Chapter 17 Thermochemistry

Section 17.1: The flow of energy Thermochemistry: Study of energy changes that occur during chemical reactions and changes in state

Section 17.1: The flow of energy energy changes can either occur through heat transfer or work heat (q), energy is transferred from a warmer object to a cooler object (always) Adding heat increases temperature

17.1: Endothermic & Exothermic processes Kinetic energy vs. potential energy potential energy is determined by the strength of repulsive and attractive forces between atoms In a chemical reaction, atoms are recombined into new arrangements that have different potential energies change in potential energy: due to absorption and release of energy to and from the surroundings

17.1: Endothermic & Exothermic processes Two parameters crucial in Thermochemistry: a) system--part of the universe attention is focused b) surroundings--everything else in the universe System + surrounding = universe Fundamental goal of Thermochem.: study the heat flow between the system and its surroundings

17.1: Endothermic & Exothermic processes If System (energy) Surrounding (energy) by the same amount the total energy of the universe does not change Law of conservation of energy

17.1: Endothermic & Exothermic processes Direction of heat flow is given from the point of view of the system So, endothermic process: system absorbs heat from the surroundings (system heats up) Heat flowing into the system = +q Exothermic process: heat is released into the surroundings (system cools down, -q)

17.1: Endothermic & Exothermic processes Example 1: A container of melted paraffin wax is allowed to stand at room temperature (r.t.) until the wax solidifies. What is the direction of heat flow as the liquid wax solidifies? Is the process exothermic or endothermic? Answer: Heat flows from the system (paraffin) to the surroundings (air) Process: exothermic

17.1: Endothermic & Exothermic processes Example 2: When solid Ba(OH)2▪8H2O is mixed in a beaker with solid NH4SCN, a reaction occurs. The beaker quickly becomes very cold. Is the reaction exothermic or endothermic? Answer: Endothermic surrounding = beaker and air System = chemicals within beaker

17.1: Units of heat flow Two units used: a) calorie (cal)—amount of heat required to raise the temperature of 1g of pure water by 1oC b) joule (j)—1 joule of heat raises the temperature of 1 g of pure water 0.2390oC Joule = SI unit of energy

17.1 Heat capacity & specific heat Heat capacity is the quantity of heat needed to raise the temperature of an object exactly 1oC Heat capacity depends on: a) mass b) chemical composition So, the greater the mass the greater the heat capacity eg.: cup of water vs. a drop of water (cup of water = greater heat capacity)

17.1: Heat capacity & specific heat Specific heat: amount of heat required to raise the temperature of 1g of a substance by 1oC Table 17.1 (p.508): List of specific heats of substances

Specific heat calculation C = q = heat (joules/calories) m * ΔT mass (g) * ΔTemp. (oC) ΔT = Tf –Ti (Tf = final temperature) (Ti = initial temperature) C = j or cal (g * oC) (g * oC)

Example 1 The temperature of a 95.4 g piece of copper increases from 25.0 oC to 48.0 oC when the copper absorbs 849 j of heat. What is the specific heat of copper? unknown: Ccu Know: mass copper = 95.4 g ΔT = Tf – Ti = (48.0 oC – 25.0 oC) = 23.0 oC q = 849 j

Example1 C = q m * ΔT C = 849 j 95.4 g * 23.0 oC C = 0.387 j/g * oC Sample problem 17.1, page 510

Example 2 How much heat is required to raise the temperature of 250.0 g of mercury (Hg) 52 oC? unknown: q Know: mass Hg = 250.0 g ΔT = 52.0 oC CHg = 0.14 j/(g * oC)

Example 2 Problem #4 page 510 C = q m * ΔT q = CHg * m * ΔT q = 0.14 (j/g * oC) * 250.0 g * 52 oC q = 1.8 x 103 j (1.8 kj) Problem #4 page 510

Measuring Enthalpy Changes Section 17.2 Measuring Enthalpy Changes

17.2: Enthalpy (measuring heat flow) Calorimetry: accurate measurement of heat flow into or out of a system in chemical and physical processes In calorimetry, heat released by a system is equal to the heat absorbed by its surroundings and vice versa Instrument used to measure absorption or release of heat is a calorimeter

17.2: Enthalpy (measuring heat flow) Two types of calorimeters: a) Constant-Pressure calorimeter (eg. foam cups) As most reactions occur at constant pressure we can say that: A change in enthalpy (ΔH) = heat supplied (q) So, a release of heat (exothermic) corresponds to a decrease in enthalpy (at constant pressure) An absorption of heat (endothermic) corresponds to an increase in enthalpy (constant pressure)

17.2: Enthalpy (measuring heat flow) b) Constant-Volume Calorimeters (eg. bomb calorimeters) Substance is burned (in the presence of O2) inside a chamber surrounded by water (high pressure) Heat released warms the water Figure 17.6 (p. 512) Bomb calorimeter.

17.2: Thermochemical equations A chemical equation that includes the enthalpy change is called a thermochemical equation Reactants and products at their usual physical state (at 25 oC) given at standard pressure (101.3 kPa) So, the heat of reaction (or ΔH) for the above equation is -65.2 kJ

17.2: Thermochemical equations So, rewrite the equation as follows: Other reactions absorb heat from the surroundings, eg.: Rewrite to show heat of reaction

17.2: Thermochemical equations Amount of heat released/absorbed during a reaction depends on the number of moles of reactants involved eg.:

Enthalpy Diagrams Enthalpy of reactants greater than of products Diagram A: Enthalpy of reactants greater than of products Diagram B: Enthalpy of reactant less than of products

17.2: Thermochemical equations Physical states of reactants and products must be stated: Vaporization of H2O(l) requires more heat (44.0 kJ)

Example1 1. Calculate the amount of heat in (kJ) required to decompose 2.24 mol NaHCO3(S) Known: Unknown: ΔH = ? 2.24 mol NaHCO3 decomposes ΔH = 129 kJ (2 mol NaHCO3) Solve: 129 kJ = ΔH 2 mol NaHCO3(s) 2.24 mol NaHCO3(s) ΔH = (129 kJ) * 2.24 mol NaHCO3(s) 2 mol NaHCO3(s) ΔH = 144 kJ Sample problem 17.3; p. 516

Example 2 2. When carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat in (kJ) absorbed when 5.66 g of carbon disulfide is formed. Known: Unknown: ΔH = ? 5.66 g CS2 is formed ΔH = 89.3 kJ (1 mol CS2(l)) Molar mass: CS2(l): C = 12.0 g/mol 2 *S = 32.1 g/mol = 64.2 g/mol 76.2 g/mol

Example 2 Solve: 1. Moles CS2(l) = 5.66g CS2 = 0.0743 mol CS2(l) 76.2 g/mol CS2(l) 2. 89.3 kJ = ΔH 1 mol CS2(l) 0.074 mol CS2(l) ΔH = (89.3 kJ) * 0.074 mol CS2(l) 1 mol CS2(l) ΔH = 6.63 kJ

17.3: Heat in changes of state Objective: -Heats of Fusion and Solidification -Heats of Vaporization and Condensation -Heat of solution

17.3: Heat of fusion and solidification The temperature remains constant when a change of state occurs via a gain/loss of energy Heat absorbed by 1 mole of a solid during melting (constant temperature) is the molar heat of fusion (ΔHfus) Molar heat of solidification (ΔHsolid) is the heat lost by 1 mole of liquid as it solidifies (constant temperature) So, ΔHfus = ΔHsolid

17.3: Heat of fusion and solidification Figure 17.9: Enthalpy changes and changes of state

17.3: Heat of fusion and solidification

17.3: Heats of Vaporization and Condensation Molar heat of vaporization (ΔHvap): Amount of heat required to vaporize one mole of a liquid at the liquid’s normal boiling point Molar heat of condensation (ΔHcond): heat released when 1 mole of vapor condenses So, ΔHvap = -ΔHcond

17.3: Heat of vaporization and condensation Figure 17.10: Heating curve of water

17.3: Heat of solution There is heat released/gained when a solute dissolves in a solvent The enthalpy change due to 1 mole of a substance dissolving: molar heat of solution (ΔHsoln)

17.3: Heat of solution Applications: hot/cold packs Hot pack: