HEAT OF NEUTRALIZATION
OBJECTIVES State what heat of neutralization is Determine D t i the th heat h t off neutralization t li ti Construct energy level diagrams for neutralization reactions Compare and explain the heat of neutralization of a strong g acid and a strong alkali with the heat of neutralization of weak acid and/or a weak alkali Solve numerical problems related to the heat of neutralization
D fi iti Definition Equation Calculation Comparison
What is heat of neutralization? The heat of neutralization is the energy change when one mole of water is formed from the neutralization t li ti b between t one mole l off hydrogen ions, H+ from an acid and one mole of hydroxide ions, OH- from an alkali
Example of neutralization reaction Hydrochloric acid reacts with sodium hydroxide to form sodium chloride and water
Chemical equation HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
Ionic equation (shows what really happens in this reaction) H+(aq) + OH-(aq) H2O(l)
one mole of and hydrogen ions
one mole of form hydroxide ions
one mole of water
Thermochemical equation q
i. From chemical equation HCl( ) + N HCl(aq) NaOH(aq) OH( ) N NaCl(aq) Cl( ) + H2O(l);Δ O(l) ΔH = - 57.3kJ 57 3kJ or ii. From ionic equation H+(aq) + OH-(aq) H2O(l);ΔH = - 57.3kJ
Ways to calculate heat of neutralization E.g g : Hydrochloric y acid,, HCl (2 ( mol dm-3, 25 cm3) Sodium hydroxide, NaOH solution (2 mol dm-3, 25 cm3) Temperature (30.0°C 41.0°C) 1. Calculate the energy change, E=mcθ
m = 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 = MV M = molarity of the solution (mol dm-3) 1000
V = volume of the solution (cm3) MV = 2 X 25 = 0.05 mol 1000 1000 3. Calculate the number of moles of hydroxide ions, OHNumber of moles of OH- = Number of moles of NaOH = MV 1000 MV = 2 X 25 = 0.05 mol 1000 1000
4. Calculate the heat given off when one mole of water is formed from one mole of hydrogen ions, ions H+ and one mole of hydroxide ions, ions OH0.05 mol H+ : 0.05 mol OH1 mol H+ : 1 mol OH0.05 mol H2O(l) formed 1 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 strength of acids and alkalis strong t alkali lk li + strong t acid id
strong alkali + weak acid
weak alkali + strong acid
weak k alkali lk li + weak k acid id
Heat of neutralization i i decreases
strong alkali + strong acid
Reaction
Energy Level Diagram
Why the heat of neutralization of the reaction is the highest?
R ti Reaction Sodium Sodi m hydroxide h dro ide reacts with ith hydrochloric h drochloric 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 57 3kJ NaCl(aq) + H2O(l)
• The reaction between sodium hydroxide, NaOH and hydrochloric acid, HCl to 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, water H2O, O the quantity of heat released is 57.3kJ. 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, chloride, water H2O. O The difference in energy is 57.3kJ. 57 3kJ • The temperature of the reaction mixture of sodium hydroxide, NaOH and hydrochloric acid, HCl will rise.
E l ti Explanation The heat given out (heat of neutralization) when a strong acid reacts with strong alkali is the highest Hydrochloric acid, acid HCl is a strong acid and sodium hydroxide, 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 acid
Reaction
Energy Level Diagram Why the heat of neutralization of the reaction decreases?
R ti Reaction Sodium Sodi m hydroxide h dro ide reacts with ith ethanoic acid to form sodium ethanoate and water
Thermochemical equation NaOH(aq) + CH3COOH(aq)
CH3COONa(aq) + H2O(l);ΔH = - 54.6kJ
Energy (kJ)
NaOH(aq) + CH3COOH(aq) ΔH = - 54.6kJ 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, CH3COONa and 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 sodium ethanoate, 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.
• Explanation 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 water in t to t produce d hydrogen h d ions, i H+ . Some of ethanoic acid still remains molecule when dissolves in 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 acid
Reaction
Energy Level Diagram Why the heat of neutralization of the reaction decreases?
R ti Reaction Ammonium Ammoni m hydroxide h dro ide reacts with ith hydrochloric h drochloric acid to form ammonium chloride and water
Thermochemical equation NH4OH(aq) + HCl(aq)
NH4Cl(aq) + H2O(l);ΔH = - 42.2kJ
Energy (kJ)
NH4OH(aq) + HCl(aq) ΔH = - 42.2kJ 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. • When Wh one mole l off ammonium i hydroxide, h d id NH4OH reacts with i h one mole l off 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.
• Explanation 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 alkali reacts with a strongg acid Ammonium hydroxide, NH4OH is a weak alkali, undergo partial ioni ation in water ionization ater to produce prod ce hydroxide h dro ide ions, 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 water. water Thus, the heat of neutralization is always less than -57.3kJ.
weak alkali + weak acid
Reaction
Energy Level Diagram Why the heat of neutralization of the reaction is the least?
R ti Reaction Ammonium Ammoni m hydroxide h dro ide reacts with ith ethanoic acid to form ammonium ethanoate and water
Thermochemical equation NH4OH(aq) + CH3COOH(aq)
CH3COONH4 (aq) + H2O(l);ΔH = - 40.8kJ
Energy (kJ)
NH4OH(aq) + CH3COOH(aq) ΔH = - 40.8kJ 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. • When one mole of ammonium hydroxide, NH4OH reacts with one mole of ethanoic acid, CH3COOH to form one mole of ammonium ethanoate, CH3COONH4 and 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 ammonium ethanoate, CH3COONH4 and one mole of water, H2O. The difference in energy i 40.8kJ. is 40 8kJ • The temperature of the reaction mixture of ammonium hydroxide, NH4OH and ethanoic acid, CH3COOH will rise.
• Explanation The 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 g ppartial ionization in water to produce hydrogen ions, H+ and hydroxide ions, OH-. Both of ethanoic acid and ammonium hydroxide still remains molecule l l when h dissolves di l in i water. t CH3COOH (aq) CH3COO- (aq) + H+ (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
Aim To determine the heat of neutralization for various types of neutralization reactions. Problem statement Is the heat of neutralization the same or different for various types of neutralization reactions? Hypothesis H th i The heat of neutralization is different for acids and alkalis of different strengths. strengths
Variables Manipulated variable : Acids and alkalis of different strengths Responding variable : Heat of neutralization Fixed variable : Volume and concentration of solution, type of container used Materials 2 moll dm d -33 dilute dil h hydrochloric d hl i acid, id HCl 2 mol dm-3 dilute ethanoic acid, CH3COOH solution 2 mol dm-3 dilute sodium hydroxide, y NaOH solution 2 mol dm-3 dilute ammonium hydroxide, NH4OH solution Apparatus Polystyrene cups Thermometer Measuring M i cylinder li d
Procedure 1. Measure 25 cm3 of 2 mol dm-3 hydrochloric acid, HCl using measuring cylinder and pour it into a polystyrene cup. 2. Put a thermometer in the polystyrene cup and left it for two minutes. Then, record the initial temperature in a table. 3. Measure 25 cm3 of 2 mol dm-3 sodium hydroxide, NaOH solution and pour it into another polystyrene cup. Measure the initial temperature of the solution as in Step 2. 2 Record the initial temperature in a table. 4. Pour the sodium hydroxide, NaOH solution quickly and carefully into the first polystyrene cup.
5. Stir the reaction mixture with the thermometer to mix the reactants. 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 of sodium chloride, NaCl solution l ti
Tabulation of data Experiment
Sodium hydroxide, NaOH solution + hydrochloric acid, HCl Sodium hydroxide, NaOH solution + ethanoic acid, CH3COOH Ammounim hydroxide, NH4OH solution + hydrochloric acid, HCl Ammounim hydroxide, hydroxide NH4OH solution + ethanoic acid, CH3COOH
Initial temperature of acid (°C)
Initial temperature of alkali (°C)
Average initial temperature of reaction mixture, θ1 ((°C))
Highest temperature of reaction mixture, θ2 (°C)
Temperature change, θ(θ2 - θ1) (°C)
Discussions 1. Why theoretical value different from value obtained? If less than… (i) Maybe some of the heat is absorbed by the container. (ii) Maybe heat loss to the surroundings. If more than… (i) Maybe M b th the concentration t ti off th the acid id and d th the alkali lk li in i this thi experiment might be more than 2.0 mol dm-3 during the preparation. (ii) Maybe have parallax error during taking the reading. 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 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.
Conclusion The h Th heatt off neutralization t li ti for f acids id and d alkalis lk li off diff differentt strengths are different.
The h End d
