chapter 7 making and breaking of bonds
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Chapter 7 Making and Breaking of Bonds. Chemical reactions are characterized by the making and breaking of chemical bonds. One possible consequence of a chemical reaction is a transfer of energy from the system to the surroundings. Energy. Hydrocarbons Carbohydrates - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 7Making and Breaking of Bonds
Chemical reactions are characterized by the making and breaking of chemical bonds.
One possible consequence of a chemical reaction is a transfer of energy from the system to the surroundings.
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Energy
● Hydrocarbons● Carbohydrates
Reaction with oxygen to release energy.
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Energy
● Kinetic Energy● Potential Energy● Transfer of energy● Conversion of energy
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Energy
● Energy is released when making a chemical bond.
● Energy is absorbed when breaking a chemical bond.
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Energy
● “This chapter is concerned with the energy transfers and conversions associated with chemical reactions. These processes are part of the area of study known as
Thermodynamics.”
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Heat
● "Heat is energy in transit."● No energy transferred means no heat
interaction.
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Heat
● Heat has units of energy.● Heat is not the same as temperature.● In order for a heat interaction to occur
between two systems, they must have different temperatures.
This heat interaction can form the basis for a definition of temperature!
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Heat and the Kinetic Molecular Theory
● System● Surroundings● Boundary
Figure 7.1
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Specific Heat
Consider the following: Figure 7.2
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Specific Heat
● Water and mercury have different specific heats.
● Specific heat is the energy required to raise the temperature of one gram of material one degree Celsius.
Units – cal/°C g– J/°C g
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Specific Heat
● Molar heat capacitySpecific heat × molecular weightQuantity of heat required to raise 1 mole of a substance 1 degree Celsius.
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Specific Heat
Table 7.1
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State Functions
● State of a systemP, T, n, V, …
● Extensive properties of a system depend on the size of the system.
● Intensive properties of a system are independent of the system size.
● State functions are independent of a system’s history.
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State Functions
● H is a state functionSo are T, P, V, E, ...
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The First Law of Thermodynamics
● Energy is conserved.● No exceptions have been observed.
Yet.That's why it's called a law.
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The First Law of Thermodynamics
● Energy can be exchanged between the system and the surroundings.
● Energy cannot appear or disappear.● Energy entering or leaving a system can
do so under two forms:Heat (q)Work (w)
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Work
● Heat transfers can be made to occur under two separate special conditions:
System held at constant volume, orSystem held at constant pressure.
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Work
● Energy transfer where the system maintains a constant volume is a heat interaction (q) only.
● Additionally, a work interaction (w) can result if the system changes its volume during the energy transfer.
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Work
● These special conditions have a profound effect on the form (q or w) in which energy can be transferred.
● Chemistry is typically done at constant pressure. Why?
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Work
● When energy is exchanged in the form of heat with a system held at constant pressure, the heat energy, qP, is described with a new term, enthalpy (H).
sysP Hq
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Work
● In chemistry, the term enthalpy of reaction (ΔH) is used.
It says how much energy is released or consumed in the form of heat if the reaction occurs under the condition of constant pressure.
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Work
Table 7.2
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The Enthalpy of a System
● Enthalpy, H, is a state function.● Calorimetry
Measures heat interactions associated with chemical and physical changes.
Figure 7.6
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The Enthalpy of a System
● Calorimeters like the one shown measure ΔE.
● A little math is used to determine ΔH once ΔE is measured.
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Enthalpies of Reaction
● ΔH = heat of reaction at constant P.May be >0 endothermic reactionMay be <0 exothermic reaction
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Enthalpies of Reaction
Enthalpy of reaction (ΔH) is an extensive property. It changes with the stoichiometric coefficients.
2H2(g) + O2(g) → 2H2O(g) ΔH = -483.64 kJ
4H2(g) + 2O2(g) → 4H2O(g) ΔH = -967.28 kJ
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Enthalpies of Reaction
The sign of ΔH changes for a reverse reaction.
2H2(g) + O2(g) → 2H2O(g) ΔH = -483.64 kJ
2H2Og) → 2H2(g) + O2(g) ΔH = +483.64 kJ
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Enthalpies of Reaction
The heat released for any amount of product or reactant can be determined from a single thermochemical equation.
2H2(g) + O2(g) → 2H2O(g) ΔH = -483.64 kJ
How much heat will be released when 50 grams of H2(g) are consumed?
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Enthalpies of Reaction
22 50464.483
HgramsreleasedkJx
HgramsreleasedkJ
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Enthalpy as a State Function
● H is a state function.● ΔH = Hfinal - Hinitial
● Independent of path:a. Convert reactant into atomsb. Make products from same atomsc. No extra or missing atoms!
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Enthalpy as a State Function
● Reactions do not proceed this way.● It doesn't matter when calculating ΔH.● H is a state function.
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Standard-State Enthalpies of Reaction
● For a reaction carried out at a pressure of 1 bar, ΔH = ΔH°.
● ΔH° is called the standard-state enthalpy of reaction.
● 1 bar is part of the definition of standard conditions.
● Most tabulated enthalpies of reaction are ΔH°.
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Calculating Enthalpies of Reaction
● Easier to calculate ΔH° than to measure it.● These are calculations, not estimates.
Break all the reactant bonds.Form all the product bonds.
ΔH° = the difference in energy between these two processes.
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Enthalpies of Atom Combination
● N(g) + 3 H(g) → NH3(g)● Notice these reactants are not in their
diatomic elemental forms.● If 1 mole of N atoms combine with 3
moles of H atoms to produce one mole of NH3 molecules, 1171.76 kJ of energy will be released.
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Enthalpies of Atom Combination
● N(g) + 3 H(g) → NH3(g)● Reaction is called atom combination.● Not meant to reflect actual mechanism.● The enthalpy associated with it is called
the enthalpy of atom combination, ΔH°ac.● In the above example, ΔH°ac = -1171.76
kJ.● Why is it negative?
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Enthalpies of Atom Combination
The reverse reactionNH3(g) → N(g) + 3 H(g)
is called atomization. The enthalpy change is called the enthalpy of atomization and in this example equals +1171.76 kJ.
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Enthalpies of Atom Combination
● Can be used to study physical processes.● How much heat is required to accomplish
the following?CH3OH(l) → CH3OH(g)
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Enthalpies of Atom Combination
● Can be used to study physical processes● How much heat is required to accomplish
the following?CH3OH(l) → CH3OH(g)
ΔH°ac = -2037.11 kJ for CH3OH(g)ΔH°ac = -2075.11 kJ for CH3OH(l)
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Enthalpies of Atom Combination
● Can be used to study physical processes● CH3OH(l) → CH3OH(g)
)(reactantsΔH(products)ΔHΔH ACAC
ΔH = -2037.11 kJ - (-2075.11 kJ) = +38.00 kJ
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Enthalpies of Atom Combination
● Can be used to study chemical processes● 2H2(g) + O2(g) → 2H2O(g) ΔH° = ?
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Enthalpies of Atom Combination
● Can be used to study chemical processes● 2H2(g) + O2(g) → 2H2O(g)
ΔH°=2×ΔH°ac(H2O(g)) -2×ΔH°ac (H2(g))-1×ΔH°ac (O2(g) )
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Enthalpies of Atom Combination
● Can be used to study chemical processes● 2H2(g) + O2(g) → 2H2O(g)
ΔH°=2×ΔH°ac(H2O(g)) – 2×ΔH°ac (H2(g)) - 1×ΔH°ac (O2(g) )
From Appendix B.13ΔH° = 2×(-926.29 kJ) - 2×(-435.30 kJ) -1×(-498.340 kJ)ΔH° = -483.64 kJ
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Enthalpies of Atom Combination
● Can be used to study chemical processes.● 2H2(g) + O2(g) → 2H2O(g) ● Established this as exothermic.● Determined that 483.64 kJ of energy will be
released for each mole of O2(g) consumed.
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Using Enthalpies of Atom Combination to Probe Chemical ReactionsIsomers
HO2C
C C
CO2H
HH
HO2C
C C
H
CO2HH
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Using Enthalpies of Atom Combination to Probe Chemical Reactions● Knowing ΔH°ac for each isomer allows for
the calculation of the enthalpy change associated with the transformation
HO2C
C C
CO2H
HH
HO2C
C C
H
CO2HH
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Using Enthalpies of Atom Combination to Probe Chemical Reactions● Knowing ΔH°ac gives insight into average
bond strengths.● These provide a microscopic interpretation
of overall ΔH° for a reaction.Endothermic or exothermic characterMagnitude of ΔH°
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Using Enthalpies of Atom Combination to Probe Chemical Reactions4HF(g) + SiO2(g) → SiF4(g) + 2H2O(g) ΔH° = -103.4 kJ
4HCl(g) + SiO2(g) → SiCl4(g) + 2H2O(g) ΔH° = +139.6 kJ
Why the difference in sign?
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Using Enthalpies of Atom Combination to Probe Chemical Reactions4HF(g) + SiO2(g) → SiF4(g) + 2H2O(g) ΔH° = -103.4 kJ
4HCl(g) + SiO2(g) → SiCl4(g) + 2H2O(g) ΔH° = +139.6 kJ
Why the difference in sign? The Si-F bond is stronger than the Si-Cl bond by an
amount greater than the difference in bond strength between H-F and H-Cl as shown on the next slide.
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Bond Length and the Enthalpy of Atom Combination
Table 7.4
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Bond Length and the Enthalpy of Atom Combination
● Longer bonds tend to be weaker bonds.● Multiple bonds tend to be stronger than
single bonds.This was also covered in section 4.8.
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Hess's Law
● An alternative method for calculating ΔH°.
● Does not use ΔH°ac.
● Takes advantage of H being a state function.
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Hess's Law
● Desired reactions are constructed from known reactions.
● ΔH° from known reactions combined in same way to calculate ΔH° for the desired reaction.
● Trial and error method!
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Hess's Law
C(s) + H2O(g) → CO(g) + H2(g) ΔH° = ?
from
C(s) + ½O2(g) → CO(g) ΔH° = -110.53 kJH2(g) + ½O2(g) → H2O(g) ΔH° = -241.82 kJ
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Enthalpies of Formation
● Enthalpy of formation, ΔH°f
Combined in same way as ΔH°ac.Used to calculate ΔH° for a reaction.Tabulated in Appendix B.16.
● Don't mix ΔH°f and ΔH°ac in a calculation.Use one set or the other.
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Enthalpies of Formation
● Defined as enthalpy change associated with the formation of one mole of a substance under standard conditions (1 bar, T, ...) from the elements in their thermodynamically stable form at T.
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Enthalpies of Formation
● At 25 °C and 1 bar:Oxygen is O2(g).Carbon is C(solid, graphite).The rest can be determined from B.16 by looking for the entry with ΔH°f = 0.
– Why is the entry with ΔH°f = 0 the thermodynamically stable form?