chapter 17 thermochemistry

Chapter 17

Thermochemistry

• 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

Section 17.1: The flow of energy

• Kinetic energy vs. potential energy

17.1: Endothermic & Exothermic processes

energy due tomotion

energy due to a substance'schemical composition

• 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

• 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


• If System (energy) Surrounding (energy) by the same amount

• the total energy of the universe does not change

Law of conservation of energy



• 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 processesExample 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 processesExample 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 11. 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

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

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

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

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)

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

CaO(s) + H2O(l) Ca(OH)2 + 65.2 kJ

Heat released

• 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


So, rewrite the equation as follows:

CaO(s) + H2O(l) Ca(OH)2(s) 65.2 kJ

• Other reactions absorb heat from the surroundings, eg.:

Rewrite to show heat of reaction

2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)+ 129 kJ

kJNaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)2

• Amount of heat released/absorbed during a reaction depends on the number of moles of reactants involved

eg.:



258 kJNaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)4

Enthalpy Diagrams

CaO(s) + H2O(l)

H = -65.2 kJ

Ca(OH)2(s)

Exothermic Reaction

Na2CO3(s) + H2O(l) + CO2(g)

H = 129 kJ

2 NaHCO3(s)

Endothermic Reaction

Diagram A:

Enthalpy of reactants greater than of products

Diagram B:

Enthalpy of reactant less than of products

Physical states of reactants and products must be stated:


H2O(l) H2(g)+ 1 O2(g)

2285.8 kJ

H2O(g) H2(g)+ 1 O2(g)

2241.8 kJ

Difference = 44.0 kJ

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:•2.24 mol NaHCO3 decomposes•ΔH = 129 kJ (2 mol NaHCO3)

Unknown:•ΔH = ?

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 kJSample problem 17.3; p. 516

Example 22. 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.

C(s) + 2 S(s) CS2(l)H = 89.3 kJ

Known:•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

Unknown:•ΔH = ?

Example 2Solve:

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

• 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


H2O(s) fus.01 kJ/molH2O(l)

H2O(l) solid 6.01 kJ/molH2O(s)


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

H2O(l) vap kJ/molH2O(g)

• Molar heat of condensation (ΔHcond): heat released when 1 mole of vapor condenses

•So, ΔHvap = -ΔHcond

H2O(l)cond kJ/molH2O(g)

• Figure 17.10: Heating curve of water

17.3: Heat of vaporization and condensation

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)

NaOH(s) soln 5.1 kJ/molNa+(aq) + OH-

(aq)

H2O(l)

• Applications: hot/cold packs

• Hot pack:

17.3: Heat of solution

CaCl2(s) soln 82.8 kJ/molCa2+(aq) + 2Cl-(aq)

H2O(l)

•Cold pack:

NH4NO3(s) soln25.7 kJ/molNH4+

(aq) + NO3-(aq)

H2O(l)

chapter 17 thermochemistry

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