International Baccalaureate Chemistry Topic 5 - Energetics.
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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.
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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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-
1
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
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