1

= X kJ mol-1

= kJ mol-1

Thus;

The heat of diplacement of copper by zinc

H = kJ mol-1

3

50 x 4.2 x 1000

50 x 4.2 x

10001 .

0.005

50 x 4.2 x

5

50 x 4.2 x

1000 .

0.005

50 x 4.2 x

1000

50 x 4.2 x

5

Volume of solution


39/75

Calculation heat of displacement

Question 1

Excess iron powder is added into 50 cm3 copper(II) chloride solution,

CuCl2 1.0 mol dm-3

, brown solid is formed and blue solution change togreen.

Iron has displace copper from its salt solution. The following data is

get from above experiment.

Initial temperature for copper(II)chloride solution = 28.0 oC

Highest temperature for mixture solution = 57.0 oC

Calculate heat changes when 1 mol of copper is displace by iron.

Solution

chemical equation;

Fe (s) + CuCl2 (aq) FeCl2 (aq) + Cu (s)

Ionic equation;Fe (s) + Cu2+ (aq) Cu (s) + Fe+2 (aq)

Step 1 : Calculate the number of mole of Cu formed

No. of mol CuCl2 = = =

No. of mol Fe = (no need to calculate because is in excess)

FBCE;


Step 2 : Calculate the heat released/given out3

MV

1000

1 X 50

10000.05 mol


40/75

Heat released/given out = mc

(exothermic) = 50 x 4.2 x J

Temperature change, = (highest temperature initial temperature)

= (57.0 28.0)o

C= 29.0 oC


= 6090 J

= 6.09 kJ

Step 3 : determine the heat of displacement of Cu by Fe

Displacement of0.05 mol Cu releasing 6.09 kJ of heat.

Therefore;

When 1 mole of copper is diplaced by zinc, the heat released is

= kJ mol-1

= 121.8 kJ mol-1Thus;

The heat of displacement of Cuby iron;

H = 121.8 kJmol-1

Draw energy level diagram

Question2

Energy

Fe + Cu2+

Fe2+ + Cu

H = -121.8 kJ mol-1

4

(determine the changes

of temperature)

1 kJ = 1000 J

6.09

0.05


41/75

Study the following equation;

Fe (s) + CuSO4 (aq) Cu (s) + FeSO4 (aq)

H = -250 kJ mol-1

If excessive iron powder is add into 100 cm3

copper(II) sulphatesolution 0.25 mol dm-3, calculate

i. Heat released

ii. Temperature rise

iii. Mass of copper that is displace

iv. Mass of salt that formed if it crystalize

v. Draw energy level diagram

If magnesium powder is use to replace iron powder, is it the energy

that release is more higher, same or lower.

[Ar = Cu, 64; Fe, 56; S, 32; O, 16]

Solution

Chemical equation;

Fe (p) + CuSO4 (ak) FeSO4 (ak) + Cu (p)

Ionic equation;

Fe (p) + Cu2+ (ak) Fe+2 (ak) + Cu (p)

i : calculate no. of mole of Cu formed

No. of mol CuSO4 = = =

No. of mol Fe = (no need to calculate because excessive)

4

MV

1000

0.25 X 100

1000

0.025 mol


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FBCE;

1 mol CuSO4 produced 1 mol Cu

Therefore;

0.025 mol CuSO4 produce 0.025 mol Cu

Thus;No. of mole of Cu formed = 0.025 mol

calculate the heat released by 0.025 mol of Cu in the exp.

[from question: H = -250 kJ mol-1]

Thats mean;

When 1 mole Cu is displaced by Fe, 250 kJ heat energy is

released/given out.

Therefore;

0.025 mol Cu is diplaced by Fe, heat released is;

= 0.025 250 kJ

= 6.25 kJ

= 6250 J

Thus;

Heat released/given out = 6.25 kJ or 6250 J

ii: determine the temperature changes during the reaction

[Heat released/given out = 6250 J]

Heat released/given out = mc6250 = 100 x 4.2 x

= 14.9 oC

Therefore;

The increases in temperature = 14.9 oC

iii: determine the mass of copper that is diplace4

Change to unit of kJ, because we

want to find


43/75

[from the previous information above]


No. of mol Cu = mass of Cu

ArCu

Mass of Cu = 0.025 64

= 1.6 g

iv: determine the mass of salt formed

FBCE;

1 mol CuSO4 produced 1 mol FeSO4Therefore;

0.025 mol CuSO4 produce 0.025 mol FeSO4Thus;

No. of mole of FeSO4 formed = 0.025 mol

No. of mol FeSO4 = mass of FeSO4MrFeSO4

Mass of FeSO4 = 0.025 [56 + 32 + 4(16)]

= 3.8 g

v : draw energy level diagram

Question 3

Energy

Fe + Cu2+

Fe2+ + Cu

H = -250 kJ mol-1

4


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100 cm3copper(II) nitrate solution 0.2 mol dm-3 is poured into plastic

container. Temperature is recorded. Then excess magnesium powder

is added to the solution. The mixture is stirred and the temperature

rises is recorded. The temperature shows an increases of 5 oC.

(a) What is the colour of the solution in the plastic container;

i. before magnesium powder is place?

ii. after magnesium powder is place?

(b) Write the total ionic equation of the reaction.

(c) How many mole of copper(II) nitrate reacts?

(d) Calculate the heat releases in this experiment.

(e) Calculate the heat energy release when one mol of copper is

formed.

(f) What is the heat of displacement of copper?

(g) Draw energy level diagram for this experiment.

(h) Why magnesium used is in form of fine powder not granulated?

(i) i. if potassium hydroxide solution is mix with the solution in

plastic beaker in the end of the experiment, what can you

observe?

ii. write the chemical equation for the reaction in (i)(i).iii. write the ionic equation for the reaction in (i)(i).

SOLUTION4


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(a) i. blue

ii. colourless

(b) Mg (aq) + Cu2+ (aq) Mg2+ (aq) + Cu (s)

(c) no. of mol Cu(NO3)2 = =

=

(d) Heat release = mc

(during reaction) = 100 x 4.2 x 5.0 J

= 2100 J

= 2.1 kJ

(e) FBCE;

1 mol CuSO4 produced 1 mol Cu

Therefore;

0.02 mol CuSO4 produce 0.02 mol Cu

Thus;


When 0.02 mol of Cu diplaced by Mg, 2.1 kJ of heat released.

Therefore;

1 mol of Cu diplaced by Mg will releases heat;

=

= 105 kJ mol

-1

(f) The heat of displacement of copper = -105 kJ mol-1

Heat change, H = -105 kJ mol-1

(g) Energy level diagram4

MV

1000

0.2 x 100

1000

0.02 mol

2.1 kJ mol-1

0.02


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(h) To increase the total surface area per volume for magnesium,

thus it will increase the rate of reaction.

(i) i. white precipitate formed

ii. Mg(NO3)2 (aq) + 2KOH (aq) Mg(OH)2 (s) + 2KNO3 (aq)

iii. Mg2+ (aq) + 2OH- (aq) Mg(OH)2 (s)

Learning task: pg 158 no. 1 & 2

Effective Practise: pg. 158 no. 3


SMKDBL

http://kemhawk.webs.com/

Energy

Mg (s) + Cu2+ (aq)

Mg2+ (aq) + Cu (s)

H = - 105 kJ mol-1

4


47/75

Thermochemistry

4.0 Heat of neutralisation

What is meant by neutralization?

Ionic equation:

Heat of neutralization can be divide into two types;

i. Reaction between strong acid with strong alkali.

ii. Reaction between weak acid with weak alkali.

4

The heat of neutralization is

the heat change when one mole of water is formed

from the reaction between an acid and an alkali.

H+ (aq) + OH- (aq) H2O (l)

Acid reacts with alkali/oxide base to produce salt and water


48/75

Heat Of Neutralization Between Strong Acid With

Strong Alkali

Example of reaction;

Neutralization reactionNeutralization heat,

H (kJ mol-1)

HCl + NaOH NaCl + H2O -57.3

HCl + KOH KCl + H2O -57.3

HNO3 + KOH KNO3 + H2O -57.3

HNO3 + NaOH NaNO3 + H2O -57.3

Therefore;

Energy level diagram,

57.3 kJ heat is released when 1 mol of water is produced from

a reaction between 1 mol of hydrogen ion, H+and1 mol of

hydroxide ion, OH- .

Energy

H+ (aq) + OH- (aq)

H2O (l)

H = - 57.3 kJmol-1

4

In neutralization reaction between strong acid and

strong alkali, the heat of neutralization is -57.3 kJmol

-1


49/75

1. Study this reaction

What is the value of heat of neutralization for the above

reaction?

Ionic equation:

H+ (aq) + OH- (aq) H2O (l) H = -57.3 kJ

- 57.3 kJ heat is released when 1 mole of water is produced

from a reaction between 1 mole of hydrogen ion, H+

and1 mole of hydroxide ion, OH- .

- Thus,the heat of neutralizationis -57.3 kJmol-1

Remember: HCl is monoprotic acid

[Monoprotic acid: When 1 mole of HCl acid dissolves in water,

it produces 1 mol of hydrogen ion]


Energy

H+ (aq) + OH- (aq)

H2O (l)

H = - 57.3 kJmol-1

4

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

H = -57.3 kJ


50/75

2. Study this reaction

What is the value of heat of neutralization for the above

reaction?

- H2SO4is diprotic acid. It produces 2 moles of hydrogen

ions when it dissolves in water.

H2SO4 (aq) 2H+ (aq) + SO4

2- (aq)

- 2 moles of H+ ions produce 2 moles of water when reacted

with alkali.

Ionic equation:

2H+ (aq) + 2OH- (aq) 2H2O (l) H = -114.6 kJ

- 114 kJ heat is released when 2 moles of water is produced

from a reaction between 2 moles of hydrogen ion, H+ and

2 mol of hydroxide ion, OH- .

- Thus,the heat of neutralizationis still -57.3 kJmol-1

5

H2SO4 (ak) + 2NaOH (ak) Na2SO4 (ak) + 2H2O (ce)

H = -114 kJ


51/75


Energy

H+ (aq) + OH- (aq)

H2O (l)

H = - 12 kJmol-1

5


52/75

Heat of neutralization between weak acid with strong

alkali

Example of reaction;

Neutralization reaction

Heat of

neutralization,

H (kJ mol-1)

CH3COOH + NaOH CH3COONa + H2O -55

HCN + KOH KCN + H2O -12

Why ?

Study the ionization equation for this weak acid;

- Weak acid only ionize partially in water. Only small amount of

hydrogen ion is produce.

- Most of the molecules of weak acid still exist as a molecule.- Therefore, some of the heat energy is use to overcome the

molecular bond in acid, so that in can be ionize in water.

- Thus, this process cause the value of heat of neutralization,

less than 57 kJmol-1

5

In neutralization between weak acid and strong alkali,

heat of neutralization is less than -57.3 kJmol-1

CH3COOH (aq) CH3COO- (aq) + H+ (aq)

Reversible rocess


53/75

To determine the heat of neutralizationbetween HCl and NaOH

Thermometer

Polystyrene

cup

Procedure ;

50 cm3 hydrochloric acid solution 2.0 mol dm-3 is measured

with measuring cylinder 50ml and pour into polystyrene cup and

the temperature is recorded using thermometer (0-110)oC.50 cm3 potassium hydroxide solution 2.0 mol dm-3 is measured

and pour into anotherpolystyrene cup and the temperature is

recorded.

Quicklythe hydrochloric acid solution is added to the

potassium hydroxide solution,the mixture is stirred using

thermometer.

highest/maximum temperature of the mixture is recorded.Repeatthe step by using different substance.

50 cm3

hidrochloric acid

2.0 mol dm-3

Thermometer

Polystyrene

cup

50 cm3 potassium

hydroxide solution

2.0 mol dm-3

5

Method to determie the heat of neutralization


54/75

Data tabulation

Initial temperature for hydrochloric acid,

HCl /oCx

Initial temperature for potassium hydroxide,NaOH /oC

y

Average temperature for the both

solution /oC

Highest/maximum temperature for the

solution /oCz

Temperature changes /

o

C z - =

Calculation heat of neutralization HCl and NaOH

Chemical equation;

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

Ionic equation;

H+ (aq) + OH (aq) H2O (l)

Calculation the heat of neutralization;

1. Calculate the number of mole of water formed

No. of mol hydrogen ions, H+ / HCl ;

= = = 0.1 mol

No. of mol of hydroxide ions, OH- / NaOH

= = = 0.1 mol

5

(x + y)

2

(x + y)2

MV

1000

2.0 X 50

1000

MV

1000

2.0 X 50

1000


55/75

FBCE;

1 mole of H+ reacts with 1 mole of OH- to form 1 mole of H2O

Therefore;

0.1 mole of H+ reacts with 0.1 mole of OH- to form 0.1 mole of H2O

Thus;No. of mole of H2O formed = 0.1 mol

2. Calculate the heat given out/releases during reaction;


Total volume of the mixture = 50 cm3 HCl + 50 cm3 NaOH

= 100 cm3


Temperature change = oC

Heat given out, H =

= 100 4.2 J

= kJ

Therefore, heat given out during the experiment is;

= kJ

5

100 4.2

1000

mc

100 4.2

1000


56/75

3. Calculate the heat of neutralization

0.01 mol of H2O produces kJ

Therefore;1 mol of H2O produces = kJ mol

-1

= kJ mol-1

= kJ mol-1

Thus;

The heat of neutralization;

H = kJ mol-1

H = - 4.2 x kJ mol-1


Energy

H+ (aq) + OH- (aq)

H2O (l)

H = - 4.2 x kJ mol-1

5

1 .

0.1

Heat given outNo. of mole

100 4.2

1000

100 4.2 100

100 4.2

1000

100 4.2

100


57/75

Question 1

When 50 cm3 potassium hydroxide solution, NaOH 1.0 mol dm-3 is

added with 50 cm3 hydrochloric acid HCl 1.0 mol dm-3, the temperature

of mixture increase 6.2 oC.

Calculate the heat of neutralization for this reaction.

Solution

Write the chemical equation for the reaction

Chemical equation;

NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)

Ionic equation;

H+(aq) + OH- (aq) H2O (l)

Calculation the heat of neutralization;

1. Calculate the number of mole of water produced

No. of mol H+ / HCl = = =

No. of mol OH- / NaOH = = =

FBCE;


Therefore;

0.05 mole of H+ reacts with 0.05 mole of OH- to form

0.05 mole of H2OThus;

No. of mole of H2O formed = 0.05 mol

5

MV

1000

1.0 X 50

10000.05 mol

MV

1000

1.0 X 50

10000.05 mol


58/75


m = (50 + 50) cm3 = 6.2 oC

= 100 g

Heat released/given out = mc(exothermic reaction) = (50+50) x 4.2 x 6.2 J

= 2604 J

= 2.604 kJ


0.05 mol of H2O formed produces 2.640 kJ

Therefore;

1 mol of H2O produces = kJ mol-1

= kJ mol-1

= 52.08 kJ mol-1

Thus;

The heat of neutralization;

H = 52.08 kJ mol-1


Energy

H++ OH-

H2O

H = -52.08 kJ mol-1

5

2.604

0.05

1 kJ = 1000 J

Heat given outNo. of mole


59/75

Question 2

When 50 cm3 potassium hydroxide solution, NaOH 1.0 mol dm-3 is

added into 50 cm3 sulphuric acid, H2SO4 0.5 mol dm-3, the

temperature of solution mixture has increase 6.2o

C?

(a) calculate the heat releases for the following reaction;

i. H2SO4 (ak) + NaOH (ak) Na2SO4 (ak) + H2O (ce)

ii. H2SO4 (ak) + 2NaOH (ak) Na2SO4 (ak) + 2H2O (ce)

SOLUTIONCalculate no. of mol of water formed

No. of mole H2SO4 = = =

Therefore;

No. of mole of H+ = 2 0.025 = 0.05 mol

No. of mole of OH- / NaOH = = =

FBCE;

1 mole of H2SO4 reacts with 2 mole of NaOH to form 2 mole of H2O

[Remember: H2SO4 is diprotic acid]


5

MV

1000

0.5 X 50

10000.025 mol

MV

1000

1.0 X 50

10000.05 mol


60/75

Therefore;

(0.025 2) mole of H+ reacts with 0.05 mole of OH- to form

0.05 mole of H2O

Thus;

No. of mole of H2O formed = 0.05 mol

Calculate the total heat release during the reaction;

Heat released/given out = mc

= (50+50) x 4.2 x 6.2 J

= 2604 J

= 2.604 kJ

Calculate the heat of neutralization

0.05 mol of H2O formed produces 2.640 kJ

Therefore;

1 mol of H2O produces = kJ mol-1

= 52.08 kJ mol-1

Thus;

The heat of neutralization;H = 52.08 kJ mol-1

Experiment (i) : H2SO4 + NaOH Na2SO4 + H2O

Ionic equation: H+ + OH- H2O [1 mole of water]

Therefore;

Heat changes in this reaction is 52.08 kJ

Experiment (ii) H2SO4 + 2NaOH Na2SO4 + 2H2O

Ionic equation: 2H+ + 2OH- 2H2O [2 moles of water]

Therefore;

Heat changes in this reaction is 2 52.08 kJ = 104.16 kJ6

2.604

0.05


61/75

Question 3

HCl (ak) + NaOH (ak) NaCl (ak) + H2O (ce)

H = -57 kJmol-1

Based on the thermochemistry equation above, answer the followingequation;

a) Write the ionic equation for the reaction above

b) When 100 cm3 excessive potassium hydroxide solution is mix

with 100 cm3 hydrochloric acid, solution temperature increase

13.6oC. calculate the concentration of hydrochloric acid use.

Solution

Ionic equation : H+ (ak) + OH- (ak) H2O (ce)

Heat release during reaction = mc

= (100 + 100) x 4.2 x 13.6 J

= 11424 J

= 11.424 kJ

[Heat changes, H = -57 kJmol-1]57 kJ heat releases by 1 mol water

1kJ heat releases by mol water

Therefore;

11.424 kJ release x 11.424 mol air,

= 0.2 mol H2O

Thus;

No. of mol of water = 0.2 mol

6

1 .

57

1 .

57


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FBCE;

1 mol water is produce by 1 mol HCl

Therefore;

0.2 mol water produce by 0.2 mol HCl

Calculate the concentration of HCl

No. of mole HCl = =

0.2 mol =

M =

Concentration of HCl = 2.0 mol dm-3

Learning Task 4.6 no. 1, 2, 3 pg. 164

Effective Practise no. 1, 2, 3

Kamal Ariffin Bin Saaim

SMKDBL

http://kemhawk.webs.com/

6

MV

1000M x 100

1000

M x 100

1000

0.2 x 1000

100


63/75

Thermochemistry

5.0 Heat of combustion

Heat of combustion of various alcohol

Name

Molecular

formula

(CnH2n+1OH)

No. of carbon

atom per alcohol

molecule

Mass of

molecula

r relative

Heat

combustion

H (kJ mol-1)

Methanol CH3OH 1 32 -725

Ethanol C2H5OH 2 46 -1376

Propan-1-ol C3H7OH 3 60 -2015

Butan-1-ol C4H9OH 4 74 -2676

What inferens can be made from the table above?

- If the number of carbon atom per molecule of alcohol is

higher, the heat of combustion also increases.

- The difference in heat of combustion between alcoholmember is almost the same because each alcohol member

difference is in one group of CH2

6

The heat of combustion is the heat change when

1 molof substance is completely burnt in oxygen

under standard conditions.


64/75

Example of reaction of heat combustion in substance ;

i. Methanol,

CH3OH (ce) + O2 (g) CO2 (g) + 2H2O (ce)

H = -725 kJmol-1

ii

Ethanol,

C2H5OH (ce) + 3O2 (g) 2CO2 (g) + 3H2O (ce)

H = -1376 kJmol-1

iii. Propan-1-ol,

C3H7OH (ce) + O2 (g) 3CO2 (g) + 4H2O (ce)H = -2015 kJmol-1

iv. Butan-1-ol,


H = -2676 kJmol-1

6

3

2

9

2

When 1 mole of methanol burnt, 725 kJ of heat is releases.

Method to determine the heat o combustion o uel


65/75

To determine the heat of combustion of methanol

250 cm3

water

Thermometer

Asbestos

screen

Copper

container

Tripod stand

Spirit lampMethanol

Wooden block

Procedure;

250 cm3 of wateris measured with measuring cylinder 100ml and

poured into copper container, the temperature of water isrecorded with thermometer (0-110)oC.

A spirit lamp is filled with methanol until half full.

A spirit lamp and are weighed with electrical balanced. and the mass is

recorded.

The spirit lamp is placed under copper container, and the wick is lighted.

The water is stirred.

When the temperature of water is increases 30 oC, the spirit lamp is

distinguished.

Spirit lamp is weighed immediately, and the mass is recorded.

The experiment is repeatedusing different alcohol.

6


66/75

Precautions steps;

- Use copper container or any suitable metal.

(metal is condustor, all heat from combustion of fuel was absorb

by water)

- Spirit flame is placed on a wooden block, so the flame contactdirectly the copper container.

(A bigger area of the flame can be in contact with the copper

container)

- Wire gauze is not used.

(to prevent wire gauze absorb heat energy)

- Asbestos screen is placed around the copper container.

(to avoid heat loss to surrounding)

- The water must always be stirred.

(temperature changes is uniform)

In this experiment, you need to get the following data;

Data tabulation;

Initial temperature of water /oC x

Highest temperature of water /oC y

Temperature rise /oC 30

Initial mass of metanol + spirit lamp /g a

Final mass of methanol + spirit lamp /g b

Mass of methanol burnt /g (a b) g

Calculation of heat of combustion

Chemical equation

6


67/75

CH3OH + O2 CO2 + 2H2O

No ionic equation!

1. Calculate the number of mole of methanol (fuel used)

Mass of methanol = (a b) g

Molar mass of methanol = 12 + 3(1) + 16 + 1

= 32 g mol-1

No. of mole of methanol = = = c mol



Heat release during reaction = mc

= 200 x 4.2 x 30 J

= kJ


c mol of methanol produces heat

Therefore, heat release by burnt of one mol of methanol,

= kJ mol-1

Thus;6

massMolar mass

(a - b) .32

100 x 4.2 x 30

1000

3

2

Volume of water

100 x 4.2 x 30

1000 .

c

100 x 4.2 x 30

1000


68/75

The heat of combustion of methanol;

H = - [ ] kJmol-1

Very easylah!

Fuel Value

Fuel value can be use to compare the energy cost..

Name

Relative

molecula

r mass

Heat of

combustion,

H (kJ mol-1)

Mass of 1

mol (g)

Burnt substance

(kJ g-1)

Metanol 32 -725 32 725/32 = 22.66

Etanol 46 -1376 46 1376/46 = 29.91

Propan-1-

ol60 -2015 60 2015/60 = 33.58

Butan-1-ol 74 -2676 74 2676/74 = 36.16

How much the price of 1 g for each burnt substance above?

Calculation for heat of combustion

Question 1

6

The amount of heat energy give out when 1 g of the fuel is

completely burnt in excess of oxygen.

100 x 4.2 x 30

1000 x c


69/75

Heat from combustion of 0.28 g oktane, C8H18 increasing temperature

of 200 cm3 water from 30 oC to 46 oC. Based on this information,

[ Ar: H, 1; C, 12; O, 16; 1 mole gas is occupy 24 dm3]

i. calculate the heat of combustion for octane

ii. value of fuel for octaneiii. volume of oxygen that need

iv. mass of water produce

Solution

Chemical equation;

C8H18 (ce) + O2 (g) 8CO2 (g) + 9H2O (ce)

Calculate no. of mol of fuel

No. of mol for octane = = = 0.0025 mol

Calculate the heat changes/total heat release in this experiment

Temperature changes = (highest temp initial temp) oC= ( 46 30) oC

= 16 oC

Heat release = mc

(exothermic react.) = 200 x 4.2 x 16 J

= 13440 J

= 13.44 kJ

Calculate the heat of combustion

Combustion 0.0025 of mol of octane releasing 13.44 kJ of heat.6

mass .

molar mass

0.28

114

m = volume of water

( 1cm3 = 1 g)

1 kJ = 1000 J

25

2


70/75

Therefore;

Combustion of 1 mol octane releases;

= kJ mol-1

= 5376 kJ mol-1

Thus;

Heat of combustion for octane = - 5376 kJ mol-1

H = - 5376 kJ mol-1

Caculate the fuel value of octane

Mr oktane, C8H18 = (8 x 12) + 18 = 114 gmol-1

Therefore;

1 mole of octane, C8H18 = 114 g

Thus;

Fuel value of octane =

= kJ g-1

= 47.15 kJ g-1

Calculate the volume of oxygen

FBCE;7

13.44

0.0025

Heat of combustion for octane

Mass of one mol of octane

5376

114Make sure the

unit is correct

25

2


71/75

1 mol octane need mol oxygen to react completely

Therefore;

0.0025 mol octane need x 0.0025 mol oxygen;

= 0.0312 mol oxygen

The number of O2 used in the reaction = 0.0312 mol oxygen

Thus;

The volume of oxygen needed = 0.0312 x 24 dm3

= 0.7488 dm3

= 748.8 cm3

Calculate the mass of water produce

FBCE;

1 mole of oktane react completely to produce 9 mole of water

Therefore;

0.0025 mol oktane produce 0.0025 x 9 mol water;

= 0.0225 mol water

The number of mole of water = 0.0225 mol

Thus;

No. of mol for water =

0.0225 mol =7

25

2

Mass of water

Mr of water

Mass of water

(2 x 1) + 16

Make sure the

unit is correct


72/75

Mass of water = 0.0225 x 18 g

= 0.405 g

Question 2

Following equation show heat of combustion for ethanol, C2H5OH


H = -1400 kJmol-1

Heat combustion for 0.46 g etanol, is use to heat200 cm3 of water, calculate the temperature rise of water .

Solution

Step 1 : calculate no. of mol for ethanol that burnt

No. of mol etanol = = = 0.01 mol

Step 2 : calculate the heat changes/total heat release in this experimen[heat combustion for etanol = -1400 kJmol-1] (from ques.)

1 mol etanol burnt releasing 1400 kJ of heat;So ;

0.1 mol etanol burnt releasing;

= 0.01 x 1400 kJ

= 14 kJ= 14000 J of heat

Step 3 : calculate the temperature rise for water


14000 J = 200 x 4.2 x

7

Make sure the

unit is correct

Mass etanol

JMR etanol

0.46

46

Change kJ to J, we wantsubstitute it into equation

mc


73/75

= oC

= 16.7 oC

Question 3

Complete heat combustion for 1 mol butanol, C4H9OH release 2600 kJ

of heat. Calculate the mass for butanol that need to burnt completely so

heat release can increase the temperature for 500 cm3 of water to 36 oC

[J.A.R: C, 12; O, 16; H, 1]

Solution

Step 1 : calculate the heat changes/total heat release in this experiment

total heat release = mc

= 500 x 4.2 x 36 J

= 75600 J

= 75.6 kJ

Step 2 : calculate no. of mol for butanol that burnt[heat combustion for butanol = -2600 kJmol-1] (from ques.)

2600 kJ heat release by 1 mol butanol that burnt;

so ;

75.6 kJ heat release by mol butanol that burnt;

= 0.03 mol

No. of mol of etanol that burnt = 0.03 mol

Step 3 : calculate the mass of butanol that burnt

No. of mol butanol =7

14000 .

200 X 4.2

mass butanol

JMR butanol

1 x 75.6 kJ

2600 kJ


74/75

0.03 mol =

Mass of butanol = 0.03 x 74

= 2.22 gQuestion 4

Energy level diagram for combuction reaction for methanol, CH3OH i

shown below.

Heat that release from the complete combustion 8.0 g of methanol, isuse to hot 1 dm3 of water. Calculate the temperature rise of the water.

Solution

Step 1 : calculate no. of mol for ethanol that is burn

Mol no for etanol = = = 0.25 mol

Step 2 : calculate the total heat released/absorb in exp.

[heat of combustion for ethanol = -720 kJmol-1] (from question)

Energy

CH3OH (ce) + O

2(g)

CO2

(g) + 2H2O (ce)

H = - 720 kJmol-1

3

2

7

mass butanol .

(4 x 12) + 9 + 16 + 1

Mass of metanol

JMR metanol

8.0 .

12+(3x1)+16+1


75/75

1 mol metanol burn releasing 720 kJ of heat;So ;

0.25 mol methanol that burn releasing;

= 0.25 x 720 kJ

= 180 kJ= 180000 J of heat

Step 3 : calculate the rising in water temperature

Volume of water = 1 dm3

= 1000 cm3

Total heat released = mc

180 000 J = 1000 x 4.2 x

= oC

= 42.85 oC

Learning Task 4.8: Problem solving pg 168 no. 1, 2, 3Effective Practise: pg 173 no. 1, 2, 3

Review Questions: pg 176 to 179


SMKDBL

180 000 .

1000 X 4.2

Change kJ to J, we want

substitute it into formula

mc

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THERMOCHEMISTRY CHA4 FORM5