HEAT OF NEUTRALIZATION

OBJECTIVESOBJECTIVES

State what heat of neutralization isState what heat of neutralization isD t i th h t f t li tiD t i th h t f t li ti Determine the heat of neutralizationDetermine the heat of neutralization

Construct energy level diagrams for Construct energy level diagrams for neutralization reactionsneutralization reactionsneutralization reactionsneutralization reactions

Compare and explain the heat of Compare and explain the heat of neutralization of a strong acid and a neutralization of a strong acid and a ggstrong alkali with the heat of strong alkali with the heat of neutralization of weak acid and/or a weak neutralization of weak acid and/or a weak alkalialkalialkalialkali

Solve numerical problems related to the Solve numerical problems related to the heat of neutralizationheat of neutralizationheat of neutralizationheat of neutralization

D fi itiDefinition

Equation

Calculation

ComparisonComparison

What is heat of neutralization?What is heat of neutralization?

The heat of neutralization is theenergy change when one mole ofenergy change when one mole of

water is formed from thet li ti b t l fneutralization between one mole of

hydrogen ions, H+ from an acid and onemole of hydroxide ions, OH- from an alkali

Example of neutralization reactionExample of neutralization reaction

Hydrochloric acid reacts with sodium hydroxideto form sodium chloride and water

Chemical equation Chemical equationHCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Ionic equation Ionic equation(shows what really happens in this reaction)H+(aq) + OH-(aq) H2O(l)

one mole of and one mole of form one mole of hydrogen ions hydroxide ions water

Thermochemical equationq

i. From chemical equationHCl( ) N OH( ) N Cl( ) H O(l) ΔH 57 3kJHCl(aq) + NaOH(aq) NaCl(aq) + H2O(l);ΔH = - 57.3kJ

oror

ii. From ionic equationH+(aq) + OH-(aq) H2O(l);ΔH = - 57.3kJ

Ways to calculate heat of neutralizationWays to calculate heat of neutralization

E.g : Hydrochloric acid, HCl (2 mol dm-3, 25 cm3)g y , ( , )Sodium hydroxide, NaOH solution (2 mol dm-3, 25 cm3)Temperature (30.0°C 41.0°C)

1. Calculate the energy change, E=mcθm = total mass of reaction mixture (g) = total volume of mixture (cm3)c = specific heat capacity of aqueous reaction mixture (4.2Jg-1°C-1)θ = change in temperature (°C)

E = [(25+25)g X 4.2Jg-1°C-1 X (41.0-30.0)°C] J= 2310 J

2. Calculate the number of moles of hydrogen ions, H+

Number of moles of H+ = Number of moles of HCl = MVNumber of moles of H = Number of moles of HCl = MV M = molarity of the solution (mol dm-3) 1000V = volume of the solution (cm3)

MV = 2 X 25 = 0.05 mol1000 1000

3. Calculate the number of moles of hydroxide ions, OH-

Number of moles of OH- = Number of moles of NaOH = MVNumber of moles of OH = Number of moles of NaOH = MV1000

MV = 2 X 25 = 0.05 mol1000 10001000 1000

4. Calculate the heat given off when one mole of water is formed fromone mole of hydrogen ions H+ and one mole of hydroxide ions OH-one mole of hydrogen ions, H and one mole of hydroxide ions, OH0.05 mol H+ : 0.05 mol OH-

1 mol H+ : 1 mol OH-

0.05 mol H2O(l) formed1 mol H2O(l) formed release = mcθ J = 2310 J

number of mol 0.05 mol= 46200 Jmol-1

5. Determine the heat of neutralization

ΔH = -46200 Jmol-1 / -46.2kJmol-1

Heat of neutralization between different Heat of neutralization between different strength of acids and alkalisstrength of acids and alkalisstrength of acids and alkalisstrength of acids and alkalis

t lk li t idstrong alkali + strong acid

strong alkali + weak acidHeat of

i i

weak alkali + strong acidneutralization

decreases

k lk li + k idweak alkali + weak acid

strong alkali + strong acidstrong alkali + strong acid

Reaction

Energy Level DiagramEnergy Level Diagram

Why the heat of neutralization of the reaction is the highest?the reaction is the highest?

R tiReaction

Sodi m h dro ide reacts ith h drochloric acid to formSodium hydroxide reacts with hydrochloric acid to form sodium chloride and water

Thermochemical equation

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l);ΔH = - 57.3kJ

Energy (kJ)

NaOH(aq) + HCl(aq)

ΔH 57 3kJ

NaCl(aq) + H2O(l)

ΔH = - 57.3kJ

• The reaction between sodium hydroxide, NaOH and hydrochloric acid, HCl to form sodium chloride, NaCl and water, H2O is exothermic.form sodium chloride, NaCl and water, H2O is exothermic.

• When one mole of sodium hydroxide, NaOH reacts with one mole of hydrochloric acid, HCl to form one mole of sodium chloride, NaCl and one mole of water H2O the quantity of heat released is 57 3kJmole of water, H2O, the quantity of heat released is 57.3kJ.

• The total energy of one mole of sodium hydroxide, NaOH and one mole of hydrochloric acid, HCl is more than the total energy of one mole of sodium chloride NaCl and one mole of water H O The difference in energy is 57 3kJchloride, NaCl and one mole of water, H2O. The difference in energy is 57.3kJ.

• The temperature of the reaction mixture of sodium hydroxide, NaOH and hydrochloric acid, HCl will rise.

E l tiExplanationThe heat given out (heat of neutralization) when a strong acid reacts withstrong alkali is the higheststrong alkali is the highest

Hydrochloric acid HCl is a strong acid and sodium hydroxide Hydrochloric acid, HCl is a strong acid and sodium hydroxide, NaOH is a strong alkali, both undergo complete ionization in water to produce hydrogen ions, H+ and hydroxide ions, OH-.

H+(aq) + OH-(aq) H2O(l) Formation of one mol of water produces 57.3kJ energy.

strong alkali + weak acidstrong alkali + weak acid

Reaction

Energy Level DiagramEnergy Level Diagram

Why the heat of neutralization of the reaction decreases?

R tiReaction

Sodi m h dro ide reacts ith ethanoic acid to formSodium hydroxide reacts with ethanoic acid to form sodium ethanoate and water

Thermochemical equation

NaOH(aq) + CH3COOH(aq) CH3COONa(aq) + H2O(l);ΔH = - 54.6kJ

Energy (kJ)

ΔH 54 6kJ

NaOH(aq) + CH3COOH(aq)

ΔH = - 54.6kJ

CH3COONa(aq) + H2O(l)

• The reaction between sodium hydroxide, NaOH and ethanoic acid, CH3COOH to form sodium ethanoate, CH3COONa and water, H2O is exothermic.

• When one mole of sodium hydroxide, NaOH reacts with one mole of ethanoic acid, CH3COOH to form one mole of sodium ethanoate, CH3COONaand one mole of water, H2O, the quantity of heat released is 54.6kJ.

• The total energy of one mole of sodium hydroxide, NaOH and one mole of ethanoic acid, CH3COOH is more than the total energy of one mole of sodiumethanoate, CH3COONa and one mole of water, H2O. The difference in energy is 54.6kJ.

• The temperature of the reaction mixture of sodium hydroxide, NaOH and ethanoic, CH3COOH will rise.

• ExplanationThe heat given out (heat of neutralization) when a strong acid reacts with The heat given out (heat of neutralization) when a strong acid reacts with strong alkali is higher than the heat given out when a weak acid reacts with a strong alkali

Ethanoic acid, CH3COOH is a weak acid, undergo partial ionization i t t d h d i H+in water to produce hydrogen ions, H+.

Some of ethanoic acid still remains molecule when dissolves in water.water.

CH3COOH (aq) CH3COO- (aq) + H+ (aq) Some of heat given out during the neutralization is absorbed to

dissociate the acid completely in water. Thus, the heat of neutralization is always less than -57.3kJ.

weak alkali + strong acidweak alkali + strong acid

Reaction

Energy Level DiagramEnergy Level Diagram

Why the heat of neutralization of the reaction decreases?

R tiReaction

Ammoni m h dro ide reacts ith h drochloric acid to formAmmonium hydroxide reacts with hydrochloric acid to form ammonium chloride and water

Thermochemical equation

NH4OH(aq) + HCl(aq) NH4Cl(aq) + H2O(l);ΔH = - 42.2kJ

Energy (kJ)

ΔH 42 2kJ

NH4OH(aq) + HCl(aq)

ΔH = - 42.2kJ

NH4Cl(aq) + H2O(l)

• The reaction between ammonium hydroxide, NH4OH and hydrochloric acid, HCl to form ammonium chloride, NH4Cl and water, H2O is exothermic.Wh l f i h d id NH OH i h l f• When one mole of ammonium hydroxide, NH4OH reacts with one mole of hydrochloric acid, HCl to form one mole of ammonium chloride, NH4Cl and one mole of water, H2O, the quantity of heat released is 42.2kJ.

• The total energy of one mole of ammonium hydroxide, NH4OH and one mole of hydrochloric acid, HCl is more than the total energy of one mole of ammonium chloride, NH4Cl and one mole of water, H2O. The difference in energy is 42.2kJ.

• The temperature of the reaction mixture of ammonium hydroxide, NH4OH and hydrochloric acid, HCl will rise.

• ExplanationThe heat given out (heat of neutralization) when a strong acid reacts with strong alkali is higher than the heat given out when a weak alkali reacts with a strong acidg

Ammonium hydroxide, NH4OH is a weak alkali, undergo partial ioni ation in ater to prod ce h dro ide ions OH-ionization in water to produce hydroxide ions, OH-.

Some of ammonium hydroxide still remains molecule when dissolves in water.

NH4OH (aq) NH4+ (aq) + OH- (aq)

Some of heat given out during the neutralization is absorbed to dissociate the alkali completely in waterdissociate the alkali completely in water.

Thus, the heat of neutralization is always less than -57.3kJ.

weak alkali + weak acidweak alkali + weak acid

Reaction

Energy Level DiagramEnergy Level Diagram

Why the heat of neutralization of the reaction is the least?

R tiReaction

Ammoni m h dro ide reacts ith ethanoic acid to formAmmonium hydroxide reacts with ethanoic acid to form ammonium ethanoate and water

Thermochemical equation

NH4OH(aq) + CH3COOH(aq) CH3COONH4 (aq) + H2O(l);ΔH = - 40.8kJ

Energy (kJ)

ΔH 40 8kJ

NH4OH(aq) + CH3COOH(aq)

ΔH = - 40.8kJ

CH3COONH4(aq) + H2O(l)

• The reaction between ammonium hydroxide, NH4OH and ethanoic acid, CH3COOH to form ammonium ethanoate, CH3COONH4 and water, H2O is exothermic., 3 4 , 2

• When one mole of ammonium hydroxide, NH4OH reacts with one mole of ethanoic acid, CH3COOH to form one mole of ammonium ethanoate, CH3COONH4and one mole of water, H2O, the quantity of heat released is 40.8kJ.

• The total energy of one mole of ammonium hydroxide, NH4OH and one mole of ethanoic acid, CH3COOH is more than the total energy of one mole of ammoniumethanoate, CH3COONH4 and one mole of water, H2O. The difference in energy i 40 8kJis 40.8kJ.

• The temperature of the reaction mixture of ammonium hydroxide, NH4OH and ethanoic acid, CH3COOH will rise.

• ExplanationThe heat given out (heat of neutralization) when a weak acid reacts with a weak acid is the least

Ethanoic acid, CH3COOH is a weak acid and ammonium hydroxide, NH4OH is a weak alkali, both undergo partial ionization in water to 4 , g pproduce hydrogen ions, H+ and hydroxide ions, OH-.

Both of ethanoic acid and ammonium hydroxide still remains l l h di l i tmolecule when dissolves in water.CH3COOH (aq) CH3COO- (aq) + H+ (aq)

NH4OH (aq) NH4+ (aq) + OH- (aq)NH4OH (aq) NH4 (aq) + OH (aq)

The heat given out during the neutralization is absorbed to dissociate the acid and alkali completely in water.

Thus, the heat of neutralization is always less than -57.3kJ.

EXPERIMENT

AimTo determine the heat of neutralization for variousTo determine the heat of neutralization for varioustypes of neutralization reactions.

Problem statementIs the heat of neutralization the same or different forvarious types of neutralization reactions?

H th iHypothesisThe heat of neutralization is different for acids and alkalisof different strengthsof different strengths.

VariablesManipulated variable : Acids and alkalis of different strengthsResponding variable : Heat of neutralizationFixed variable : Volume and concentration of solution,

type of container usedtype of container used

Materials2 l d 3 dil h d hl i id HCl 2 mol dm-3 dilute hydrochloric acid, HCl

2 mol dm-3 dilute ethanoic acid, CH3COOH solution 2 mol dm-3 dilute sodium hydroxide, NaOH solutiony 2 mol dm-3 dilute ammonium hydroxide, NH4OH solution

ApparatusApparatus Polystyrene cups Thermometer

M i li d Measuring cylinder

Procedure

1. Measure 25 cm3 of 2 mol dm-3 hydrochloric acid, HCl usingmeasuring cylinder and pour it into a polystyrene cup.measuring cylinder and pour it into a polystyrene cup.

2. Put a thermometer in the polystyrene cup and left it for twominutes. Then, record the initial temperature in a table.

3. Measure 25 cm3 of 2 mol dm-3 sodium hydroxide, NaOHsolution and pour it into another polystyrene cup. Measure theinitial temperature of the solution as in Step 2 Record theinitial temperature of the solution as in Step 2. Record theinitial temperature in a table.

4. Pour the sodium hydroxide, NaOH solution quickly andcarefully into the first polystyrene cup.

5. Stir the reaction mixture with the thermometer to mix the reactantsthe reactants.

6. Record the highest temperature reached in the table.7. Repeat Step 1 to 6 by using combination of acid and alkali

as listed in the table below.

Figure 1.1 Determining the heat of neutralization ofsodium chloride, NaCl

l tisolution

Tabulation of dataInitial Initial Average initial Highest Temperature

ExperimentInitial

temperature of acid (°C)

Initial temperature of alkali (°C)

Average initialtemperature of

reactionmixture,θ1 (°C)

Highest temperature of reaction mixture, θ2

(°C)

Temperaturechange,θ(θ2 - θ1)

(°C)1 ( )

Sodium hydroxide,NaOH solution +hydrochloric acid,

HCl

Sodium hydroxide,NaOH solution +ethanoic acid,

CH3COOH

Ammounim hydroxide,

NH OH solution +NH4OH solution +hydrochloric acid,

HCl Ammounim hydroxidehydroxide,

NH4OH solution +ethanoic acid,

CH3COOH

Discussions

1. Why theoretical value different from value obtained? If less than…

(i) Maybe some of the heat is absorbed by the container.(i) Maybe some of the heat is absorbed by the container.(ii) Maybe heat loss to the surroundings.

If more than…(i) M b th t ti f th id d th lk li i thi(i) Maybe the concentration of the acid and the alkali in this

experiment might be more than 2.0 mol dm-3 during thepreparation.

(ii) Maybe have parallax error during taking the reading.

2 Why is cup made of polystyrene used in this experiment?2. Why is cup made of polystyrene used in this experiment?To prevent heat loss to the surroundings because it is a heat

insulator.

3. State two precautions that can increase the accuracy of results in this experiment.p(i) Mix the reactants quickly so that the reaction can be

completed faster.(ii) The reading of the thermometer should be observed until(ii) The reading of the thermometer should be observed until

the highest temperature is recorded.

4. Why the heat of neutralization has a negative sign?The reaction gives out heat that results in the increase of temperature of the products formed.

ConclusionTh h t f t li ti f id d lk li f diff tThe heat of neutralization for acids and alkalis of differentstrengths are different.

h dThe End