Waja Thermochemistry Edited

7/27/2019 Waja Thermochemistry Edited

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13 THERMOCHEMISTRY

CONCEPT MAP

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ENERGY LEVEL DIAGRAM

Energyproduct

H : positiveReactant

Energy

ReactantH : negative

product

Endothermic reactionHeat energy absorbs

Exothermic reactionHeat energy releases

Energy change

THERMOCHEMISTRY

types of heat of reaction

Heat of precipitationMeaning,Example of chemical reaction andequation

Heat of displacementMeaning,Example of chemical reaction and equation

Heat of neutralizationMeaning,Example of chemical reaction andequation

Heat of precipitationMeaning,Example of chemical reaction and equation


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13.1 ENERGY CHANGES IN CHEMICAL REACTION

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( Refer to the form five text book page : 143 - 144 )

1 What is exothermic reaction ?

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

Heat energy given out from the reaction is to the surroundings

The temperature of the surroundings .

2 What is endothermic reaction ?

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

The reactants . heat energy from the surroundings. The temperature of the

surroundings .

3 Identify the following reactions as exothermic or endothermic reaction

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

4 The amount of heat energy released or absorbed during a cemical reaction is called the heat

of reaction. It is given a symbol . and the unit is ..

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

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

products formed have less energy than the reactants, Therefore, H is .

Endothermic reaction : The reactants absorbs heat energy to form the products . Thus the

products formed have energy than the reactant,Therefore, H is positive sign.

6 Energy level diagram ( Refer to the text book page 145 - 146 )(i) Energy level diagram forexothermic reaction :

The products have less energy than the reactants,

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(ii)

Energy

H = negative

Construct energy level diagram based on the chemical equation,

Mg + H2SO4 ZnSO4 + H2 H = -467 kJ

(reactants) ( products )

Energy level diagram forendothermic reaction :

The products have more energy than the reactants,

Energy

H = negative

Construct energy level diagram based on the chemical equation

CaCO3 CuO + CO2 H = + 178 kJ

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reactants

products

reactants

products


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(iii)

Information that can be obtained from the energy level diagram

Energy

H = - 190 kJ

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

sulphate,

a) The reaction between and . is an ..reaction.

b) During the reaction, the temperature of the mixture .

c) The total energy of one mole of. and one mole of ..

is . than the total energy of one mole of copper and one mole of zinc

sulphate by . kJ

d) When one mole of .. reacts with one mole of ..

to form .. .. mole of copper and .. mole of zinc sulphate,

.. kJ of heat is

Energy

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Zn + CuSO4

Cu + ZnSO4

2HI


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H = +53 kJ

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

iodine

a) The reaction between and . is an ..

reaction.

b) During the reaction, the temperature of the mixture .

c) The total energy of one mole of. and one mole of ..

is . than the total energy of two moles of hydrogen iodide by kJ

d) When one mole of .. reacts with one mole of ..to form .mole of hydrogen iodide..kJ of heat is

7 Explain the application of exothermic and endothermic reaction in our every day life,

( Refer to the text book page 147 - 148 )(i)

(ii)

Instant cold packs :

Instant cold packs are used to treat ,

have separate compartments of and in a plastic

bag. When the barrier between the two is broken by squeezing the outer bag, the

.. dissolve in .endothermically to provide

instant coldness.

First-aid chemical hot pack :

13.2 HEAT OF REACTION ( refer to the text book page : 149 - 169 )

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H2 + I2

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

Symbol : .The


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Complete the table

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HEAT OF PRECIPITATIONMeaning :

....

HEAT OF DISPLACEMENTMeaning :

....

HEAT OF NEUTRALIZATIONMeaning : ....

HEAT OF COMBUSTIONMeaning : ....

The heat change in a reaction can be calculated using the formula , H = mcm = mass ofsolution, g

c = specific heat capacity of the solution, J g-1 o

C-1

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

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

Example :

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

H = mc m = 200 g

= 200 g x 4.2 J g-1 o

C

x 120

C c = 4.2 J g-1 o

C-1

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

Chemical equation Type of reaction Heat of reaction

Pb(NO3)2 + 2KI PbI2 + 2KNO3 Precipitation reaction Heat of precipitationDisplacement reaction

Heat of neutralizationC2H5OH + 3O2 2CO2 + 3H2O

Precipitation reaction

Guidelines for the calculation of the heat of reactionSteps to follow:Step 1 - Calculate the heat change using the formula, H =mcStep 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 or

molar mass

Number of moles = MV M : Molarity of the solution1000 V : volume of the solution in cm3

Step 4 - Link the number of moles of reactants ( step 3 ) with the heat change ( step 1 )

Calculate the heat of reaction, H, by using the following formula :

Heat of reaction , H = heat change ( answer from step 1 )Number of moles ( answer from step 3 )


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NUMERICAL PROBLEMS RELATED TO HEAT OF REACTION

1. HEAT OF PRECIPITATION

Aim : To determine the heat of precipitation of silver chloride

Procedure :1. Measure 25 cm3 0.5 mol dm-3 silver nitrate solution and pour it into the polystrene 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.5. Stir the solution mixture with the thermometer and record the highest temperature achieved.

Result :Initial temperature of silver nitrate solution = 290CInitial temperature of silver nitrate solution = 290CHighest temperature of the mixture = 320C

Calculation

Step 1 :Calculate the heatchange using theformula H = mc

Changes of temperature, = 320C - 290C = 30C

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

= 30CH = .. J

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25 cm3 0.5 mol dm-3

sodium chloride solution

25 cm3 0.5 mol dm-3

silver nitrate solution

Average initial temperature = 290C


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Step 2:Write chemicalequation or ionicequation for thereaction that occurs

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

Ionic equation : .

Deduce the mole ratio from the ionic equation

mol silver ion, Ag+ react with mol of chloride ion, Cl-

to produce . mol of silver chloride , AgCl

Step 3 :Calculate the numberof moles of reactantthat reacts

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

= .. mol

Number of moles of chloride ion = the number of moles of sodium chloride

= .. mol

Number of moles of silver chloride formed = .. mol

Step 4Calculate the heat ofprecipitation of silverchloride, H

Heat of reaction , H = heat change ( answer from step 1 )Number of moles of silver chloride

= J

H = - kJ mol - 1

Draw an energy level diagram for the reaction that occurs in this

experiment

2. HEAT OF DISPLACEMENT

Aim : To determine the heat of displacement of copper metal from a copper (ll) sulphate solutionby zinc metal

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Procedure :1. Measure 25 cm3 0.2 mol dm-3 copper(ll) sulphate solution and pour into a polystirene cup.2. Put the thermometer into the silver nitrate 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 :Initial temperature of silver nitrate solution = 300CHighest temperature of the mixture = 400C

Calculation

Step 1 :Calculate the heatchange using theformula H = mc

Changes of temperature, = 400C - 300C = 100C

Heat change , H = mc m = 25 gc = 4.2 J g-1 oC-1

= 100CH = .. J

Step 2:Write chemicalequation or ionicequation for thereaction thatoccurs

Chemical equation : Zn (s) + CuSO4(aq) Cu (s) + ZnSO4 (aq)copper displaced

Deduce the mole ratio from the equation

mol copper metal, Cu is displaced from .. mol ofcopper(ll) sulphate solution, CuSO4 by zinc

Step 3 :Calculate thenumber of molesof reactant thatreacts

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

Number of moles of copper = .. mol

Step 4Calculate the heatof displacement ofcopper, H

Heat of reaction , H = heat change ( answer from step 1 )Number of moles copper

= J

H = - kJ mol - 1

Draw an energy level diagram for the reaction that occurs in thisexperiment

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zinc powder

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


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3. HEAT OF NEUTRALISATION

Aim : To determine the heat of neutralisation between a strong acid ( hydrochloric acid solution )and a strong alkali ( sodium hydroxide )

Procedure :1. Measure 50 cm3 2.0 mol dm-3 sodium hydroxide solution and pour it into the polystrene 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 silver nitrate solution = 290CInitial temperature of silver nitrate solution = 290CHighest temperature of the mixture = 420C

Calculation

Step 1 :Calculate the heat

change using theformula H = mc

Changes of temperature, = 420C - 290C = .0C

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

= .0CH = .. J

Step 2:Write chemicalequation or ionicequation for thereaction that occurs

Chemical equation : .Ionic equation : .

Deduce the mole ratio from the ionic equation :

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50 cm3 2.0 mol dm-3

sodium hydroxide solution

50 cm3 2.0 mol dm-3

hydrochloric acid solution

Average initial temperature = 290C


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mol hydrogen ion, H+ react with mol of hydroxide ion, OH-

to produce . mol of water , H2O

Step 3 :

Calculate thenumber of moles ofreactant that reacts

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

= .. mol

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

= .. mol

Number of moles of water formed = .. mol

Step 4Calculate the heat

of neutralisation ofhydrochloric acidand sodiumhydroxide, H

Heat of reaction , H = heat change ( answer from step 1 )Number of moles water

= J

H = - kJ mol - 1

Draw an energy level diagram for the reaction that occurs in thisexperiment

4. HEAT OF COMBUSTION

Aim : To determine the heat of combustion of ethanol

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Experiment to show 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 weight 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.

Result :Mass of lamp before burning = 46.50 gMass of lamp after burning = 46.15 gInitial temperature of water = 29.00CHighest temperature of water = 59.00C

CalculationStep 1 :Calculate the heatchange using theformula H = mc

Changes of temperature, = 59.00C - 29.00C = 300CMass of water, m = 200 g

Heat change , H = mcc = 4.2 J g-1 oC-1

= 300CH = .. J

Step 2:Write a balancechemical equationfor the combustionof ethanol

C2H5OH + O2

Step 3 :Calculate thenumber of molesof ethanol that isused in theexperiment,

Mass of ethanol burnt/used = 46.50 - 46.15 = 0.35 g

Number of moles of ethanol burnt = 0.35 = 0.35molar mass ..

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[ Relative atomicmass : H, 1 : C,12;O,16 ]

.. mol

Step 4Calculate the heat

of combustion ofethanol, H

Heat of reaction , H = heat change ( answer from step 1 )

Number of moles of ethanol

= J

H = - kJ mol - 1

Draw an energy level diagram for the combustion of ethanol in thisexperiment

EXTRA ACTIVITY ( Refer to the Chemistry Form Five text book page 149 - 168 )

1 The thermochemical equation for the precipitation of calcium carbonate is given below,Ca2+ (aq) + CO32- (aq) CaCO3 (s) H = +12.6 kJ mol - 1

In an experiment, calcium carbonate, CaCO3 is precipitated when 100 cm3 of 0.5 mol dm-3

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calcium nitrate, Ca(NO3)2 solution is added to 100 cm3 of 0.5 mol dm-3 sodium carbonate,Na2CO3 solution.What is the temperature change in the reacting mixture ?[ Specific heat capacity of solution : 4.2 J g-1 oC-1 . Density of solution : 1 g cm -3 ]

SolutionStep 1 : Calculate the number of moles of precipitate formed :

Number of moles of calcium ion, Ca2+ = Number of moles of calcium nitrate, Ca(NO3)2solution

= . mol

Number of moles of carbonate ion, CO32- = Number of moles of sodium carbonate,Na2CO3 solution

= .mol

Based on the ionic equation : Ca2+ (aq) + CO32- (aq) CaCO3 (s)1 mol 1 mol 1 mol

1 mol of calcium ion, Ca2+ reacts with 1 mol of carbonate ion, CO32- to

form . mol of calcium carbonate, CaCO3

From the calculation : . mol of calcium ion, Ca2+ reacts with . mol of

carbonate ion, CO32- to form . mol of calcium carbonate, CaCO3

Step 2 : Calculate the heat change by using the heat of precipitation,Given that H = +12.6 kJ mol - 1 , this means that when 1 mol of calcium carbonate,

CaCO3 is precipitated, the heat absorbed is

Therefore when ..mol (from step 1 ) of calcium carbonate, CaCO3 pricipitated,

the heat absorbed = kJ

= ..J --------- (i)

Step 3: Calculate the temperature change by using the heat changeMass of solution , m ..g

Heat change (absorbed ) ,H = J ( from (i) )

H = mc = . x . x = J

= 0C

2 In an experiment, excess magnesium powder is added to 50 cm3 of 0.5 mol dm-3iron(ll) sulphate solution at 29.0 0C. The thermochemical equation is shown below,

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

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What is the highest temperature reached in this experiment ?

Solution

Step 1 : Calculate the number of moles of iron, Fe displaced

The number of moles of iron(ll) ions , Fe2+ = the number of moles of iron(ll) sulphatesolution

=. molBased on the ionic equation :

Mg(s) + Fe2+ (aq) Mg2+ (aq) + Fe (s)1 mol 1 mol

1 mol of iron(ll) ion, Fe2+ produces 1 mol of iron, Fe

mol of iron(ll) ion, Fe2+ produces .mol of iron, Fe

Step 2 : Calculate the amount of heat given out from the heat of displacement

Given that H = -80.6 kJ mol - 1

This means that when 1 mol of iron, Fe is displaced, the heat given out is . kJ

Therefore when mol of iron, Fe is displaced in the experiment,

The heat given out , H = .kJ

= .J --------- (i)

Step 3 : Calculate the highest temperature reached using the heat given :

Mass of the solution, m = .g

The heat given out , H = . J ( from (i) )

H = mc = . x . x = J

= 0C ---------- (ii)

Therefore, the highest temperature reached :

= 29.00C + = 0C

3 The thermochemical equation for the reaction between ethanoic acid and sodium hydroxideis given below,

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

Calculate the heat given out when 200 cm3 of ethanoic acid 0.5 mol dm-3 is added to200 cm3 of sodium hydroxide 0.5 mol dm-3

SolutionStep 1: Calculate the number of moles of water produced

Number of moles of hydrogen ion, H+ = Number of moles of ethanoic acid solution

= . mol

Number of moles of hydrogen ion, OH- = Number of moles of sodium hydroxide solution

= . mol

The ionic equation : H+(aq) + OH- (aq) H2O (l)1 mol 1 mol 1 mol

Based on ionic equation, 1 mol of hydrogen ion, H+ reacts with 1 mol of hydroxide ion, OH-to produce 1 mol of water, H2O,

Therefore,

mol of hydrogen ion, H+ reacts with mol of hydroxide ion, OH-

to produce mol of water, H2O --------- (i)

Step 2 : Calculate the heat given out using the heat of neutralisation,

Given that H = -55 kJ mol - 1

This means that when 1 mol of water is produced , the heat given out is 55 kJ

Therefore, when .. ( from (i) ) is produced,

The heat given out = .kJ

= J

EXTRA EXERCISE

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1 The table below shows the heat of combustion for alcohols .

Type of alcohol Molecular formula Relative molecularmass

Heat of combustion/kJ mol-1

Methanol CH3OH 32 - 715Ethanol C2H5OH 46 - 1376

Propanol C3H7OH 60 - 2017Butanol C4H9OH 74 - 2679

Based on information given in the table answer the following questions,

(a) Draw a graph of the magnitude of the heat of combustion against the number of carbonatoms in an alcohol

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

.

.

(c) Predict the heat of combustion of hexanol

(d)

By using the formula below, calculate the fuel value of the following alcohols

(i) Methanol

(ii) Butanol

(e) Which alcohol is more effective to be used as fuel ? Give one reason for your answer

..

..2 Table shows three thermochemical equations for experiment l, ll and lll

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

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


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Experiment Thermochemical equation

l HCl (aq) + NaOH ( aq) NaCl (aq) + H2O(l) H = -57 kJ mol -1

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

lll H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) H = -114 kJ mol -1

Based on the thermochemical equations, answer the following questions

(a) Explain why there is a difference in the heat of reaction for(i) Experiment l and ll(ii) Experiment l and lll

[ 12 marks ](b) If experiment lll is repeated by replacing the sodium hydroxide solution with potassium

hydroxide solution,

(i) predict the heat of reaction for the experiment(ii) Explain your answer in (b) (i)[ 5 marks ]

(c) Based on thermochemical equation in experiment ll, calculate the heat change when100 cm3 2 mol dm-3 ethanoic acid solution react with 100 cm3 2 mol dm-3 sodiumhydroxide solution

[ 3 marks ]

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Waja Thermochemistry Edited