International Baccalaureate Chemistry Topic 5 - Energetics.

International International Baccalaureate Baccalaureate

ChemistryChemistryTopic 5 - EnergeticsTopic 5 - Energetics

ThermochemistryThermochemistry• Thermochemistry is the study of

energy changes associated with chemical reactions. – Studies the amount of energy in a

chemical reaction– Energy is measured in Joules (J)

• Energy evolved or absorbed in a chemical reaction has nothing to do with the rate of the reaction (how fast or slow the reaction takes place.

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Potential EnergyPotential EnergyEnergy an object possesses by virtue of its position or chemical composition.

Kinetic EnergyKinetic EnergyEnergy an object possesses by virtue of its motion.

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KE = mv2

EnergyEnergy• The ability to do work or transfer

heat.– Work: Energy used to cause an object

that has mass to move.– Heat: Energy used to cause the

temperature of an object to rise.

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EnthalpyEnthalpy• Also called heat content• Energy stored by the reactants and

products • Units kJ mol-1• Bonds are broken and made in chemical

reactions but the energy absorbed in breaking bonds is almost never exactly the same as the energy that is released in making new bonds. – Bonds broken = required energy– Bonds formed = released energy *

Enthalpy ChangeEnthalpy Change• All reactions have a change in

potential energy of the bonds. • This change in potential energy is

known as an enthalpy change. • Enthalpy changes can only be

measured for chemical reactions, not for state changes of substances.

• Given the symbol ΔH

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Endothermic and Endothermic and ExothermicExothermic

• A process is endothermic when H is positive.

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Endothermicity and Endothermicity and ExothermicityExothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

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Endothermic ReactionsEndothermic Reactions• Chemical reaction where the total enthalpy of the reactants is less than

the total enthalpy of the products • Positive H values• Heat energy is absorbed from surroundings• They get cooler or external heat must be added to make the reaction

work. • Will not occur as a spontaneous reaction. • Bonds broken are stronger than bonds formed

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Exothermic ReactionsExothermic Reactions• Chemicals lose energy as products are formed. • Negative H values• Heat energy is lost and surroundings get warmer. • Most spontaneous reactions are exothermic. • Products are more stable than the reactants.• Bonds made are stronger than bonds broken.

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SummarySummaryType of

ReactionHeat

Energy Change

Temperature Change

Relative Enthalpies

Sign of ΔH

Exothermic Heat energy evolved

Becomes Hotter

Hp < Hr Negative

Endothermic Heat energy

absorbed

Becomes colder

Hp > Hr Positive

Enthalpies of ReactionEnthalpies of ReactionThis quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Enthalpies of ReactionEnthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants *

Enthalpy Conditions Enthalpy Conditions • By definition, an enthalpy change must

occur at constant pressure. • Thermochemical standard conditions have

been defined as a temperature of 25°C, a pressure of 101.3 kPa, and all solutions have a concentration of 1 mol dm-3.

• Thermochemical quantities that relate to standard conditions are indicated with a standard sign (θ).

• ΔH θ – Standard thermochemical conditions.

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Types of EnthalpiesTypes of Enthalpies• ΔHrxn – Heat produced in a chemical reaction.• ΔHcomb – Heat produced by a combustion

reaction.• ΔHneut – Heat produced in a neutralization

reaction.• ΔHsol – Heat produced when a substance

dissolves. • ΔHfus – Heat produced when a substance melts. • ΔHvap – Heat produced when a substance

vaporizes. • ΔHsub – Heat produced when a substance

sublimes.

Calculation of Enthalpy Calculation of Enthalpy ChangesChanges

• Temperature:– A measure of the average kinetic energy

of the molecules.• Heat:

– The amount of energy exchanged due to a temperature difference between two substances.

• An exchange of heat causes a change in temperature.

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CalorimetryCalorimetrySince we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

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Heat Capacity and Specific Heat Capacity and Specific HeatHeat

• The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

• We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

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Heat Capacity and Specific Heat Capacity and Specific HeatHeat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

c =q

m T

*q = mC T

Constant Pressure Constant Pressure CalorimetryCalorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Pressure Constant Pressure CalorimetryCalorimetry

Because the specific heat for water is well known (4.184 J/g K), we can measure H for the reaction with this equation:q = mC T *

ExampleExample• Calculate the heat that would be required

an aluminum cooking pan whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is 0.902 J g-1 oC-

1.• Solution

• q = mCT

• = (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)

• = 64,944 J

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ExampleExample• What is the final temperature when 50 grams of

water at 20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-

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Solution: q (Cold) = -q (hot) mCT= -mCT Let T = final temperature

(50 g) (4.184 J g-1 oC-1)(T- 20oC) = -(80 g) (4.184 J g-1 oC-1)(T-60oC) (50 g)(T- 20oC) = -(80 g)(T-60oC) 50T -1000 = – 80T + 4800 130T = 5800 T = 44.6 oC

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Bomb CalorimetryBomb CalorimetryReactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Bomb CalorimetryBomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

First Law of First Law of ThermodynamicsThermodynamics

• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Use Fig. 5.5

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Hess’s LawHess’s Law H is well known for many

reactions, and it can be inconvenient to measure H for every reaction in which we are interested.

• However, we can estimate H using H values that are published and the properties of enthalpy.

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Hess’s LawHess’s Law

Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

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Hess’s LawHess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

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Standard Enthalpies of Standard Enthalpies of FormationFormation

Standard enthalpies of formation, Hθf, are

measured under standard conditions (25°C and 1.00 atm pressure).

Hess’s Law CalculationsHess’s Law Calculations• If a chemical reaction involves two or more

steps, the overall change in enthalpy equals the sum of the enthalpy changes of the individual steps.

• If a reaction is reversed, then the sign of H must be reversed.

• If a reaction is multiplied by an integer, then H must also be multiplied by the integer.

• Attention must be paid to physical states.

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Example 1:Example 1:• Calculate the heat of combustion (H comb) of

C(s) to CO(g)

Given: H (kJ)

C(s) + O2(g) → CO2(g) -393.5

CO(g) + ½O2(g) → CO2(g -283

C(s) + ½O2(g) → CO(g) ????

(-110.5 kJ)

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Example 2Example 2• Calculate ΔH for the reaction:

2C(s) + H2(g) → C2H2(g)

Given: ΔH (kJ)

C2H2(g) +5/2O2(g) → 2CO2(g) + H2O(l) -1299.6

C(s) + O2(g) → CO2(g) -393.5

H2(g) + ½O2(g) → H2O(l) -285.9

(+226.7 kJ)

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Example 3Example 3• Given:

ΔH(kJ)NO(g) + O3(g) → NO2(g) + O2(g) -198.9

O3(g) → 3/2 O2(g) -142.3

O2(g) → 2O(g) +495.0

• Find ΔH for the reaction:NO(g) + O(g) → NO2(g)

(-304.1 kJ)

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Try This!Try This!• Calculate ΔH for the synthesis of diborane from its elements

according to the equation:

2B(s) + 3H2(g) → B2H6(g)Given:

ΔH(kJ)2B(s) + 3/2O2(g) → B2O3 -1273B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) -2035H2(g) + ½O2(g) → H2O(l) -286H2O(l) → H2O(g) +44

(+36 kJ)

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Bond EnthalpiesBond Enthalpies• Bond enthalpy is the amount of energy required to convert

a gaseous molecule into gaseous atoms.HCl(g) → H(g) + Cl(g)

note: Cl2 and H2 are not formed.• Limitations:

– Can only be used if all the reactants and products are gaseous

– Values have been obtained from a number of similar compounds and therefore, will vary slightly in different compounds.

• Enthalpy changes (ΔH) can be calculated using bond enthalpies.

• Bond formation: negative (exothermic)• Bond breaking: positive (endothermic)• ΔH = H bonds broken – H bonds formed

Example 1Example 1

• C2H4(g) + H2(g) → C2H6(g)Bonds Broken (kJ mol-1) Bonds Formed(kJ mol-1)

C=C 612 C-C 3484(C-H) 4(412) 6(C-H) 6(412)H-H 436

Total: 2696 kJ 2820 kJ

ΔH = 2696 - 2820 = -125 kJ mol-1

Example 2Example 2• Estimate the ΔH for the following reaction from bond

energies.

C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g)

Bonds Broken (kJ mol-1) Bonds Formed (kJ mol-1)6 C-H 6(413) = 2478 4 C=O 4(745) = 29807/2 O=O 7/2(495) = 1732.5 6 O-H 6(467) = 28021 C-C 1(347) = 347

Total = 4557.5 Total = 5782

ΔH = 4557.5 – 5782 = -1224.5 kJ mol-1