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TOPIC 4 THERMOCHEMISTRY - Copy.pptx

This document discusses various types of thermochemistry and enthalpy changes in chemical reactions. It begins by defining exothermic and endothermic reactions, and describes how enthalpy (ΔH) is the heat released or absorbed during a reaction. It then explains how to calculate enthalpy changes using specific heat and temperature change. The main types of enthalpy covered are heat of precipitation, neutralization, displacement, and combustion. Examples are provided for calculating enthalpy changes using thermochemical equations and experimental temperature changes. In summary, the document provides an overview of thermochemistry concepts and calculations involving enthalpy for different reaction types.

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TOPIC 4 THERMOCHEMISTRY
LEARNING OUTCOME • Describe the basic concept of endothermic and exothermic • Solve numerical problems involving enthalpy change. • Explains the exothermic and endothermic reactions with the aids of simple energy level diagram. • Explains the energy change during formation and breaking of bonds. • Solve numerical problems involving enthalpy change during a simple chemical reaction
ENERGY CHANGES IN CHEMICAL REACTIONS
INTRODUCTION THE LAW OF CONVERSION states that energy may neither be created nor destroyed. However, energy can be changed from one form to another.
CHEMICAL ENERGY • All chemical substance contain chemical energy • During chemical reactions, the chemical energy in the reactants will be changed into other form of energy; heat or thermal energy, mechanical energy, electrical energy, light energy and sound energy. • Unit for measuring energy is Joule (J). 1kJ = 1000J
THERMOCHEMISTRY •Thermochemistry is the study of changes in heat energy during chemical reactions. •There are two types of chemical reactions: a) Exothermic reaction b) Endothermic reaction
a) Exothermic Reaction • Exothermic reaction is a chemical reaction that gives out heat to the surroundings • In exothermic reaction, i. Chemical energy in reactant is change into heat energy ii. The heat energy is transferred to the surroundings iii. Temperature of the surroundings increases
b) Endothermic Reaction • Endothermic reaction is a chemical reaction that absorbs heat from the surroundings. • In an endothermic reaction, i. Heat energy is absorbed from the surroundings ii. The heat energy is changed into chemical energy iii. Temperature of the surroundings decreases
ENTHALPY, H • Every chemical substance has a certain amount of chemical energy. This energy is known as enthalpy or heat energy content with symbol of H and is different for different substances. • Heat of reaction or enthalpy is the amount of heat energy released or absorbed. It is given the symbol ΔH and is usually measured in kilojoules (kJ). • Only ΔH can be measured, not H for the initial or final state of a system ΔH = total energy content of products – total energy content of reactant ΔH = Hproducts - Hreactant
•In exothermic reaction, the product formed has less energy than the reactant, Hproduct < Hreactants. Therefore, the ΔH is negative.
•In endothermic reaction, the products formed have more energy than reactants, Hproducts> Hreactants. Therefore, ΔH is positive.
CALCULATION OF ENTHALPY CHANGES
Factors affecting heat quantities • The amount if heat contained by an object depends primarily on three factors: i. The mass of material ii. The temperature iii. The kind of material and its ability to absorb or retain heat.
Calculating enthalpy changes ΔH = total energy content of products – total energy content of reactant ΔH = Hproducts - Hreactant Q = mcθ
Specific Heat • Specific heat: the amount of heat energy required to raise the temperature of 1g of a substance of 1oC. • Units are: J g-1 oC-1 Specific heat of water (or aqueous solutions) = 4.184 J = 1 cal
Heat Quantities • The heat required to raise the temperature of 1.00 g of water 1 oC is known as a calorie • Calorie (with a capital “C”): dietary measurement of heat. Food has Food has potential energy stored in the chemical bonds of food.1 Cal = 1 kcal = 1000 cal. The SI unit for heat is the joule. It is based on the mechanical energy requirements.1.00 calorie = Joules
Q = ΔH = heat energy change (J or calories) m = mass (g) c = specific heat (J g1 oC-1) Specific heat of water = 4.184 J g1 oC-1 θ = change in temperature Q = mcθ
Example 1 How much heat in joules will be absorbed when 32.0g of water is heated from 25.0 oC to 80.0oC? Q= mcθ Q= ? M=32.0g C=4.18 J g-1 oC-1 Θ= 80-25=55 Q= 32 X 4.18 X 55 =7360 J
Example 2 When 435 J of heat is added to 3.4g of olive oil at 21oC, the temperature increases to 85oC. What is the specific heat of olive oil? Q= mcθ Q= 435 J M=3.4g C= ? θ = 85-21 = 64 435 = 3.4 X c X 64 C=(3.4 X 64)/435 =0.5 J g-1 OC-1
Enthalpy Change During Breaking and Formation of Bonds •Chemical reaction involves bond breaking and bond formation •Bond breaking need energy  endothermic •Bond formation releases energy  exothermic •The balance of heat release to the surroundings.
• In an exothermic reaction, the heat energy released during the formation of the bonds in the products is greater than the heat energy absorbed to break the bonds in the reactants. Bond enthalpy reactants< bond enthalpy products • In an endothermic reaction, the heat energy absorbed to break the bonds in the reactants is greater than the heat energy released during the formation of the bonds in the products. Bond enthalpy reactants> bond enthalpy products
Enthalpy Change of a Reaction Using Thermochemical Equations Mg(s) + ½ O2(g)  MgO(s) + heat Equation 1 Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) Equation A ∆H = -A MgO(s) + 2HCl(aq)  MgCl2(aq) +H2O(l) Equation B ∆H = -B H2(g) + ½ O2(g)  H2O(l) Equation C ∆H = +C The final ΔH = -A + (-B) + C
Example: Using Thermochemical Equations Calcium oxide combines with water to produce calcium hydroxide and heat (exothermic reaction). CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ OR CaO(s) + H2O(l)  Ca(OH)2(s) ΔH = -65.2 kJ mol-1 These Δ H values assume 1 mole of each compound (based on the coefficients) How many kJ of heat are produced when 7.23 g of CaO react? 1. Write out and balance equation: already balanced 2. Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129mol 3. Multiply: 0.129 mol CaO x 65.2 kJ mol-1 * Notice that mol / mol cancel out and you’re left with kJ 4. Solve = 8.41 kJ
LETS TRY! Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide. How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)? 1. Write out and balance the equation 2. Determine the number of moles of NaHCO3(s) 3. Set up the ratio 4. Solve for x 5. State, with justification, whether the reaction is endothermic or exothermic.
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide. 2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) OR 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) ΔH = 129 kJ How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)? 1. Balanced equation 2. Moles NaHCO3(s) = 2.24 mol 3. Ratio: x kJ / 2.24 NaHCO3(s) = 129 kJ / 2mol 4. Solve for x = 144 kJ 5. The reaction is endothermic because energy is being absorbed / the ΔH is positive.
TYPES OF ENTHALPY
TYPES OF ENTHALPY TYPE OF REACTION HEAT OF REACTION Precipitation Heat of precipitation Displacement Heat of displacement Neutralization Heat of neutralization Combustion Heat of combustion
Calculate Enthalpy for Heat of Reaction
Since most of the chemical reactions carried out to determine the heat of reaction involve aqueous solutions. Some assumptions are made: • Density of any aqueous solution is equal to density of water, 1g cm-3 • The specific heat of any aqueous solution is equal to water, 4.2 J g-1 oC-1
HEAT OF PRECIPITATION Heat change when one mole of the precipitation is formed from their ions in aqueous solution •Unit: kJ mol-1
Example: • 100 cm3 of 1.0 mol dm−3 of lead(II) nitrate, Pb(NO3)2 solution is added to 100 cm3 of 1.0 mol dm−3 of sodium sulphate, Na2SO4 solution. The temperature of the mixture rises from 30.0 °C to 33.0 °C. Calculate the heat of precipitation of lead(II) sulphate, PbSO4 • [Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution = 1 g cm−3]
Lets try! The following equation shows the precipitation of silver chloride, AgCl If 20cm3 of 0.5 mol dm-3 of silver nitrate, AgNO3 solution is added to 20cm3 of 0.5 mol dm-3 of potassium chloride, KCl solution, calculate the rise in temperature of the mixture. [Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution = 1 g cm−3] Answer: 3.9 °C
HEAT OF NEUTRALIZATION Heat of neutralization is defined as the change of heat or enthalpy during the formation of one mole of water by the neutralisation reaction of required amounts of acid and a base. HCl + NaOH  NaCl + H2O ΔH = -57.34 kJ mol-1
Example: 60 cm3 of 2.0 mol dm−3 of sodium hydroxide, NaOH solution is added into 60 cm3 of 2.0 mol dm−3 of ethanoic acid,CH3COOH.The highest temperature of the mixture is 40.5 °C. The initial temperature of sodium hydroxide, NaOH solution is 28.0 °C, and ethanoic acid, CH3COOH solution is 28.0 °C. Calculate the heat of neutralisation. [Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution = 1 g cm−3]
HEAT OF DISPLACEMENT • Heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal. • Unit kJ mol-1 Example: Zn + CuSO4  ZnSO4 + Cu • Zinc is more electropositive than copper • Zinc can replace copper from the solution
Example: An excess of magnesium powder, Mg is added into 50 cm3 of 0.25 mol dm−3of iron(II)sulphate, FeSO4 solution. The temperature of the mixture increases by 4.0 °C. Calculate the heat of displacement of iron, Fe from its salt solution. [Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution = 1 g cm−3]
HEAT OF COMBUSTION •Reaction between a substance and oxygen to form carbon dioxide and water Example: CH4 + 2O2  CO2 + 2H2O The heat is given out (exothermic reaction) •The heat of combustion is the heat released when 1 mole of a substance is completely burnt in excess oxygen, O2. •ΔH = negative (exothermic) •Unit = kJ mol-1
Example: The thermochemical equation for the complete combustion of ethanol, C2H5OH is shown below. Calculate the mass of ethanol, C2H5OH needed to burn completely in excess oxygen, O2 in order to raise the temperature of 200 cm3 of water, H2O by 50.0 °C. (Assume that no heat is lost to the surroundings)
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