Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy:
(i)Kinetic Energy(KE) ;the energy in motion.
(ii)Potential Energy(PE); the stored/internal energy.
Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb.
Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat
The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the:
(i) quantity of energy transferred when a force of one newton acts through a distance of one metre.
(ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.
All thermochemical reactions should be carried out at standard conditions of:
(i) 298K /25oC temperature
(ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure.
2.Exothermic and endothermic processes/reactions
Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic.
Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise)
Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease)
To demonstrate/illustrate exothermic and endothermic processes/reactions
- Dissolving Potassium nitrate(V)/ammonium chloride crystals
Procedure:
Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals.
Sample results
| Temperture (oC) | Using Potassium nitrate(V) crystals | Using Ammonium chloride crystals |
| T2(Final temperature) | 21.0 | 23.0 |
| T1 (Initial temperature) | 25.0 | 26.0 |
| Change in temperature(T2 –T1) | 4.0 | 3.0 |
Note:
(i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2)
(ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or“-3.0”.
(iii)Dissolution of both potassium nitrate(V) and ammonium chloride crystals isan endothermic process becauseinitial(T1) temperatureis higher than the final temperature(T2) thus causes a fall/drop in temperature.
- Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals
Procedure:
Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals.
CAUTION:
(i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin.
(ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin.
Sample results
| Temperture (oC) | Using Sodium hydroxide pellets | Using Concentrated sulphuric(VI) acid |
| T2(Final temperature) | 30.0 | 32.0 |
| T1 (Initial temperature) | 24.0 | 25.0 |
| Change in temperature(T2 –T1) | 6.0 | 7.0 |
Note:
(i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).
(ii)Dissolution of both Sodium hydroxide pellets and concentrated sulphuric (VI) acid isan exothermic process becausefinal (T2) temperatureis higher than the initial temperature (T1) thus causes a rise in temperature.
The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperature/thermometer readings.
Dissolving both potassium nitrate(V) and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer reading.
Dissolving both Sodium hydroxide pellets and concentrated sulphuric (VI) acid causes heat loss to the surrounding that causes rise in thermometer reading.
At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H.
Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e.
Enthalpy/energy/ change in heat content ∆H = Hfinal – Hinitial
For chemical reactions:
∆H = Hproducts – Hreactants
For exothermic reactions, the heat contents of the reactants is more than/higher than the heat contents of products, therefore the ∆H is negative (-∆H)
For endothermic reactions, the heat contents of the reactants is less than/lower than the heat contents of products, therefore the ∆H is negative (+∆H)
Graphically, in a sketch energy level diagram:
(i)For endothermic reactions the heat content of the reactants should be relatively/slightly lower than the heat content of the products
(ii)For exothermic reactions the heat content of the reactants should be relatively/slightly higher than the heat content of the products
Sketch energy level diagrams for endothermic dissolution
3.Energy changes in physical processes
Melting/freezing/fusion/solidification and boiling/vaporization/evaporation are the two physical processes. Melting /freezing point of pure substances is fixed /constant. The boiling point of pure substance depend on external atmospheric pressure.
Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid.
Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e
A (s) ========A(l)
Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour. Condensation/liquidification is the physical change of gas/vapour to liquid. Boiling/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e
B (l) ========B(g)
Practically
(i) Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together. Solids are made up of very strong bonds holding the particles very close to each other (Kinetic Theory of matter).On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom. Melting/freezing/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding.
(ii)Freezing/fusion/solidification involves cooling a a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter). Freezing /fusion / solidification is an exothermic (–∆H)process that require particles holding the liquid together to lose energy to the surrounding.
(iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together. Gaseous particles have high degree of freedom (Kinetic Theory of matter). Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding.
(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process.
The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.
H2O(s) -> H2O(l) ∆H = +6.0kJ mole-1 (endothermic process)
H2O(l) -> H2O(s) ∆H = -6.0kJ mole-1 (exothermic process)
The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. e.g.
H2O(l) -> H2O(g) ∆H = +44.0kJ mole-1 (endothermic process)
H2O(g) -> H2O(l) ∆H = -44.0kJ mole-1 (exothermic process)
The following experiments illustrate/demonstrate practical determination of melting and boiling
- To determine the boiling point of water
Procedure:
Measure 20cm3 of tap water into a 50cm3 glass beaker. Determine and record its temperature.Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minutes.
Sample results
| Time(seconds) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 |
| Temperature(oC) | 25.0 | 45.0 | 85.0 | 95.0 | 96.0 | 96.0 | 96.0 | 97.0 | 98.0 |
Questions
1.Plot a graph of temperature against time(y-axis)
Sketch graph of temperature against time
2.From the graph show and determine the boiling point of water
Note:
Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.
3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O)
Working:
Mass of water = density x volume => (20 x 1) /1000 = 0.02kg
Quantity of heat produced
= mass of water x specific heat capacity of water x temperature change
=>0.02kg x 4.2 x ( 96 – 25 ) = 5.964kJ
Heat of vaporization of one mole H2O = Quantity of heat
Molar mass of H2O
=>5.964kJ = 0.3313 kJ mole -1
18
To determine the melting point of candle wax
Procedure
Weigh exactly 5.0 g of candle wax into a boiling tube. Heat it on a strongly Bunsen burner flame until it completely melts. Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes.
Sample results
| Time(seconds) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 |
| Temperature(oC) | 93.0 | 85.0 | 78.0 | 70.0 | 69.0 | 69.0 | 69.0 | 67.0 | 65.0 |
Questions
1.Plot a graph of temperature against time(y-axis)
4.Energy changes in chemical processes
Thermochemical reactions measured at standard conditions of 298K(25oC) and 101300Pa/101300Nm2/ 1 atmospheres/760mmHg/76cmHg produce standard enthalpies denoted ∆Hᶿ.
Thermochemical reactions are named from the type of reaction producing the energy change. Below are some thermochemical reactions:
- Standard enthalpy/heat of reaction ∆Hᶿr
- Standard enthalpy/heat of combustion ∆Hᶿc
- Standard enthalpy/heat of displacement ∆Hᶿd
- Standard enthalpy/heat of neutralization ∆Hᶿn
- Standard enthalpy/heat of solution/dissolution ∆Hᶿs
- Standard enthalpy/heat of formation ∆Hᶿf
(a)Standard enthalpy/heat of reaction ∆Hᶿr
The molar standard enthalpy/heat of reaction may be defined as the energy/heat change when one mole of products is formed at standard conditions
A chemical reaction involves the reactants forming products. For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e.
A–B + C-D -> A-C + B-D
Old Bonds broken A-B and C-D on reactants
New Bonds formed A-C and B-D on products
The energy required to break one mole of a (covalent) bond is called bond dissociation energy. The SI unit of bond dissociation energy is kJmole-1
The higher the bond dissociation energy the stronger the (covalent)bond
Bond dissociation energies of some (covalent)bonds
| Bond | Bond dissociation energy (kJmole-1) | Bond dissociation energy (kJmole-1) | |
| H-H | 431 | I-I | 151 |
| C-C | 436 | C-H | 413 |
| C=C | 612 | O-H | 463 |
| C = C | 836 | C-O | 358 |
| N = N | 945 | H-Cl | 428 |
| N-H | 391 | H-Br | 366 |
| F-F | 158 | C-Cl | 346 |
| Cl-Cl | 239 | C-Br | 276 |
| Br-Br | 193 | C-I | 338 |
| H-I | 299 | O=O | 497 |
| Si-Si | 226 | C-F | 494 |
The molar enthalpy of reaction can be calculated from the bond dissociation energy by:
(i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H).
(ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H).
Practice examples/Calculating ∆Hr
1.Calculate ∆Hr from the following reaction:
- H2(g) + Cl2(g) -> 2HCl(g)
Working
Old bonds broken (endothermic process/+∆H )
= (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ
New bonds broken (exothermic process/-∆H )
= (2(H-Cl ) => (- 428 x 2)) = -856kJ
∆Hr =(+ 670kJ + -856kJ) = 186 kJ = -93kJ mole-1
2
The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.
The thermochemical reaction is thus:
½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ
- CH4(g) + Cl2(g) -> CH3Cl + HCl(g)
Working
Old bonds broken (endothermic process/+∆H )
= (4(C-H) + Cl-Cl)
=> ((4 x +413) + (+ 239)) = + 1891kJ
New bonds broken (exothermic process/-∆H )
= (3(C-H + H-Cl + C-Cl)
=> (( 3 x – 413) + 428 + 346) = –2013 kJ
∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1
The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.
The thermochemical reaction is thus:
CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ
- CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
Working
Old bonds broken (endothermic process/+∆H )
= (4(C-H) + Cl-Cl + C=C)
=> ((4 x +413) + (+ 239) +(612)) = + 2503kJ
New bonds broken (exothermic process/-∆H )
= (4(C-H + C-C + 2(C-Cl) )
=> (( 3 x – 413) + -436 +2 x 346 = –2367 kJ
∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1
The above reaction has negative +∆H enthalpy change and is therefore practically endothermic.
The thermochemical reaction is thus:
CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ
Note that:
(i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products.
(ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products.
(b)Standard enthalpy/heat of combustion ∆Hᶿc
The molar standard enthalpy/heat of combustion(∆Hᶿc) is defined as the energy/heat change when one mole of a substance is burnt in oxygen/excess air at standard conditions.
Burning is the reaction of a substance with oxygen/air. It is an exothermic process producing a lot of energy in form of heat.
A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use. A fuel may be solid (e.g coal, wood, charcoal) liquid (e.g petrol, paraffin, ethanol, kerosene) or gas (e.g liquefied petroleum gas/LPG, Water gas-CO2/H2, biogas-methane, Natural gas-mixture of hydrocarbons)
To determine the molar standard enthalpy/heat of combustion(∆Hᶿc) of ethanol
Procedure
Put 20cm3 of distilled water into a 50cm3 beaker. Clamp the beaker. Determine the temperature of the water T1.Weigh an empty burner(empty tin with wick).
Record its mass M1.Put some ethanol into the burner. Weigh again the burner with the ethanol and record its mass M2. Ignite the burner and place it below the clamped 50cm3 beaker. Heat the water in the beaker for about one minute. Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3
Sample results:
| Volume of water used | 20cm3 |
| Temperature of the water before heating T1 | 25.0oC |
| Temperature of the water after heating T2 | 35.0oC |
| Mass of empty burner M1 | 28.3g |
| Mass of empty burner + ethanol before igniting M2 | 29.1g |
| Mass of empty burner + ethanol after igniting M3 | 28.7g |
Sample calculations:
1.Calculate:
(a) ∆T the change in temperature
∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC
(b) the mass of ethanol used in burning
mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g
(c) the number of moles of ethanol used in burning
moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3moles
molar mass of ethanol 46
2. Given that the specific heat capacity of water is 4.2 kJ-1kg-1K-1,determine the heat produced during the burning.
Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T
=> 20 x 4.2 x 10 = 840 Joules = 0.84 kJ
1000
3.Calculate the molar heat of combustion of ethanol
Molar heat of combustion ∆Hc = Heat produced ∆H
Number of moles of fuel
=> 0.84 kJ = 96.5517 kJmole-1
0.0087 /8.7 x 10-3 moles
4.List two sources of error in the above experiment.
(i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol.
A draught shield tries to minimize the loss by protecting wind from wobbling the flame.
(ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc
5.Calculate the heating value of the fuel.
Heating value = molar heat of combustion => 96.5517 kJmole-1 = 2.0989 kJg-1
Molar mass of fuel 46 g
Heating value is the enrgy produced when a unit mass/gram of a fuel is completely burnt
6.Explain other factors used to determine the choice of fuel for domestic and industrial use.
(i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost.
(ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky.
(iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic.
(iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point.
7.Explain the methods used to reduce pollution from common fuels.
(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere.
(ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes.
Further practice calculations
1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C-12.0,H=1.0 O=16.0).
Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles
Molar mass of methanol 32
Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T
=> 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ
1000
Molar heat of combustion ∆Hc = Heat produced ∆H
Number of moles of fuel
=> 14.7 kJ = 540.8389 kJmole-1
0.02718 moles
Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1
Molar mass of fuel 32 g
2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).
Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles
Molar mass of carbon 12
Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T
=> 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ
1000
Molar heat of combustion ∆Hc = Heat produced ∆H
Number of moles of fuel
=> 30.24 kJ = 363.0252 kJmole-1
0.0833 moles
Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1
Molar mass of fuel 12 g
(c)Standard enthalpy/heat of displacement ∆Hᶿd
The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.
A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ /negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ /positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.
(i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)
Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq)
(ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq)
Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq)
(iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s)
This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.
(iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq)
Ionically: Cl2(g)+ 2Br– (aq) -> Br2(aq) + 2Cl– (aq)
Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.
To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper
Procedure
Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter. Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 . Repeat the experiment to complete table 1 below
Table 1
| Experiment | I | II |
| Final temperature of solution(T2) | 30.0oC | 31.0oC |
| Final temperature of solution(T1) | 25.0oC | 24.0oC |
| Change in temperature(∆T) | 5.0 | 6.0 |
