What is heat of precipitation?

What is heat of precipitation?

 

Heat of precipitation:

  • When two aqueous solutions are added together and a precipitate is formed, this reaction is called a precipitation reaction or double decomposition. This reaction is used to prepare insoluble salts.
  • The heat given out in a precipitation reaction is called the heat of precipitation.
  • The heat of precipitation is the heat change when one mole of a precipitate is formed from its ions in aqueous solution under standard conditions.

Some examples of precipitation reactions are:

  • When barium chloride solution is added to sodium sulphate solution, a white precipitate, barium sulphate, is formed.
    BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
    Ba2+(aq) + SO42-(aq) → BaSO4(s)
  • When silver nitrate solution is added to hydrochloric acid, a white precipitate, silver chloride, is formed.
    AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
    Ag+(aq) + Cl(aq) → AgCl(s)
  • When lead(II) nitrate solution is added to potassium iodide solution, a yellow precipitate, lead(II) iodide, is formed.
    Pb(NO3)2(aq) + 2KI(aq) → PbI+ 2KNO3(aq)
    Pb2+(aq) + 2I(aq) → PbI2(s)

People also ask

Heat of precipitation of silver chloride experiment

Aim: To determine the heat of precipitation of silver chloride.
Materials: 0.5 mol dm-3 silver nitrate solution, 0.5 mol dm-3 sodium chloride solution.
Apparatus: Thermometer, plastic cups with covers, 50 cm3 measuring cylinders.
Caution:
Add the two solutions together as quickly as possible.
Stir the mixture throughout the activity.
Safety measures:
Handle the chemicals with caution. Wear eye protection.
Silver nitrate solution may cause dark stains on your skin and clothing.
Procedure:
What is heat of precipitation 1

  1. 20 cm3 of 0.5 mol dm-3 silver nitrate solution is measured and poured into a plastic cup.
  2. The initial temperature of the silver nitrate solution is measured after a few minutes.
  3. 20 cm3 of 0.5 mol dm-3 sodium chloride solution is measured.
  4. The initial temperature of the sodium chloride solution is measured after a few minutes.
  5. The sodium chloride solution is added quickly and carefully to the silver nitrate solution. The mixture is stirred with the thermometer.
  6. The highest temperature of the reacting mixture is measured and recorded.

Results:
What is heat of precipitation 2

Interpreting data:
What is heat of precipitation 3

Discussion:

  1. When silver nitrate solution and sodium chloride solution are added, a white precipitate is formed. The precipitate is silver chloride.
  2. The chemical equation for the reaction is:
    AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
  3. The ionic equation for the reaction is:
    Ag+(aq) + Cl(aq) → AgCl(s)
  4. The thermochemical equation for the precipitation of silver chloride is:
    AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)    ΔH = -0.1 y kJ mol-1
    or
    Ag+(aq) + Cl(aq) → AgCl(s)    ΔH = -0.1 y kJ mol-1
  5. The energy level diagram for the precipitation of silver chloride is:
    What is heat of precipitation 4
  6. The theoretical value for the heat of precipitation for silver chloride is -65.5 kJ mol-1.
  7. The value for the heat of precipitation of silver chloride obtained from this activity is always less than the theoretical value. This is because in the computation of the theoretical value of the heat of precipitation, it is assumed that no heat is lost to the surroundings and the plastic cup does not absorb heat.
  8. The following are some precautions that need to be taken in this activity.
    (a) The two solutions are mixed quickly to avoid too much heat loss to the surroundings.
    (b) A plastic cup is used because plastic is a good insulator of heat. This will reduce heat loss to the surroundings.
    (c) The initial temperatures of the silver nitrate solution and sodium chloride solution are taken after a few minutes to let the solutions achieve a uniform temperature.
    (d) The reacting mixture is stirred slowly throughout the activity to ensure the temperature of the mixture is uniform.
    (e) The reading of the thermometer should be observed until the highest temperature is recorded.
  9. If the activity is repeated using twice the volume of silver nitrate solution and sodium chloride solution, but of the same concentration, the change in temperature will be the same. This is because:
    (a) When the volumes of both reactants are doubled, the number of moles of both reactants is also doubled. This makes the heat change double, 2H.
    (b) The mass of the solution is also doubled, 2m.
    (c) Heat change, H = mcθ
    When the volumes of both reactants are doubled, heat change, 2H = 2mcθ.
    Cancelling the number 2 from both sides of the equation gives H = mcθ.
    Thus, θ remains the same.
    If the sodium chloride solution is replaced with hydrochloric acid of the same concentration, the heat of precipitation is still the same. This is because both sodium chloride solution and hydrochloric acid provide the same amount of chloride ions. Sodium ions, Na+ and hydrogen ions, H+ are spectator ions. The reaction involves Ag+ ions and Cl ions.

Conclusion:
The heat of precipitation for silver chloride is -0.1 y kJ mol-1.

How to calculate heat of precipitation example problems with solutions

1. 50 cm3 of 0.5 mol dm-3 silver nitrate solution at 29.5°C is added to 50 cm3 of 0.5 mol dm-3 potassium chloride solution which is at a temperature of 28.5°C. The mixture is stirred and the highest temperature reached is 32.0°C. Calculate the heat of precipitation for silver chloride.
[Specific heat capacity of solution: 4.2 J g-1 °C-1. Density of solution: 1 g cm-3]
Solution:
What is heat of precipitation 5
What is heat of precipitation 6

2. The thermochemical equation for the precipitation of lead(II) sulphate is given below.
Pb2+(aq) + SO42-(aq) → PbSO4(s)     ΔH = -50.4 kJ
When 100 cm3 of 0.5 mol dm-3 lead(II) nitrate solution is added to 100 cm3 of 0.5 mol dm-3 sodium sulphate solution, lead(II) sulphate is precipitated. What is the temperature change in the reacting mixture in the experiment?
[Specific heat capacity of solution = 4.2 J g-1 °C-1. Density of solution = 1 g cm-3]
Solution:
What is heat of precipitation 7

3. When 50 cm3 of 2.0 mol dm-3 lead(II) nitrate solution is added to 50 cm3 of 2.0 mol dm-3 potassium sulphate solution, there is an increase of 10°C in the temperature.
What is the change in temperature if
(a) 100 cm3 of 2.0 mol dm-3 lead(II) nitrate solution is added to 100 cm3 of 2.0 mol dm-3 potassium sulphate solution?
(b) 50 cm3 of 1.0 mol dm-3 lead(II) nitrate solution is added to 50 cm3 of 1.0 mol dm-3 potassium sulphate solution?
(c) 50 cm3 of 2.0 mol dm-3 lead(II) nitrate solution is added to 50 cm3 of 1.0 mol dm-3 potassium sulphate solution?
(d) 50 cm3 of 2.0 mol dm-3 potassium sulphate solution is replaced by 50 cm3 of 2.0 mol dm-3 sodium sulphate solution?
Solution:
(a)
The volumes of both reactants are doubled.
By calculation:
When 50 cm3 of 2.0 mol dm-3 of lead(II) nitrate solution and 50 cm3 of 2.0 mol dm-3 potassium sulphate solution are used,
What is heat of precipitation 8
When 100 cm3 of 2.0 mol dm-3 of lead(II) nitrate solution and 100 cm3 of 2.0 mol dm-3 potassium sulphate solution are used,
What is heat of precipitation 9
By deduction:
When the volumes of both reactants are doubled, the number of moles of both reactants is also doubled. This makes the heat change double, 2H.
The mass of the solution is also doubled, 2m.
What is heat of precipitation 10

(b) The concentration of both reactants are halved.
By calculation:
When 50 cm3 of 1.0 mol dm-3 lead(II) nitrate solution and 50 cm3 of 1.0 mol dnr3potassium sulphate solution are used,
What is heat of precipitation 11
By deduction:
When the concentrations of both reactants are halved, the number of moles of both reactants is also halved.
What is heat of precipitation 12

(c) The concentration of potassium sulphate solution is halved. When 50 cm3 of 2.0 mol dm-3 lead(II) nitrate solution and 50 cm3 of 1.0 mol dm-3 potassium sulphate solution are used,
What is heat of precipitation 13
Potassium sulphate is the limiting substance, thus the calculation is based on 0.05 mole of SO42- ions.
Since the number of moles of substances is halved, the heat produced is also halved.
Mass of solution is the same, thus
What is heat of precipitation 14

(d) 50 cm3 of 2.0 mol dm-3 of potassium sulphate solution is replaced by 50 cm3 of 2.0 mol dm-3 sodium sulphate solution.
This reaction involves the reaction between lead(II) ions and sulphate ions to form lead(II) sulphate as shown below.
Pb2+(aq) + SO42-(aq) → PbSO4(s)
Both potassium sulphate solution and sodium sulphate solution provide the same number of moles of sulphate ions.
Thus, the temperature change is the same, θ = 10°C

What is the heat of combustion?

What is the heat of combustion?

 

What is the definition of enthalpy of combustion?

Heat of Combustion:

  • A fuel is a chemical substance that burns in oxygen to produce heat energy.
  • Combustion is a chemical reaction between a fuel and oxygen to release heat. Combustion is always an exothermic reaction.
  • The heat of combustion is the heat produced when one mole of a substance is completely burnt in oxygen under standard conditions.
  • The substances can be elements or compounds.
    (a) Combustion of elements
    C(s) + O2(g) → CO2(g)    ΔH = -394 kJ
    When 1 mole of carbon burns completely in oxygen to form carbon dioxide, 394 kJ of heat is released. The heat of combustion of carbon is -394 kJ mol-1. The energy level diagram for the combustion of carbon is as shown below.
    What is the heat of combustion 1
    (b) Combustion of compounds
    CH4(g) + O2(g) → CO2(g) + 2H2O(l)
    When 1 mole of methane burns completely in oxygen to form carbon dioxide and water, 890 kJ of heat is released. The heat of combustion of methane is -890 kJ mol-1. The energy level diagram for the combustion of methane is as shown in Figure.
    What is the heat of combustion 2
  • Excess oxygen is necessary during combustion to make sure that the combustion is complete.
  • Incomplete combustion due to insufficient oxygen may result in different products being formed, and thus the heat given out would be different.
    (a) In excess oxygen, 1 mole of carbon burns completely to form carbon dioxide and produce 394 kJ of heat.
    C(s) + O2(g) → CO2(g)    ΔH = -394 kJ
    (b) In limited supply of oxygen, 1 mole of carbon burns to form carbon monoxide and produce 108 kJ of heat.
    C(s) + ½ O2(g) → CO2(g)    ΔH = -108 kJ
    Note: The heat of combustion of carbon is -394 kJ mol-1 and not -108 kJ mol-1.
  • When writing thermochemical equations, we can use fractions to represent the number of moles of oxygen required so that the number of moles of the substance is always 1 mole.
  • Different fuels have different values of heat of combustion.
  • The value of heat of combustion is an important factor for us to decide what fuel to use for a certain purpose.

What is the heat of combustion 3

People also ask

Determination of heat combustion by bomb calorimeter

Determining the heat of combustion:

  1. The heat of combustion of a fuel can be determined accurately by using a bomb calorimeter.
  2. An insulated container is filled with a known quantity of water. A small sample of known mass in the steel bomb is ignited electrically and burnt completely. The heat given out in the reaction is determined from the temperature rise in the water that surrounds the bomb.
  3. The bomb calorimeter is specifically designed to minimise heat loss to the surroundings.
    What is the heat of combustion 4

In the laboratory, the heat of combustion of a fuel can be determined as follows.

  • A liquid fuel is burnt in excess oxygen.
  • The mass and the number of moles of the fuel are determined.
  • The heat produced is used to heat a known mass of water.
  • The increase in the temperature of the water is measured.
  • The heat absorbed by the water is calculated using the formula H = mcθ.
  • The heat given out by the fuel is calculated. We assume that all the heat given out during the combustion of the fuel is absorbed by the water.
    Heat given out during combustion of the fuel = heat absorbed by the water
  • The heat of combustion (heat given out by 1 mole of the fuel) is calculated.

Heat of combustion of various alcohols

One example of liquid fuel is alcohol. Different members of the alcohol family have different heat of combustion. The table below shows the heat of combustion of some alcohols.

AlcoholMolecular
formula
Number of carbon atoms per moleculeNumber of hydrogen atoms per moleculeRelative molecular massHeat of combustion (kJ mol-1)
MethanolCH3OH1432-728
EthanolC2H5OH2646-1376
Propan-1-olC3H7OH3860-2016
Butan-1-olC4H9OH41074-2678
Pentan-1-olC5H11OH51288-3332
Hexan-1-olC6H13OH614102-3981

Table shows that the heat of combustion of alcohol increases as the

  • number of carbon atoms per molecule increases
  • number of hydrogen atoms per molecule increases
  • relative molecular mass increases

During the combustion of alcohol:

  • Carbon atom is burnt to form carbon dioxide.
    [C] + O2(g) → CO2(g) + heat
  • Hydrogen atom is burnt to form water.
    [H] + ½ O2(g) → H2O(g) + heat
  • Both the reactions are exothermic. Therefore, more heat is produced when more carbon and hydrogen atoms are burnt.
  • Relative molecular mass is proportional to the number of carbon and hydrogen atoms per molecule. Thus, the heat of combustion of alcohol increases as the relative molecular mass increases.

The difference in the heat of combustion of successive members is almost the same, that is, about 650 kJ.

Pair of alcoholsDifference in the heat of combustion (kJ)
Methanol and ethanol1376 – 728 = 648
Ethanol and propan-1-ol2016 – 1376 = 640
Propan-1-ol and butan-1-ol2678 – 2016 = 662
Butan-1-ol and pentan-1-ol3332 – 2678 = 654
Pentan-1-ol and hexan-1-ol3981 – 3332 = 649

(a) Successive members of the homologous series of alcohols differ from each other by a -CH2 group.
(b) The constant increase in the heat of combustion of the successive members of alcohol is due to the extra heat given out by the extra one carbon atom and two hydrogen atoms in the -CH2 group.

When the heat of combustion is plotted against the number of carbon atoms per alcohol molecule, the graph in Figure is obtained.
What is the heat of combustion 5

Heat of combustion of alcohols experiment

Aim: To investigate whether an alcohol with a higher number of carbon atoms per molecule has a higher heat of combustion.
Problem statement: Does an alcohol with a higher number of carbon atoms per molecule have a higher heat of combustion?
Hypothesis: The higher the number of carbon atoms per alcohol molecule, the higher is the heat of combustion.
Variables:
(a) Manipulated variable : Different types of alcohols/Number of carbon atoms per molecule of alcohol
(b) Responding variable : Heat of combustion
(c) Controlled variables : Volume of water, copper can, thermometer
Materials: Methanol, ethanol, propan-1-ol, butan-1-ol, water.
Apparatus: Copper can, tripod stand, thermometer (0 – 100°C), 100 cm3 measuring cylinder, spirit lamps, electronic balance, pipe-clay triangle, windshield, wooden block.
Procedure:

  1. 200 cm3 of water is measured using a measuring cylinder and poured into a copper can.
  2. The copper can is placed on a tripod stand.
  3. The initial temperature of the water is measured and recorded.
  4. A windshield is placed as shown in Figure.
    What is the heat of combustion 6
  5. About 50 cm3 of methanol is poured into a spirit lamp and the mass of the lamp and its contents is recorded.
  6. The lamp is put under the copper can as shown in Figure and the wick of the lamp is lighted immediately.
  7. The water is stirred throughout the experiment.
  8. When the temperature of the water increases by 30°C, the flame is put off and the highest temperature reached by the water is recorded.
  9. The mass of the lamp and its contents is weighed immediately and recorded.
  10. Steps 1 to 9 are repeated using ethanol, propan-1-ol and butan-1-ol to replace methanol with other factors remain unchanged.

Results:
What is the heat of combustion 7

Interpreting data:
1. Heat of combustion of methanol
What is the heat of combustion 8
What is the heat of combustion 9
2. Similarly, the heats of combustion of ethanol, propan-1 -ol and butan-1 -ol can be calculated.
Discussion:

  1. Combustion of alcohol is an exothermic reaction, thus heat is given out.
  2. An alcohol is a compound containing carbon, hydrogen and oxygen.
  3. The heat of combustion of each alcohol depends on the number of carbon and hydrogen atoms in the molecular formula of the alcohol molecule.
  4. The higher the number of carbon and hydrogen atoms per alcohol molecule, the higher is the heat of combustion.
  5. The heat of combustion obtained from the experiment is always less than the theoretical value. This is because:
    (a) The combustion of alcohol is not always complete. Instead of carbon dioxide and water being formed, carbon and carbon monoxide might be formed due to the incomplete combustion.
    (b) Some heat is lost to the surroundings during combustion.
    (c) A small quantity of the alcohol has been evaporated.
  6. The following precautions need to be taken in order to obtain a more accurate result.
    (a) The flame from the burning of alcohol must always touch the base of the copper can.
    (b) A thin copper can is used. Copper is a good conductor of heat, thus it can transfer the heat given out during the combustion of alcohol to the water.
    (c) A wire gauze is not used in the experiment because it might absorb some of the heat given out during the combustion.
    (d) When the flame has been put out, the spirit lamp must be weighed immediately because alcohols evaporate easily.
    (e) The water in the copper can must be stirred throughout the whole experiment to ensure that the temperature of the water is uniform.
    (f) A windshield is used to shield the flame from air currents. Air currents accelerate heat loss to the surroundings.

Conclusion:
The heat of combustion increases as the number of carbon and hydrogen atoms per alcohol molecule increases. Thus, the hypothesis is accepted.

How to calculate heat combustion example problems with solutions

1. Complete combustion of 1 mole of butan-l-ol, C4H9OH produces 2678 kJ of heat. Calculate the mass of butan-l-ol needed to burn completely in excess oxygen in order to raise the temperature of 500 cm3 of water by 35°C.
[Specific heat capacity of water: 4.2 J g-1 °C-1; density of water: 1 g cm-3]
Solution:

What is the heat of combustion 10
What is the heat of combustion 11

2. An experiment is carried out to determine the heat of combustion of ethanol, C2H5OH. The results of the experiment are shown below.
Volume of water used = 200 cm3
Initial temperature of water = 29.0°C
Highest temperature of water = 59.0°C
Mass of spirit lamp and ethanol before combustion = 245.85 g
Mass of spirit lamp and ethanol after combustion = 244.95 g
Based on the results, calculate the heat of combustion of ethanol and construct the energy level diagram for the complete combustion of ethanol.
[Specific heat capacity of water: 4.2 J g-1 °C-1; density of water: 1 g cm-3; relative atomic mass: H, 1; C, 12; O,16]
Solution:
What is the heat of combustion 12
What is the heat of combustion 13

What is the enthalpy of neutralization?

What is the enthalpy (heat) of neutralization?

 

  • Neutralisation is the reaction between an acid and a base to form a salt and water.
  • Some examples of neutralisation reaction are as follows.
    What is the enthalpy of neutralization 1
    What is the enthalpy of neutralization 2
  • During neutralisation reaction, hydrogen ions from acid react with hydroxide ions from alkali to form water.
    H+(aq) + OH(aq) → H2O(aq)
  • Since water is formed during neutralisation, heat of neutralisation is defined based on the formation of water.
  • The heat of neutralisation is the heat produced when one mole of water is formed from the reaction between an acid and an alkali.
  • Neutralisation is an exothermic reaction. Neutralisation always produces heat. Therefore, heat of neutralisation, AH is always negative.
  • The energy level diagram for a neutralisation reaction is as shown below.
    What is the enthalpy of neutralization 3

People also ask

Determine heat of neutralization of strong acid and strong base

Determining heat of neutralisation:

The heats of neutralisation between strong acids and strong alkalis are always the same.
What is the enthalpy of neutralization 4

  • The energy level diagram for neutralisation between a strong acid and a strong alkali is as shown below.
    What is the enthalpy of neutralization 5
  • The value of the heat of neutralisation depends on:
    (a) The basicity of the acid
    (b) The strength of the acid
    (c) The strength of the alkali
  • Basicity of the acid
    (a) Complete neutralisation of a strong diprotic acid with an alkali produces double amount of heat as compared to a strong monoprotic acid.
    What is the enthalpy of neutralization 6
    (b) This is because a diprotic acid like sulphuric acid produces two moles of hydrogen ions when it dissociates in water.
    What is the enthalpy of neutralization 7
    (c) Two moles of hydrogen ions produce two moles of water when reacted with hydroxide ions from an alkali.
    What is the enthalpy of neutralization 8
    (d) Sulphuric acid is also a strong acid and the heat of neutralisation is equal to -57.3 kJ mol-1.
    What is the enthalpy of neutralization 9
  • Strength of the acid
    (a) The heat given out when a strong acid reacts with a strong alkali is higher than the heat given out when a weak acid reacts with a strong alkali.
    What is the enthalpy of neutralization 10
    (b) Ethanoic acid is a weak acid that dissociates partially in water. Most of the ethanoic acid still exists as molecules when it dissolves in water.
    What is the enthalpy of neutralization 11
    (c) Some of the heat given out during the neutralisation is used to dissociate the acid completely in water, thus the heat given out is always less than 57.3 kJ.
  • Strength of the alkali
    (a) The heat given out when a strong acid reacts with a strong alkali is higher than the heat given out when a strong acid reacts with a weak alkali.
    What is the enthalpy of neutralization 12
    (b) Ammonia solution is a weak alkali which dissociates partially in water. Most of the ammonia still exists as molecules.
    (c) Some of the heat given out during the neutralisation is used to dissociate the alkali completely in water, thus the heat given out is always less than 57.3 kJ.
  • The heat of neutralisation between a weak acid and a weak alkali is the least. This is because more energy is needed to dissociate both the weak acid and the weak alkali completely to produce hydrogen ions and hydroxide ions which then react together to form one mole of water.
    What is the enthalpy of neutralization 13

Heat of Neutralisation Experiment

Aim: To determine the heat of neutralisation.
Materials: 2.0 mol dm-3 hydrochloric acid, 2.0 mol dm-3 sodium hydroxide solution, 2.0 mol dm-3 nitric acid, 2.0 mol dm-3 potassium hydroxide solution.
Apparatus: 50 cm3 measuring cylinders, thermometer, plastic cups with covers.
Safety measure:
Handle the chemicals with caution.
Wear eye protection.
Caution: Stir the mixture throughout the experiment.
Procedure:

What is the enthalpy of neutralization 14

  1. 50 cm3 of 2.0 mol dm-3 sodium hydroxide acid is measured using another measuring cylinder and poured into a plastic cup. The initial temperature of the solution is measured after a few minutes.
  2. 50 cm3 of 2.0 mol dm-3 hydrochloric acid is measured using another measuring cylinder and poured into a plastic cup. The initial temperature of the solution is measured after a few minutes.
  3. The hydrochloric acid is then poured quickly and carefully into the sodium hydroxide solution.
  4. The mixture is stirred using the thermometer and the highest temperature reached is recorded.
  5. Steps 1 to 4 are repeated using nitric acid and potassium hydroxide solution to replace hydrochloric acid and sodium hydroxide solution with other factors remain unchanged.

Results:
What is the enthalpy of neutralization 15

Interpreting data:
1. Reaction between sodium hydroxide solution and hydrochloric acid
What is the enthalpy of neutralization 16
What is the enthalpy of neutralization 17
2. Reaction between potassium hydroxide solution and nitric acid
What is the enthalpy of neutralization 18
Discussion:

  1. It is found that the heat of neutralisation between sodium hydroxide and hydrochloric acid and the heat of neutralisation between potassium hydroxide and nitric acid are the same. This is because both the reactions are between a strong monoprotic acid and a strong alkali.
    H+(aq) + OH(aq) → H2O(l)    ΔH = -57.3 kJ
  2. It is found that the value of heat of neutralisation obtained in the experiment is less than the theoretical value, ΔH = -57.3 kJ mol-1. This is because some heat is lost to the surroundings.
  3. It is necessary to mix the acid and the alkali quickly to reduce heat loss to the surroundings.
  4. A plastic cup is used to reduce heat loss to the surroundings.

Conclusion:
The heat of neutralisation between a strong monoprotic acid and a strong alkali is -57.3 kJ mol-1.

Determine heat of neutralization of between acid and base experiment

Aim: To determine and compare the heats of neutralisation between acids and alkalis of different strength.
Materials: 2.0 mol dm-3 hydrochloric acid, 2.0 mol dm-3 sodium hydroxide solution, 2.0 mol dm-3 ethanoic acid, 2.0 mol dm-3 ammonia solution.
Apparatus: 50 cm3 measuring cylinders, thermometer, plastic cups with covers.
Procedure:

  1. 50 cm3 of 2.0 mol dm-3 sodium hydroxide solution is measured using a measuring cylinder and poured into a plastic cup. The initial temperature of the solution is measured after a few minutes.
  2. 50 cm3 of 2.0 mol dm-3 hydrochloric acid is measured using another measuring cylinder and poured into a plastic cup. The initial temperature of the solution is measured after a few minutes.
  3. The hydrochloric acid is then poured quickly and carefully into the sodium hydroxide solution.
  4. The mixture is stirred using a thermometer and the highest temperature reached is recorded.
  5. Steps 1 to 4 are repeated using
    • Sodium hydroxide solution and ethanoic acid
    • Ammonia solution and hydrochloric acid
    • Ammonia solution and ethanoic acid

Results:

1. The results of the experiment are shown in the table below.
What is the enthalpy of neutralization 19
2. The temperatures of all the resulting mixtures increase.
3. The increase in the temperature of the reacting mixture is in the order of θ1 > θ2 > θ3 > θ4.

Discussion:

  1. The heat of neutralisation for the reaction between a strong acid and a strong alkali is the highest, whereas the heat of neutralisation for the reaction between a weak acid and a weak alkali is the lowest.
  2. The heat of neutralisation for the reactions between acids and alkalis decreases in the order:
    What is the enthalpy of neutralization 20
  3. Ethanoic acid is a weak acid and ammonia solution is a weak alkali, they both dissociate partially when dissolved in water. Most of the ethanoic acid and ammonia solution still exist as molecules.
  4. Some of the heat given out during the neutralisation reaction is used to dissociate the weak acid or the weak alkali completely in water.
  5. For the reaction between ethanoic acid and ammonia solution, the heat of neutralisation is the lowest. This is because much more energy is needed to dissociate both the weak acid and the weak alkali completely to produce hydrogen ions and hydroxide ions which then react together to form one mole of water.
  6. The equations for the neutralisation reactions are as follows.
    What is the enthalpy of neutralization 21
  7. It is necessary to mix the acid and the alkali quickly to reduce heat loss to the surroundings.
  8. A plastic cup is used in this experiment to reduce heat loss to the surroundings.

Conclusion:
The heat of neutralisation is the highest for the reaction between a strong acid and a strong alkali, and is the lowest for the reaction between a weak acid and a weak alkali.

How to calculate heat of neutralization problems with solutions

1. In an experiment to determine the heat of neutralisation, 50 cm3 of 1.0 mol dm-3 sulphuric acid at 28.5°C is added to 50 cm3 of 2.0 mol dm-3 potassium hydroxide solution which is also at 28.5°C in a plastic cup with a cover. The mixture is then stirred and the highest temperature reached is 41.5°C. Calculate the heat of neutralisation.
[Specific heat capacity of solution: 4.2 J g-1 °C-1; density of solution: 1 g cm-3]
Solution:
What is the enthalpy of neutralization 22
The heat of neutralisation between sulphuric acid and potassium hydroxide solution is -54.6 kJ mol-1.

2. A student carried out an experiment to investigate the change in temperature during a titration between sodium hydroxide solution and hydrochloric acid.
5.0 cm3 of m mol dm-3 hydrochloric acid is added to 50.0 cm3 of 2.0 mol dm-3 sodium hydroxide solution. The mixture is stirred and the highest temperature is then recorded. Another 5.0 cm3 of hydrochloric acid is quickly added and the process is repeated until a total of 50.0 cm3 of the acid is added. The results of the experiment are shown in Figure.
What is the enthalpy of neutralization 23
(a) (i) What is the initial temperature of the sodium hydroxide solution?
(ii) What is the highest temperature of the mixture?
(b) What is the volume of hydrochloric acid at the end point?
(c) What is the value of m?
(d) What is the heat of neutralisation?
[Specific heat capacity of solution: 4.2 J g-1 °C-1; density of solution: 1 g cm-3]
Solution:
What is the enthalpy of neutralization 24
The heat of neutralisation between hydrochloric acid and sodium hydroxide solution is -49.98 kJ mol-1.

3. The thermochemical equation for the reaction between nitric acid and sodium hydroxide solution is as shown below.
HNO3,(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)    ΔH = -57.3 kJ
When 250 cm3 of 1.0 mol dm-3 nitric acid is added to 200 cm3 of 2.0 mol dm-3 sodium hydroxide solution, what is the change in temperature?
[Specific heat capacity of solution: 4.2 J g-1 °C-1; density of solution: 1 g cm-3]
Solution:
What is the enthalpy of neutralization 25

What is heat of displacement?

What is heat of displacement?

 

  • When a more electropositive metal displaces a less electropositive metal from a solution of its salt, heat change occurs.
  • 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.
  • The thermochemical reaction for the displacement reaction of copper by zinc can be represented as follows.
    Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)    ΔH = -210 kJ
    (a) When one mole of copper is displaced from its salt solution by zinc, 210 kJ of heat is given out.
    (b) The heat of displacement of copper by zinc is -210 kJ mol-1.
    (c) The energy level diagram for the displacement reaction of copper by zinc is shown in Figure.
    What is heat of displacement 1
    (d) As the copper is displaced by zinc, the intensity of the blue solution decreases until it becomes colourless.
  • The heat of displacement of a metal can be determined by measuring the heat change occurred when a more electropositive metal in excess is added to a known quantity of a salt solution of a less electropositive metal.
  • The heat of displacement of a metal is different when it is displaced by different metals in the electrochemical series.
  • For example, the heat of displacement of silver by magnesium is higher than the heat of displacement of silver by zinc. This is because magnesium is more electropositive than zinc.

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Heat of Displacement Experiment

Aim: To investigate whether the heat of displacement of copper by magnesium is higher than the heat of displacement of copper by zinc.
Problem statement: Is the heat of displacement of copper by magnesium higher than the heat of displacement of copper by zinc?
Hypothesis: The heat of displacement of copper by magnesium is higher than the heat of displacement of copper by zinc.
Variables:
(a) Manipulated variable : Different metals used to displace copper
(b) Responding variable : Heat of displacement of copper
(c) Controlled variables : Volume and concentration of copper(II) sulphate solution, plastic cup, mass of metals
Materials: 0.2 mol dm-3 copper(II) sulphate solution, magnesium powder, zinc powder.
Apparatus: Thermometer, plastic cup with cover, 50 cm3 measuring cylinder, electronic balance, weighing bottle.
Procedure:
What is heat of displacement 2

  1. 50 cm3 of 0.2 mol dm-3 copper(II) sulphate solution is measured and poured into a plastic cup.
  2. The initial temperature of the solution is measured and recorded after a few minutes.
  3. About 2 g of magnesium powder is weighed in a weighing bottle.
  4. The magnesium powder is then added quickly and carefully into the copper(II) sulphate solution.
  5. The mixture in the plastic cup is stirred using the thermometer and the highest temperature reached is recorded.
  6. Steps 1 to 5 are repeated using zinc powder to replace the magnesium powder with other factors remain unchanged.

Results:
What is heat of displacement 3

Interpreting data:
1. Heat of displacement of copper by magnesium
What is heat of displacement 4
What is heat of displacement 5

2. Heat of displacement of copper by zinc
What is heat of displacement 6

Discussion:

  1. Magnesium and zinc are more electropositive than copper, therefore both magnesium and zinc can displace copper from copper(II) sulphate solution.
  2. In both the reactions, brown solids (copper metal) are formed. The intensity of the blue solutions decreases until they become colourless.
  3. The equations for the reactions are shown below.
    (a) Displacement of copper by magnesium:
    Chemical equation : Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)
    Ionic equation : Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
    (b) Displacement of copper by zinc:
    Chemical equation : Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
    Ionic equation : Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
  4. The theoretical values of heat of displacement of copper for both the reactions are shown below.
    Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)    ΔH = -352 kJ
    Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)     ΔH= -210 kJ
  5. The energy level diagrams for both reactions are shown below.
    What is heat of displacement 7
  6. Excess magnesium and zinc are used to make sure all the copper(II) ions are displaced to form copper.
  7. The amount of heat absorbed by the remaining unreacted magnesium and zinc is very little and can be neglected in the calculation of the heat of displacement.
  8. The volume and concentration of copper(II) sulphate solution used in both reactions are the same so that the same number of moles of copper is formed.
  9. The following precautions need to be taken during the experiment to get more accurate results.
    (a) Metals in the powder form are used so that the reactions will take a shorter time to complete, and thus less heat is lost to the surroundings.
    (b) The initial temperature of the copper(II) sulphate solution is taken after a few minutes to let the solution achieve a uniform temperature.
    (c) The metals are added quickly to the solution to reduce heat loss to the surroundings.
    (d) The mixture is stirred slowly throughout the experiment to make sure the temperature of the solution is uniform.
    (e) The reading of the thermometer is observed carefully so that the highest temperature of the solution can be recorded. This is done to ensure that the reactions are completed and all the heat has been given out.

Conclusion:
The heat of displacement of copper by magnesium is higher than the heat of displacement of copper by zinc.
Hence, the hypothesis is accepted.

Heat of Displacement Calculation – Problems with Solutions

1. In an experiment to determine the heat of displacement of iron by magnesium, excess magnesium powder is added to 50 cm3 of 0.2 mol dm-3 iron(II) sulphate solution. The results of the experiment are shown below.
Initial temperature of iron(II) sulphate solution = 30.5°C
Highest temperature of the mixture = 40.0°C
Calculate the heat of displacement of iron.
[Specific heat capacity of solution: 4.2 J g-1 °C-1; density of solution: 1 g cm-3]
Solution:
What is heat of displacement 8
What is heat of displacement 9

2. 1.4 g of iron powder is added to 200 cm3 of 1.5 mol dm-3 copper(II) sulphate solution at an initial temperature of 29.0°C. The thermochemical equation is as follows.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)    ΔH = -150 kJ
What is the highest temperature of the mixture?
[Specific heat capacity of solution: 4.2] g-1 °C-1. Density of solution: 1 g cm-3. Relative atomic mass: Fe, 56]
Solution:
What is heat of displacement 10
What is heat of displacement 11