Ch 17 Thermochemistry(First Class)

7/27/2019 Ch 17 Thermochemistry(First Class)

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Chapter 17

Thermochemistry


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Thermochemistry:

Study ofenergy changes that occur

during chemical reactions andchanges in state

Section 17.1: The flow of energy


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energy changes can either occurthrough heat transfer or work

heat (q), energy is transferredfrom a warmer object to a coolerobject (always)

Adding heat increasestemperature

Section 17.1: The flow of energy


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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


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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



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IfSystem (energy) Surrounding(energy) by the same amount

the total energy of the universe does notchange

Law of conservation of energy



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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 thesurroundings (system cools down, -q)


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Example 1:

A container of melted paraffin wax is allowed tostand at room temperature (r.t.) until the waxsolidifies. What is the direction of heat flow asthe liquid wax solidifies? Is the process

exothermic or endothermic?

Answer: Heat flows from the system (paraffin) tothe surroundings (air)

Process: exothermic


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Example 2:

When solid Ba(OH)28H2O is mixed in a beakerwith solid NH4SCN, a reaction occurs. Thebeaker quickly becomes very cold. Is thereaction exothermic or endothermic?

Answer: Endothermic

surrounding = beaker and air

System = chemicals within beaker


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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 by1oC

b) joule (j)1 joule of heat raises thetemperature of 1 g of pure water 0.2390oC

Joule = SI unit of energy


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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)


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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


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Specific heat calculation

C = q = heat (joules/calories)

m * T mass (g) * Temp. (oC)

T = TfTi (Tf= final temperature)

(Ti = initial temperature)

C = j or cal

(g * oC) (g * oC)


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Example 1

1. The temperature of a 95.4 g piece of copper

increases from 25.0 oC to 48.0 oC when thecopper absorbs 849 j of heat. What is the

specific heat of copper?

unknown: Ccu

Know:

mass copper = 95.4 g

T = TfTi = (48.0 oC25.0 oC)

= 23.0 oC

q = 849 j


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Example1

C = q

m * T

C = 849 j

95.4 g * 23.0 oC

C = 0.387 j/g * oC

Sample problem 17.1, page 510


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Example 2

2. How much heat is required to raise the

temperature of 250.0 g of mercury (Hg) 52oC?

unknown: q

Know:

mass Hg = 250.0 gT = 52.0 oC

CHg = 0.14 j/(g *oC)


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Example 2

C = qm * T

q = CHg

* m * T

q = 0.14 (j/g * oC) * 250.0 g * 52 oC

q = 1.8 x 103 j (1.8 kj)

Problem #4 page 510


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Section 17.2

Measuring Enthalpy Changes


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Two types of calorimeters:

a) Constant-Pressure calorimeter (eg. foam cups)

As most reactions occur at constant pressure we can saythat:

A change in enthalpy (H) = heat supplied (q)

So, a release of heat (exothermic) corresponds to adecrease in enthalpy (at constant pressure)

An absorption of heat (endothermic) corresponds to anincrease in enthalpy (constant pressure)

17.2: Enthalpy (measuring heat flow)


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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.


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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 (orH) for the above

equation is -65.2 kJ


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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

kNaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)2


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Amount of heat released/absorbed during a

reaction depends on the number of moles of

reactants involved

eg.:


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

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


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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


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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)

l 1


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Example1

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

1. Calculate the amount of heat in (kJ) required to decompose 2.24 mol

NaHCO3(S)

Known:

2.24 mol NaHCO3 decomposes

H= 129 kJ (2 mol NaHCO3)

Unknown:

H= ?

Solve:

129 kJ = H

2 mol NaHCO3(s) 2.24 mol NaHCO3(s)

H = (129 kJ) * 2.24 mol NaHCO3(s)2 mol NaHCO3(s)

H = 144 kJ

Sample problem 17.3; p. 516

E l 2


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Example 22. When carbon disulfide is formed from its

elements, heat is absorbed. Calculate theamount 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

H= 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:

H= ?


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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 = H

1 mol CS2(l) 0.074 mol CS2(l)

H = (89.3 kJ) * 0.074 mol CS2(l)

1 mol CS2(l)

H = 6.63 kJ


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17.3: Heat in changes of state

Objective:

-Heats of Fusion and Solidification

-Heats of Vaporization and Condensation

-Heat of solution


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The temperature remains constant when achange of state occurs via a gain/loss of energy

Heat absorbed by 1 mole of a solid during

melting (constant temperature) is the molarheat of fusion (Hfus)

Molar heat of solidification (Hsolid) is the heat

lost by 1 mole of liquid as it solidifies(constant temperature)

So, Hfus = Hsolid

17.3: Heat of fusion and solidification


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Figure 17.9: Enthalpy changes and changes of

state



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

H2O(l) solid6.01 kJ/molH2O(s)



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17.3: Heats of Vaporization and Condensation

Molar heat of vaporization (Hvap): Amount of

heat required to vaporize one mole of a liquid

at the liquids normal boiling point

H2O(l) vap kJ/molH2O(g)

Molar heat of condensation (Hcond): heat released

when 1 mole of vapor condenses

So, Hvap = -Hcond

H2O(l) cond kJ/molH2O(g)


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Figure 17.10: Heating curve of water

17.3: Heat of vaporization and

condensation


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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

(Hsoln)

NaOH(s) soln5.1 kJ/molNa+

(aq) + OH-(aq)

H2O(l)


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Applications: hot/cold packs

Hot pack:

17.3: Heat of solution

CaCl2(s) soln82.8 kJ/molCa2+

(aq) + 2Cl-(aq)

H2O(l)

Cold pack:

NH4NO3(s) soln25.7 kJ/molNH4+

(aq) + NO3-(aq)

H2O(l)

Ch 17 Thermochemistry(First Class)

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Transcript of Ch 17 Thermochemistry(First Class)