CHEMISTRY CHAPTER 4
2010
THERMOCHEMISTRY
Concept Map
Energy Changes in Chemical Reactions 1. Energy changes occur in almost all chemical reactions. 2. There are two kinds of chemical reactions that undergo changes in heat. They are: (a) Exothermic reactions (b) Endothermic reactions Exothermic Reaction 1. Chemical reaction that release heat into its surroundings. 2. The total energy contained in the product is less than that of the reactants. 3. Heat energy is released from the reactant into the surroundings to increase. 4. The reaction mixture becomes hot, so does the container that holds it. 5. The rise in temperature can be felt by hand. A thermometer can be used to measure the rise in temperature. 6. Example: (a) Reaction between metal carbonates with dilute acid: Na2CO3(s) 2NaCl(aq) + H 2O( ) + + 2HCl(aq) CO2(g) (b) Neutralisation reaction between acid and alkali: NaOH(aq) + HCl(aq) NaCl(aq) + H2O( ) (c) Reaction between water and reactive metal oxide: K2O(s) + H2O( ) 2KOH(aq) (d) Reaction between alkali metals (Group 1 element) and water: 2Na(s) + 2H2O( ) 2NaOH(aq) + H2(g) (e) Dilution of concentrated acid: H2SO4( ) H2SO4(aq) (f) Reaction between reactive metals (Mg/Al/Zn/Fe) and diluted acid: Mg(s) + H2SO4(aq) MgSO4(aq) + H2 (g) (g) Combustion of fuel in excess oxygen: 4.1
(h)
C2H4(g) + 3O2 2CO2(g) + 2H2O( ) Rusting of iron: 4Fe(OH)2(s) + O2(g) 4Fe(OH)3(s) + 2H2O( )
(i)
Action of water on anhydrous copper(II) sulphate: CuSO4(s) + 5H2O( ) CuSO4.5H2O(s) (j) Crystallisation of molten sodium thiosulphate: Na2S2O3( ) Na2S2O3(s) (k) Displacement reaction of metal drom its salt solution: Zn(s) + CuCl2(aq) ZnCl2(aq) + Cu(s) (l) Dissolving Group 1 hydroxides in water: 2Na(s) + 2H2O( ) 2NaOH(aq)+H2(g) Endothermic Reaction 1. An endothermic reaction is a chemical reaction that absorbs heat from the surroundings. 2. The total energy contained in the products is more than that of the reactants. 3. Heat energy is absorbed by the reactants form the surroundings, causing the surroundings temperature to drop. 4. The reaction mixture becomes cold, so does the container that holds it. 5. The reduction in temperature can be detected by hand or by using a thermometer. 6. Example: (a) Reaction between sodium hydrogen carbonate and dilute hydrochloric acid: NaHCO 3(s) + HCl(aq) NaCl(aq) + H2O( ) + CO2(g) (b) Dissolving ammonium salts in water: NH4NO3(s) NH4NO3(aq) (c) Dissolving potassium salts like potassium nitrate in water: KNO3(s) KNO3(aq) (d) Decomposition of carbonate salts by heat:
= Energy of the products ± Energy of the 5.
6.
7.
8.
reactants If the energy content in the reactants exceed the energy content in the products, then is negative. Negative occurs in exothermic reactions. Exothermic reaction negative Exothermic reactions have negative values because the reactants have lost (given off) energy to the surroundings. The products contain less energy as compared to the reactants.
If the energy content of the reactants is less than the energy content of the products, then positive. Positive occurs in endothermic reactions. Endothermic reaction positive Endothermic reactions have positive values because the reactants have absorbed energy from the surroundings. The products have higher energy content as compared to the reactants.
(e)
CaCO3(s) CaO(s) + CO2(g) Decomposition of of nitrate nitrate salts by heat
(f)
2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g) Action of heat on hydrated copper(II) sulphate:
CuSO4.5H2O(s) CuSO4(s) + 5H2O( ) (g) Dissolving sodium thiosulphate crystals in water: Na2S2O3(s) Na2S2O3(aq) Energy Level Diagram 1. The energy level diagram is a graph that shows the energy change of a chemical reaction. 2. The energy level diagram shows the energy content of the reactants and products of a reaction. 3. The energy change that occurs in a reaction depends on the energy c ontent within the reactants and its products. Energy change can be represented by the symbol (Delta H). 4. is the difference between the energy of the reactants and the energy of the products.
9.
The information that can be obtained from an energy level diagram includes: (a) Heat energy change value. (b) Energy content of the reactants which which is higher or lower than the energy content of the products. (c) The reactions that occur occur will result in a rise or a fall in temperature. 10. The chemical equation equation that includes includes is called the thermochemical equation. Example:
Zn(s) + Cu(NO 3)2(aq) Zn(NO3)2(aq) + Cu(s) =-220 kJ
Relationship between energy change with the formation and breaking of chemical bonds 1. When chemical reactions occur, heat energy might be absorbed or released. 2. Heat energy is absorbed when chemical bonds in a compound are broken. 3. On the other hand, heat energy is released when chemical bonds in a compound are formed. 4. In an exothermic reaction, the energy that is released from the formation of bonds
Form 5
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CHEMISTRY
2010
exceeds the energy required to break the bonds. 4. 5. 6.
Endothermic reaction ( positive)
7.
8.
Exothermic reaction ( negative) In an endothermic reaction, the energy that is needed to break the bonds exceeds the energy that is released during bond formation (negative ). Application of exothermic and endothermic reactions. 1. Exothermic and endothermic reactions are applied in the making of hot and cold packs. 2. Figure below briefly shows the principle behind the making of hot and cold packs. 5.
Making of hot packs and cold packs
y y
y
y
The hot or cold packs contain two plastic bags. The outer plastic bag is made of a strong thick plastic and is filled with chemical powder. The inner plastic bag is made of weaker thin plastic that contains water. When the beg is punched, the inner plastic bag will break open. The water inside will then mix with and dissolve the chemical powder, causing the solution to become either very hot or cold.
Hot packs The chemicals that are suitable for use to produce an exothermic reaction when mixed with water are: Dry copper(II) sulphate Dry potassium chloride magnesium chloride calcium chloride magnesium sulphate y
y
Cold packs The chemicals that are suitable for use to provide an endothermic reaction when mixed with water are: ammonium nitrate potassium nitrate sodium thiosulphate y y y
y y y
3.
4.
4.2
Hot packs are widely used by mountain climber or explorer to areas with a cold climate to warm their hands and feet. Cold packs are widely used by athletes to help them reduce pain in swellings that results from sporting injuries. Heat of Precipitation
Heat of reaction 1. All reactions that occur are followed by heat energy changes. 2. Energy changes in a chemical reaction are called the heat of reaction. The symbol for the heat of reaction is . 3. The heat of reaction is defined defined as as heat heat change
(a) when 1 mol of of reactants react; or (b) when 1 mol of of products products are formed. The unit used to measure the heat of reaction is joule(J). The heat of reaction, , that is marked negative (-) shows an exothermic reaction. The heat of reaction, , that is marked positive (+) shows and endothermic reaction. The heat of reaction is usually named according to the type of reaction that occurs. Examined the displacement reaction below.
Zn(s) + CuSO4(aq) ZnSO 4(aq) + Cu(s) -1 = -190 kJ mol The thermochemical equation above sh ows that: 190 kJ of heat energy is released when 1 mol of copper, Cu, is displaced from its salt solution by zinc, Zn. y
y
The heat of displacement of copper by zinc -1 () is -190 kJ mol .
Determining heat change 1. Heat change is the amount of heat that is released or absorbed in a reaction. 2. The change of heat of a reaction depends depends on the (a) mass of the substance used. (b) temperature change ( ) that is the difference between the final temperature (T 2) and initial temperature (T 1) of the reaction. = T 2 ± T1 (c) specific heat capacity (c) of of a substance is the heat that is needed to raise the temperature of 1 g of O substance by 1 C. 3. Heat change in a reaction can be determined by using the following formula. Heat change = solution mass specific heat capacity temperature change = 4. The unit used to measure heat change is joule (J). 5. The above formula can be used to calculate the heat change that occurs in a reaction by making four assumptions, namely: (a) The density of of the aqueous solution that is used is the same as the -3 3 density of water that is 1 g cm . 1 cm of aqueous solution is assumed to have mass of 1 g. Assumption: Density of aqueous solution = density of water -3 = 1 g cm
(c)
(d)
Therefore, 3 1 cm of solution has a mass of 1 g. Chemicals in solid form are are not not assumed to absorb heat energy. Ignore the mass of chemicals in solid form. No heat is lost to the surroundings.
Calculation on heat change There are two types of question about heat change calculations: (a) involving volume of solution and temperature change only (b) involving a thermochemical equation Calculation involving volume of solution and temperature change only Heat change occurring in a reaction that does not involve thermochemical equation can be calculated by the following guidelines: Step 1: 1: Write the formula of heat change if the volume of solution and the temperature change are given in the question. Heat change = Where, = the total volume of solution is the m ass (g) of the solution -2 O -1 = Specific heat capacity (4.2 J g C ) O = Temperature change ( C) Step 2: Replace the values of , , and in the formula. The answer that is obtained is written in the unit of joule, J. Example: Heat change = J
Calculation involving thermochemical equations 1. If a substance mass (or the volume and molarity of solution) is given, and you are required to calculate heat change, the following guidelines can be followed: Step 1: 1: First, calculate the number of moles. (a) If a substance substance mass is given, given, use the following formula to determine its number of moles: Number of moles =
mass mass molar
(b) If a solution¶s solution¶s volume and molarity (concentration) are given, use the following formula to determine the number of moles: Number of moles =
MV 1000
Step 2: Multiply the number of moles that was obtained with the value. Heat change = Number of moles
2.
If the heat change value value is given given and you are required to calculate the mass of the substance used, you can use the following guidelines.
Step 1: 1: First, calculate the number of moles. Because the heat change value is given in the question, thus, The number of moles is determined by using the following formula: Number of moles =
heat change value
Step 2: Multiply the number of moles that was obtained with the substance molar mass. Heat change = Number of moles molar mass
3.
If the heat of reaction ( ), volume and concentration of solution are given and you are required to calculate the temperature change, you can use the following guidelines:
Step 1: 1: If the heat of reaction ( ), volume and concentration of solution are given in th e question, you have to first determine the number of m oles of solution using the following formula: Number of moles =
Form 5
MV 1000
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CHEMISTRY
2010 3.
Step 2: Determine the heat change that occurs by adjusting the following formula: Number of moles =
heat change value
Therefore, Heat change = number of moles value
4. Step 3: Use the formula and adjust it to form a new formula: Heat change = =
5.
Determining the heat of reaction 1. The heat of reaction such as heat of precipitation, heat of displacement, heat of neutralisation, and heat of combustion can be determined if the following information is known: (a) The substance mass (b) Temperature change (c) Specific heat capacity (d) The number of of moles of substance or solution (i) The number of moles of a fuel such as alcohol liquid can be determined by using the formula below: Number of moles = (ii)
2.
MV 1000
You can refer to the following guidelines to make it easier to determine the heat of reaction ( ) of certain reactions.
Step 1: 1: Calculate heat change using the formula: Heat change= Step 2: 2: Calculate the number of moles of reactant(s) which reacts, using either of the following formulae: Number of moles =
mass mass molar
or Number of moles =
MV 1000
If there are two t ypes of solution present, calculate the number of moles for each one.
4.3
3
-3
25 cm of 1 mol dm silver nitrate, AgNO3 3 -3 solution is mixed with 25 cm of 1 mol dm sodium chloride, NaCl solution. The O O temperature changes form 30 C to 38 C. Calculate the heat of precipitation of silver chloride, AgCl. Then draw an energy level diagram for the reaction above. -1 [Specific heat energy of solution = 4.2 J g O -1 -3 C , Density of solution = 1 g cm ] Solution Step 1: Calculate the heat change Heat change = = 50 4.2 8 = 1 680 J Step 2: Calculate the number of moles of reactants
number of moles
Heat of precipitation 1. The heat of precipitation is the heat change when 1 mol of precipitate is formed. 2. For example, the reaction between silver nitrate AgNO3, solution and sodium chloride, NaCl, solution forms the white precipitate silver chloride, AgCl. AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
MV 1000 1 = 1000
+
The heat of displacement is the heat that is released when 1 mol of metal is displaced from its salt solution by a more electropositive metal. 2. For example, 1 mol of copper, Cu, is displaced from a copper(II) sulphate, CuSO4, solution by zinc, Zn, to release 210 kJ of heat. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) -1 =-210 kJ mol H =-210 2+ 2+ Zn(s) + Cu (aq) Zn (aq) + Cu(s) -1 =-210 kJ mol H =-210 3. When 1 mol of copper is displaced from its its solution by zinc, up to 210 kJ of heat is released. Therefore, the heat of displacement of copper by zinc is -210 kJ -1 mol . 4. The heat of displacement of of copper by zinc can be defined as follows: The heat of displacement of copper by zinc is the heat that is released when 1 mol of copper metal is displaced by zinc.
Number of moles of Ag ion =
5.
= 0.025 mol -
Heat of reaction =
Heat of Displacement
1.
Step 3: 3: Write chemical equation or ionic equation for the reaction that occurs. Step 4: 4: Link the number of moles of r eactants with the heat change which occurs. Calculate the heat of reaction, by using the following formula:
Therefore, the heat of precipitation of silver -1 chloride, = -67.2 kJ mol Energy level diagram for the precipitation of silver chloride, AgCl
The precipitation reaction of silver chloride, AgCl, is an exothermic reaction because the temperature of the reaction mixture increases. Example below shows the heat of precipitation ( ) of silver chloride can be calculated.
mass mass molar
The number of moles of a solution can be determined by using the formula: Number of moles =
6.
The actual reaction that occurs is the + combination of 1 mol of silver ion, Ag with 1 mol of chloride ion, Cl , to form 1 mol of silver chloride, AgCl. The ionic equation that represents the reaction that occurs is: + Ag (aq) + Cl (aq) AgCl(s) The heat of precipitation of silver chloride can be defined as follows: The heat of precipitation of silver chloride is the heat that is released when 1 mol of silver chloride is formed.
MV 1000 1 = 1000
Number of moles of Cl ion =
4.4
1680 J 0.025
= 67 200 J = 67.2 kJ
Heat of Neutralisation
1.
= 0.025 mol Step 3: Write down the ionic equation + Ag (aq) + Cl (aq) AgCl(s) Step 4: Link the number of moles of reactants with the heat change, and calculate the heat of reaction. + 0.025 mol of Ag ion reacts with 0.025 mol of Cl ion to produce 1 680 J of heat. + Therefore, 1 mol of Ag ion reacts with 1 mol of Cl ion to produce =
The displacement reaction of copper by zinc is an exothermic reaction because the temperature of reaction increases.
The heat of neutralisation is is the heat that is + released when 1 mol of hydrogen ion, H , neutralizes 1 mol of hydroxide ion, OH , to form 1 mol of water, H 2O. 2. For example, 1 mol of dilute hydrochloric acid, HCl, react with 1 mol of sodium hydroxide, NaOH, to release 57 kJ of heat. HCl(aq) + NaOH(aq) NaCl(aq) + H2O( ) -1 = -57 kJ mol +
-
H (aq) + OH (aq) H2O( ) = -57 kJ mol 3. The neutralisation reaction between acid and alkali is an exothermic reaction because the heat that is released causes the reaction temperature to increase.
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Comparison of the heat of neutralization for reactions between monoprotic acid and alkali, and diprotic acid and alkali. Case
A
B
C
1.
2.
Chemical equation
Ionic equation
HCl(aq) + NaOH(aq) NaCl(aq) + H2O( HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O( ) H2SO4(aq) + 2NaOH(aq) 2Na2SO4(aq) + 2H2O()
H (aq) + OH (aq) H2O( )
+
4.
5. 6.
7.
8.
9.
57
+
H (aq) + OH (aq) H2O( )
57
+
2H (aq) + 2OH (aq) 2H2O( )
114
Hydrochloric acid, HCl, and nitric aicd, HNO3 are strong monoprotic acid (monobasic) acid. 1 mol of a strong monopotric acid + produces 1 mol of hydrogen ion, H , in an aqueous solution. + HCl(aq) H (aq) + Cl (aq) 1mol 1mol +
3.
Heat of neutralisation, H (kJ)
Reaction between a strong acid and a strong alkali. 1. Strong acid such hydrochloric acid, HCl, dissociate completely in water to produce a + high concentration of hydrogen ions, H . + HCl(aq) H (aq) + Cl (aq) 1mol 1mol 2. Strong alkalis such as sodium hydroxide solution contain a high concentration of hydroxide ions, OH . + NaOH(s) Na (aq) + OH (aq) 1mol 1mol 3. As such, the reaction between strong acids acids and strong alkalis release a lot of heat. 4. The heat of neutralisation between strong acids and strong alkalis is a constant, that -1 is -57 kJ mol .
3.
4.
5.
6.
About 393.5 kJ of heat is released when 1 mol of carbon C, burns completely in excess oxygen, O 2 to produce 1 mol of carbon dioxide, CO 2. This heat is known as the heat of combustion of carbon. The combustion reaction is an exothermic reaction because the heat that is released causes the reaction temperature to increase. The heat of combustion of of fuel such as alcohol and hydrocarbons can be determined if the following information is known ± (Fuel mass, water volume and temperature increase) The determination of the heat of combustion is different from that of the other heat reactions.
The general method of determining the heat of reaction Example: Determining the heat of neutralisation
-
HNO3(aq) H (aq) + NO3 (aq) 1mol 1mol The actual reaction that occurs in neutralisation is between 1 mol of + hydrogen ion, H , with 1 mol of hydroxide ion, OH , to form 1 mol of water, H 2O. + H (aq) + OH (aq) H2O( ) 1mol 1mol 1mol As such, the heat of neutralisation for a reaction between alkali is the same. In case A and case B, the heat of neutralisation is the same that is 57 kJ -1 mol . Sulphuric acid, H2SO4, is a strong diprotic (dibasic) acid. 1 mol of a strong diprotic diprotic acid acid produces produces 2 + mol of hydrogen ions, H , in an aqueous solution. + 2H2SO4 2H (aq) + SO4 (aq) 1mol 2mol When sulphuric acid, H 2SO4, reacts with sodium hydroxide, NaOH, the equation of neutralisation reaction is: H2SO4(aq) + 2NaOH(aq) 2Na2SO4(aq) + 2H2O( ) or + 2H (aq) + OH (aq) 2H2O( ) 2mol 2mol 2mol -1 H = -114kJ mol As such, the heat of neutralisation value for a reaction that involves diprotic acid is two times the heat of neutralisation value for a reaction that involves monopotric acid. In case C, the heat of neutralisation value -1 is -57 x 2 = -144 kJ mol .
Comparison of the heat of neutralisation for reactions between strong acids and strong alkalis, and between weak acids and strong acids.
Reaction between a weak acid and a strong alkali
1.
2.
3.
4.5 1.
2.
Weak acid such as ethanoic acid, CH3OOH, partially ionise in water to produce a low concentration of hydrogen + ions, H . + CH3COOH(aq) CH3COO (aq) + H (aq) + To produce hydrogen ions, H , heat must be absorbed to break the bonds in the ethanoic acid molecule. This enables the ethanoic acid to fully ionise to produce free + moving hydrogen ions, H . As such, the heat of neutralisation between 1 mol of weak acid and 1 mol of strong alkali to produce 1 mol of water is lower than 57 kJ. CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O( ) -1 H = -55kJ mol
Heat change = mc Heat of neutralisation = The method of determining of the heat of combustion Example: Determining the heat of combustion if ethanol
Alcohol heat up the water. The heat is absorbed by a 3 certain volume of water (cm ) Heat change = mc Heat of combustion =
Heat of Combustion The heat of combustion is the heat that released when 1 mol of substance burns completely in excess oxygen. For example, the heat of carbon, C is -1 -393.5 kJ mol . This reaction can be represented by the following thermochemical equation: -1 C(s) + O2(g) CO2(g) H = -393.5kJ mol Form 5
The mass of alcohol (fuel) that is burned can be obtained by weighing the mass of lamp before and after the experiment. Alcohol mass = Initial lamp mass ± final lamp mass. Then, the number of moles
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Comparison of the heat of combustion of various types of alcohol 1. Different types of alcohol have different values of heat of combustion. 2. Table below shows the heat of combustion for a few
Form 5
5
