waja chp.4 thermochemistry

WAJA F5 Chemistry 2010 Chapter 4: Thermochemistry

CHAPTER 4 : THERMOCHEMISTRY

A ENERGY CHANGES IN CHEMICAL REACTION

Activity 1:

(a) What is an exothermic reaction?

An exothermic reaction is a chemical reaction that gives out heat to the surroundings.

Heat energy given out from the reaction is (1)…………………………to the surroundings

The temperature of the surroundings (2)……………………….

(b) What is an endothermic reaction?

An endothermic reaction is a chemical reaction that absorbs heat from the surroundings .

The reactants (3) …………………. heat energy from the surroundings. The temperature of the

surroundings (4) ……………………….

Activity 2

(a) Identify the following reactions as exothermic or endothermic reaction by writing a ‘’ at the appropriate box as shown in Question (a).

Reaction Exothermic Endothermic

(a) Combustion of ethanol √

(b) Burning of magnesium

(c) Neutralisation between acid and alkali

(d) Adding water to concentrated sulphuric acid

(e) Photosynthesis

(f) Reaction between acid and magnesium

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Learning Outcomes :

You should be able to: State what exothermic reaction is, State what endothermic reaction is, Identify exothermic reactions, Identify endothermic reactions, Give examples of exothermic reactions, Give examples of endothermic reactions, Construct energy level diagrams for exothermic reactions, Construct energy level diagrams for endothermic reactions, Interpret energy level diagram, Interrelate energy change with formation and breaking of bonds, Describe the application of knowledge of exothermic and endothermic reactions in everyday life.


(g) Reaction between acid and calcium carbonate

(h) Dissolving ammonium salt in water

(i) Thermal decomposition of copper(ll) carbonate

(j) Thermal decomposition of zinc nitrate

(b) The amount of heat energy released or absorbed during a chemical reaction is called the heat of

reaction.

It is given a symbol (1)………………. and the unit is (2) …………………..

(c)

(d)

The heat of reaction , Δ H = H products - H reactants

(a) Exothermic reaction : The reactants lose heat energy to form the products . Thus the products

formed have less energy than the reactants, Therefore, Δ H is (3)…………………

(b) Endothermic reaction : The reactants absorb heat energy to form the products . Thus the

products formed have (4) ……………energy than the reactant. Therefore, Δ H is (5)…………….

List two other examples of exothermic and endothermic reaction

Exothermic reaction Endothermic reaction

1)

2)

Activity 3 :

(a) Energy level diagram

(i) Energy level diagram for exothermic reaction :

The products have less energy than the reactants,

Energy

ΔH = negative

Construct the energy level diagram based on the given chemical equation,

Mg(s) + H2SO4 (aq) MgSO4 (aq) + H2 (g) ΔH = -467 kJ

(reactants) ( products )

2

reactants

products


(ii)

(iii)

Energy level diagram for endothermic reaction :

The products have more energy than the reactants,

Energy

ΔH = positive

Construct energy level diagram based on the given chemical equation

CaCO3(s) CaO(s) + CO2 (g) ΔH = + 178 kJ

Information that can be obtained from the energy level diagram ,

Energy

ΔH = - 190 kJ

Figure 1

Figure 1 shows the energy level diagram for the reaction between zinc and copper(ll)

sulphate,

The reaction between (1) ……………and (2)…………………. is an (3)

…………………….. reaction.

During the reaction, the temperature of the mixture (4) ………………………

The total energy of one mole of (5)……………. and one mole of (6) …………… is (7)

…………. than the total energy of one mole of copper and one mole of zinc sulphate

by (8)……………………. kJ

When one mole of (9)………………….. reacts with one mole of (10)…………to form

(11)…………….. .. mole of copper and (12)……………….. mole of zinc sulphate, (13)

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reactants

products

Cu + ZnSO4

Zn + CuSO4


…………………….. kJ of heat is (14) …………………………

Energy

ΔH = +53 kJ

Figure 2

Figure 2 shows the energy level diagram for the reaction between hydrogen gas and iodine

The reaction between (15)……………and (16)…………………. is an (17)

……………… reaction.

During the reaction, the temperature of the mixture (18)……………………

The total energy of one mole of (19)……………. and one mole of (20)…………….

is (21) ………than the total energy of two moles of hydrogen iodide by (22) ……kJ

When one mole of (23)………………….reacts with one mole of (24) ………………

to form (25) …………….mole of hydrogen iodide (26)………..kJ of heat is (27)

…………………

(b) (i) A chemical reaction involves (1) ....……… (2) ............... of the reactants and

(3)…………….. (4) ........................of the product

(ii) Bonds breaking always(1) …………. energy and bonds formation always

(2) ………………… energy

Bond breaking Bond formation

(iii) If the energy absorbed during bonds breaking is less than the energy released during

bonds formation ,energy is (1)………………….. to the surrounding.

The reaction is an (2) ……………….. reaction

(iv) If the energy absorbed during bonds breaking is more than the energy released

during bonds formation, energy is(1)…………………………. from the surrounding.

Hence it is an (2) ……………………reaction

(c) Explain the application of exothermic and endothermic reaction in our daily lives.

(i) Instant cold packs :

Instant cold packs are used to treat (1)…………………………………………………, have

separate compartments of (2) …………………and (3)……………………in a plastic bag. When

the barrier between the two is broken by squeezing the outer bag, the (4)…………..

…………………… dissolves in the(5) ……………….endothermically to provide instant

coldness.

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H2 (g) + I2(g)

2HI (g)


(ii) A reusable heat pack:

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

B HEAT OF PRECIPITATION

HEAT OF REACTION

Activity 4: Match the chemical equation with the type of reaction below : Chemical equation Type of reaction

HNO3 + KOH KNO3 + H2O Precipitation C2H5OH + 3O2 2CO2 + 3H2O Displacement Mg + CuSO4 MgSO4 + Cu Neutralization Pb(NO3)2 + 2KI PbI2 + 2KNO3 Combustion

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HEAT OF REACTIONThe change in the amount of heat in a chemical reaction

Symbol : ΔHThe

HEAT OF PRECIPITATION HEAT OF DISPLACEMENT

HEAT OF NEUTRALIZATION

HEAT OF COMBUSTION

The heat change in a reaction can be calculated using the formula , H = mcӨ m = mass of solution, g c = specific heat capacity of the solution, J g-1 oC-1

Ө = temperature change , 0C

Assumption : i) The solution is dilute, it has the same density as water , 1 g cm-3

( 1 cm3 = 1 g ) ii) The solution has the same specific heat capacity as water, 4.2 J g-1 oC-1

Learning Outcomes You should be able to: State what heat of reaction is, State what heat of precipitation is, Determine the heat of precipitation for a reaction, Construct an energy level diagram for a precipitation reaction, Solve numerical problems related to heat of precipitation


Activity 5 : (a) What is meant by heat of precipitation ?

…………………………………………………………………………………………………………..

…………………………………………………………………………………………………………..

(b) To determine the heat of precipitation of silver chloride, AgCl

Procedure :1. Measure 25 cm3 0.5 mol dm-3 silver nitrate solution and pour it into the polystyrene cup,2. Put the thermometer into the silver nitrate solution. Record the initial temperature,3. Measure 25 cm3 0.5 mol dm-3 sodium chloride solution and record the initial temperature,4. Pour the sodium chloride solution quickly into the silver nitrate solution in the polystyrene cup.

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Example :

Calculate the amount of heat change when the temperature of 200 cm3 of water is raised from 280C to 400C,

Q = mcӨ m = 200 g = 200 g x 4.2 J g-1 oC-1 x 12 0C c = 4.2 J g-1 oC-1

= 10080 J Ө = 40 - 28 = 12 0C = 10.08 kJ

Guidelines for the calculation of the heat of reaction Steps to follow:Step 1 - Calculate the heat change using the formula, H = mcӨ ( p Joule )Step 2 - Write chemical equation or ionic equation for the reaction that occursStep 3 - Calculate the number of moles of reactant that reacts using either the following formulae : Number of moles = mass = q mol or molar mass

Number of moles = MV ( q mol) M : Molarity of the solution (Solution) V : volume of the solution in dm3

Step 4 - Link the number of moles of reactants ( step 3 ) with the heat change ( step 1 ) q mol of reactants react heat lost/gain is p J ∴ 1 mol of reactants react heat lost/gain is p J q Heat of reaction , ΔH = +/- p kJ mol-1

1000 x q = +/- r kJ mol-1 Note : “ + “ is used for endothermic reaction and “-“ is used for exothermic reaction

Step 5 - Draw the energy level diagram

25 cm3 0.5 mol dm-3

sodium chloride solution

25 cm3 0.5 mol dm-3

silver nitrate solution


5. Stir the solution mixture with the thermometer and record the highest temperature achieved.

Result : Initial temperature of silver nitrate solution = 28.5 0CInitial temperature of sodium chloride solution = 29.5 0CHighest temperature of the mixture = 32.0 0C

Calculation

Step 1 : Calculate the heat change using the formula H = mcӨ

Temperature change, Ө = 32.0 0C - 29.0 0C = 3.0 0C

Heat change , H = mcӨ m = ( 25 + 25 ) g = 50 g c = 4.2 J g-1 oC-1

Ө = 3.0 0C H =(1) ……….........J

Step 2:Write balanced chemical equation or ionic equation for the reaction that occurs

Chemical equation : AgNO3 (aq) + NaCl(aq) AgCl (s) + NaNO3 (aq) Precipitate

Ionic equation : (2)……………………………………………………….

Deduce the mole ratio from the ionic equation :

(3)…………mol silver ion, Ag+ react with (4) ……… mol of chloride ion,

Cl- to produce (5)……………. mol of silver chloride , AgCl.

Step 3 :Calculate the number of moles of reactant that reacts,

Use the formula : n = MV ( V in dm3 )

Number of moles of silver ion = the number of moles of silver nitrate

= (6) …………………….. molNumber of moles of chloride ion = the number of moles of sodium chloride

= (7)…………………….. mol

Number of moles of silver chloride formed = (8) ………………….. mol

Step 4Calculate the heat of precipitation of silver chloride, ΔH Step 5Energy level diagram

When (9) ….. mol of silver chloride formed, heat released is (10) .……..

When 1 mol of silver chloride formed, heat released is (11)......................

Heat of precipitation of silver chloride, ΔH = (12) - ……………kJ mol – 1

Draw the energy level diagram for the reaction that occurs in this experiment (13)

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Average initial temperature = 29.00C


(c)

(d)

Calculate the heat change when 200 cm3 of 0.5 mol dm-3 calcium chloride, CaCl2 solution is added to 200 cm3 of 0.5 mol dm-3 sodium carbonate, Na2CO3 solution if the heat of precipitation of calcium carbonate, CaCO3 is +12.6 kJ mol-1

[ Specific heat capacity of solution : 4.2 J g-1 0C-1 . Density of solution : 1 g cm-3 ]

The thermochemical equation for the precipitation of silver chloride is as follows :

Ag+ (aq) + Cl- (aq) AgCl ΔH = –65.5 kJ mol-1

Calculate the temperature change when 100 cm3 of 0.5 mol dm-3 silver nitrate, AgNO3 ,

solution is added to 100 cm3 of 0.5 mol dm-3 potassium chloride, KCl solution

C HEAT OF DISPLACEMENT

Activity 6

(a) What is meant by the heat of displacement ?

…………………………………………………………………………………………………………..

…………………………………………………………………………………………………………..

(b) To determine the heat of displacement of copper from a copper (ll) sulphate solution by zinc.

Procedure :1. Measure 25 cm3 0.2 mol dm-3 copper(ll) sulphate solution and pour into a polystyrene cup.2. Put the thermometer into the copper(ll) sulphate solution. Record the initial temperature,3. Add half a spatula of zinc powder (in excess) quickly into copper(ll) sulphate solution.5. Stir the mixture with the thermometer and record the highest temperature achieved.

Result :

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zinc powder

25 cm3 0.2 mol dm-3 copper(ll) sulphate solution

Learning Outcomes,You should be able to: State what heat of displacement is, Determine the heat of displacement, Construct an energy level diagram for a displacement reaction, Solve numerical problems related to heat of displacement,


Initial temperature of copper(II) sulphate solution = 30.0 0CHighest temperature of the mixture = 40.0 0C

Calculation

Step 1 : Calculate the heat change using the formula H = mcӨ

Changes of temperature, Ө = 40.0 0C - 30.0 0C = 10.0 0C

Heat change , H = mcӨ m = 25 g c = 4.2 J g-1 oC-1

Ө = 10.0 0CH = (1)…………………………….. J


Chemical equation : Zn (s) + CuSO4(aq) Cu (s) + ZnSO4 (aq) copper displacedDeduce the mole ratio from the equation :

(2)……… mol copper metal, Cu is displaced from (3) …………….. mol of

copper(ll) sulphate solution, CuSO4 by zinc

Step 3 :Calculate the number of moles of reactant that reacts

Number of moles copper(ll) sulphate = (4)…………………….. mol

Number of moles of copper = (5)…………………….. mol

Step 4Calculate the heat of displacement of copper, ΔH Step 5Energy level diagram

When (6) ….. mol of copper is displaced, the heat released is (7) .…….. ∴ When 1 mol of copper is displaced, the heat released is (8).................

Heat of reaction , ΔH = (9) ……………… kJ mol – 1


(c) Figure shows an experiment carried out to determine the heat of displacement of silver from silver nitrate by copper metal.

Based on figure above, calculate the heat of displacement for the reaction.

In an experiment, excess magnesium powder is added to 50 cm3 of 0.25 mol dm-3

iron(ll) sulphate solution at 29.0 0C. The thermochemical equation is shown below,9


(d)

Mg(s) + Fe2+ (aq) Mg2+ (aq) + Fe (s) ΔH = -80.6 kJ mol - 1

What is the highest temperature reached in this experiment ?

D HEAT OF NEUTRALIZATION

Activity 7 (a) What is meant by the heat of neutralization …………………………………………………………………………………………………………..

……………………………………………………………………………………………………………(b) To determine the heat of neutralisation between a strong acid ( hydrochloric acid ) and a strong alkali ( sodium hydroxide )

Procedure :1. Measure 50 cm3 2.0 mol dm-3 sodium hydroxide solution and pour it into the polystyrene cup,2. Put the thermometer into the sodium hydroxide solution. Record the initial temperature,3. Measure 50 cm3 2.0 mol dm-3 hydrochloric acid solution and record the initial temperature,4. Pour the hydrochloric acid solution quickly into the sodium hydroxide solution in the polystyrene cup.5. Stir the mixture with the thermometer and record the highest temperature achieved.

Result : Initial temperature of sodium hydroxide solution = 29.0 0CInitial temperature of hydrochloric acid solution = 29.0 0CHighest temperature of the mixture = 42.0 0C

Calculation

Step 1 : Changes of temperature, Ө = 42.0 0C - 29.0 0C = (1) ……………….0C

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Learning Outcomes :You should be able to: State what heat of neutralization is, Determine the heat of neutralization, Construct energy level diagrams for various types of neutralization reactions, Compare the heat of neutralization for the reactions between a strong acid and a strong alkali with the heat of neutralization for a reaction between a weak acid and a weak alkali, Explain the difference of the heat of neutralization for a strong acid and a strong alkali with heat of neutralization for a reaction involving a weak acid and/or a weak alkali. solve numerical problems related to heat of neutralization,

50 cm3 2.0 mol dm-3

sodium hydroxide solution

50 cm3 2.0 mol dm-3

hydrochloric acid solution

Average initial temperature = 29.0 0C


Calculate the heat change using the formula H = mcӨ

Heat change , H = mcӨ m = ( 50 + 50 ) g = 100 g c = 4.2 J g-1 oC-1

Ө = (2) ……………….0CH = (3)…………………………….. J


Chemical equation (4) …………………………………………………………. Ionic equation (5) ……………………………………………………….

Deduce the mole ratio from the ionic equation :

(6)……mol hydrogen ion, H+ react with (7)……mol of hydroxide ion, OH-

to produce (8)……………. mol of water , H2O Step 3 :Calculate the number of moles of reactant that reacts,

Use , n = MV ( V in dm3 )

Number of moles of H+ = the number of moles of hydrochloric acid

= (9)…………………….. mol

Number of moles of OH- = the number of moles of sodium hydroxide

= (10)…………………….. mol

Number of moles of water formed = (11) ………………….. molStep 4Calculate the heat of neutralisation of hydrochloric acid and sodium hydroxide, ΔH

Step 5 : Energy level diagram

When (12) ….. mol of water formed, heat released is (13) .……..

∴ When 1 mol of water formed, heat released is (14) .................

Heat of neutralisation , ΔH = (15) ……………… kJ mol - 1


(c)

(d)

The thermochemical equation for the reaction between ethanoic acid and sodium hydroxide is given below,

CH3COOH (aq) + NaOH (aq) NaCH3COO (aq) + H2O (l) ΔH = -55 kJ mol - 1

Calculate the heat given out when 200 cm3 of ethanoic acid 0.5 mol dm-3 is added to 200 cm3 of sodium hydroxide 0.5 mol dm-3

The energy level diagram of a neutralization reaction is shown in figure below

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Energy

H2SO4 + 2NaOH ΔH = - 114 kJ Na2SO4 + 2H2OWhen 100 cm3 of 1.0 mol dm-3 sulphuric acid is added to 100 cm3 of 1.0 mol dm-3 sodium hydroxide solution. What is the change in temperature?

E : HEAT OF COMBUSTION

Activity 8 :

(a) What is meant by the heat of combustion ………………………………………………………………………………………………………. (b) To determine the heat of combustion of ethanol

Experiment to determine the heat of combustion.Procedure :1. Measure 200 cm3 of water and pour it into a copper can. Record the initial temperature of the water and place the copper can on a tripod stand.2. Fill a lamp with ethanol and weigh it. Record the mass of the lamp together with its content.3. Light up the wick of the lamp immediately. Stir the water continuously until the temperature of the water increases by about 30 0C.5. Put off the flame and record the highest temperature reached by the water

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Learning OutcomesYou should be able to :

state what heat of combustion is, determine heat of combustion for a reaction, construct an energy level diagram for a combustion reaction, compare the heats of combustion of various alcohols, state what fuel value is, describe the difference between heats of combustion of various alcohols, describe the applications of fuel value, compare and contrast fuel values for various fuels, solve numerical problems related to heat of combustion.


6. Weigh again the lamp and record the mass of the lamp. Result : Mass of lamp and ethanol before burning = 190.55 gMass of lamp and ethanol after burning = 189.80 gInitial temperature of water = 29.00CHighest temperature of water = 59.00C

Calculation

Step 1 : Calculate the heat change/heat absorbed by the water using the formula, H = mcӨ

Changes of temperature, Ө = 59.00C - 29.00C = 30.0 0CMass of water, m = 200 g

Heat change/ heat absorbed by water , H = mcӨ c = 4.2 J g-1 oC-1

Ө = 30.0 0CH = (1)…………………………….. J

Step 2:Write a balanced chemical equation for the combustion of ethanol

C2H5OH + (2)……………= (3)……………+ (4) ……………………….

Step 3 :Calculate the number of moles of ethanol that is used in the experiment,[ Relative atomic mass : H, 1 : C,12; O,16 ]

Mass of ethanol burnt/used = (5) ………………. g

Number of moles of ethanol burnt = (6) __________ molar mass of ethanol

= …………………….. mol

Step 4Calculate the heat of combustion of ethanol, ΔH Step 5Energy level diagram

When(7) ….. mol of ethanol is burnt in oxygen heat released is(8) .……

∴ When 1 mol of ethanol is burnt in oxygen, heat released is (9) ..........

Heat of combustion of ethanol , ΔH = (10) ……………… kJ mol - 1

Draw an energy level diagram for the combustion of ethanol in this experiment (11)

(c) The following results are obtained by a student in an experiment to determine the heat of combustion of alcohols. Complete the table and calculate the heat of combustion of methanol, propanol and butanol based on the data given in the table below . [ Relative atomic mass : H,1 ; C,12 ; O,16 . Specific heat capacity of water , 4.2 J g-1 0C-1 ]

Alcohol Methanol Propanol Butanol

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Volume of water/cm3 200 200 200

Initial temperature of water / 0C

28.0 28.0 28.0

Final temperature of water / 0C

60.0 58.0 59.0

Mass of alcohol burnt/g 1.184 0.750 0.719

Changes in temperature / 0C

(1) (2) (3)

Molecular formula (4) (5) (6)

Number of carbon atoms

(7) (8) (9)

Molar mass (10) (11) (12)

Number of moles of alcohol burnt

(13) (14) (15)

Heat change/ absorbed by the

water / J

(16) (17) (18)

Heat of combustion of alcohol /

kJ mol-1

(19) (20) (21)

(i) State the relationship between the number of carbon atoms in an alcohol and the heat of combustion,

……………………………………………………………………………………………………………….

……………………………………………………………………………………………………………….

(ii) Predict the heat of combustion for pentanol

……………………………………………………………………………………………………………….

(d)

(e)

When 1 mole of butanol, C4H9OH is burnt in excess of oxygen, 2600 kJ of heat is produced. Calculate the mass of butanol needed to burn completely in oxygen in order to raise the temperature of 500 cm3 of water by 300C( Relative atomic mass : H , 1 ; C , 12 ; O , 16, Specific heat capacity of water , 4.2 J g-1 0C-1)

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The fuel value of a fuel is the amount of heat energy given out when one gram of the fuel is completely burnt in excess of oxygen.

Fuel value ( kJ g-1 ) = heat of combustion of alcohol ( kJ mol -1 ) molar mass


The heat of combustion of propanol, C3H7OH is -2016 kJ mol -1 What is its fuel value ?Solution :

(i) Calculate the molar mass of propanol, C3H7OH.

(ii) Calculate the fuel value of propanol, C3H7OH.

ANSWERActivity 1(a) (1) transferred , (2) increases , (b) (3) absorbs , (4) decreases

Activity 2

(a) Identify the following reactions as exothermic or endothermic reaction

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Reaction Exothermic Endothermic

(a) Combustion of ethanol √

(b) Burning of magnesium √

(c) Neutralisation between acid and alkali √

(d) Adding water to concentrated sulphuric acid √

(e) Photosynthesis √

(f) Reaction between acid and magnesium √

(g) Reaction between acid and calcium carbonate √

(h) Dissolving ammonium salt in water √

(i) Thermal decomposition of copper(ll) carbonate √

(j) Thermal decomposition of zinc nitrate √

(b) (1) ΔH , (2) kJ, (3) negative sign , (4) higher , (5) positive sign (c) Two other examples of exothermic and endothermic reaction

Exothermic reaction Endothermic reaction

1) Physical changes : gas liquid

liquid solid

Physical changes : solid liquid ,

liquid gas

2) Reaction between reactive metals

(sodium and potassium) with water //

Rusting of iron

Dissolving crystalline salts in water

example : CuSO4.5H2O

Activity 3

(a) (i) Mg + H2SO4 MgSO4 + H2 ΔH = -467 kJ

(ii) CaCO3 CaO + CO2 ΔH = + 178 kJ

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Mg(s)+H2SO4(aq)

MgSO4(aq) + H2(g)

ΔH = -467 kJ

Energy

Energy

CuCO3

CuO + CO2

ΔH = + 178 kJ


(iii) (1) zinc , (2) copper(II) sulphate , (3) exothermic , (4) increases , (5) zinc , (6) copper(II) sulphate , (7) more , (8) 190 kJ , (9) zinc , (10) copper(II) sulphate, (11) one , (12) one , (13) 190 , (14) produced // released (15) hydrogen gas, (16) iodine gas, (17) endothermic , (18) decreases, (19) hydrogen gas (20) iodine gas, (21) less , (22) 53 , (23) hydrogen gas , (24) iodine gas , (25) two , (26) 53 (27) absorbed.

(b) (i) (1) bond(2) breaking , (3) bond , (4) formation (ii) (1) requires / absorbs (2) releases

(iii) (1) releasing , (2) exothermic

(iv) (1) absorbing , (2) endothermic

(c) (i) (1) sports injuries , (2) water , (3) solid ammonium nitrate , (4) ammonium nitrate , (5) water

(ii) Uses a sodium acetate crystallization . By bending the metal disc in the bag, the sodium acetate crystallizes and gives off heat.Activity 4Match the chemical equation with the type of reaction below : Chemical equation Type of reactionHNO3 + KOH KNO3 + H2O Precipitation reactionC2H5OH + 3O2 2CO2 + 3H2O Displacement reactionMg + CuSO4 MgSO4 + Cu Neutralization Pb(NO3)2 + 2KI PbI2 + 2KNO3 Combustion

Activity 5

(a). The heat of precipitation is the heat change when one mole of a precipitate is formed from their ions in aqueous solution.

(b) (1) 630 J , (2) Ag+ + Cl- AgCl (3) one , (4) one , (5) one , (6) 0.0125 , (7) 0.0125 , (8) 0.0125 , (9) 0.0125 , (10) 630 J , (11) 630 = 50400 J = 50.4 kJ 0.0125 (12) - 50.4 kJ mol -1 (13) Energy level diagram

(c) 1.26 kJ // 1260 J

(d) 3.9 0C

Activity 6

(a) The heat of displacement is the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal.

(b) (1) 1050 J , (2) one , (3) one , (4) 0.2 x 25 = 0.005, (5) 0.005 mol , (6) 0.005 1000

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Energy

Ag+ + Cl-

ΔH = - 50.4 kJ mol -1

AgCl


(7) 1050 J , (8) 1050 = 210000J = 210 kJ , (9) - 210 kJ mol-1

0.005

(10) Energy level diagram

(c) ΔH = - 16. 8 kJ mol -1

(d) 33.8 0C

Activity 7

(a) The heat of neutralization is the heat change when one mole of water is formed from the reaction between an acid and an alkali.

(b) (1) 13.0 , (2) 13.0 , (3) 5460 J , (4) HCl + NaOH NaCl + H2O , (5) H+ + OH- H2O (6) 1 , (7) 1 , (8) 1 , (9) 0.1 mol , (10) 0.1 mol , (11) 0.1 mol , (12) 0.1 mol , (13) 5460 J, (14) 5460 = 54600J = 54.6 kJ , (15) - 54.6 kJ mol-1

0.1

(16) Energy level diagram

(c) 5.5 kJ(d) 6.8 0 C

Activity 8

(a) The heat of combustion is the heat change when one mole of a substance is completely burnt in oxygen under standard conditions.

(b) (1) 25200 J, (2) 3O2 , (3) 2CO2 , (4) 3H2O , (5) 0.75 , (6) 0.75 = 0.0163 mol , (7) 0.0163 46 (8) 25200 J, (9) 25200 = 1546012.2 J = 1546 kJ , 0.0163 (10) Energy level diagram

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Energy

Zn + CuSO4

ΔH = - 210 kJ mol -1

Cu + ZnSO4

Energy

HCl + NaOH ΔH = - 54.6 kJ mol -1

NaCl + H2O

Energy

C2H5OH +3O2

ΔH = - 1546 kJ mol -1

2CO2 + 3H2O


(c) (1) 32.0 , (2) 30.0 , (3) 31.0 , (4) CH3OH , (5) C3H7OH , (6) C4H9OH , (7) 1 , (8) 3 , (9) 4 (10) 32 , (11) 60 , (12) 74 , (13) 0.037 , (14) 0.0125 , (15) 0.0097 , (16) 26880 J (17) 25200 J , (18) 26040 J , (19) 726.5 kJ , (20) 2016 kJ , (21) 2684.5 kJ

(i) The more carbon atoms in the molecular formula of an alcohol, the higher is the heat of combustion // The heat of combustion is proportional to the number of carbon atom per molecule in alcohol ,

(ii) 3300 kJ mol-1 to 3400 kJ mol-1

(d) 1.793 g

(e) (i) 60g mol-1

(ii) 33.6 kJ g-1

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waja chp.4 thermochemistry

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