Chapter 7 Making and Breaking of Bonds

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.

Energy

● Hydrocarbons● Carbohydrates

Reaction with oxygen to release energy.

Energy

● Kinetic Energy● Potential Energy● Transfer of energy● Conversion of energy

Energy

● Energy is released when making a chemical bond.

● Energy is absorbed when breaking a chemical bond.

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.”

Heat

● "Heat is energy in transit."● No energy transferred means no heat

interaction.

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!

Heat and the Kinetic Molecular Theory

● System● Surroundings● Boundary

Figure 7.1

Specific Heat

Consider the following: Figure 7.2

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

Specific Heat

● Molar heat capacitySpecific heat × molecular weightQuantity of heat required to raise 1 mole of a substance 1 degree Celsius.

Specific Heat

Table 7.1

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.

State Functions

● H is a state functionSo are T, P, V, E, ...

The First Law of Thermodynamics

● Energy is conserved.● No exceptions have been observed.

Yet.That's why it's called a law.

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)

Work

● Heat transfers can be made to occur under two separate special conditions:

System held at constant volume, orSystem held at constant pressure.

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.

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?

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

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.

Work

Table 7.2

The Enthalpy of a System

● Enthalpy, H, is a state function.● Calorimetry

Measures heat interactions associated with chemical and physical changes.

Figure 7.6

The Enthalpy of a System

● Calorimeters like the one shown measure ΔE.

● A little math is used to determine ΔH once ΔE is measured.

Enthalpies of Reaction

● ΔH = heat of reaction at constant P.May be >0 endothermic reactionMay be <0 exothermic 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


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


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?


22 50464.483

HgramsreleasedkJx

HgramsreleasedkJ

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!

Enthalpy as a State Function

● Reactions do not proceed this way.● It doesn't matter when calculating ΔH.● H is a state function.

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°.

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.

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.


● 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?


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.


● Can be used to study physical processes.● How much heat is required to accomplish

the following?CH3OH(l) → CH3OH(g)


● 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)


● 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


● Can be used to study chemical processes● 2H2(g) + O2(g) → 2H2O(g) ΔH° = ?


● 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) )


● 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


● 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.

Using Enthalpies of Atom Combination to Probe Chemical ReactionsIsomers

HO2C

C C

CO2H

HH

HO2C

C C

H

CO2HH

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

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°

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?

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.

Bond Length and the Enthalpy of Atom Combination

Table 7.4

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.

Hess's Law

● An alternative method for calculating ΔH°.

● Does not use ΔH°ac.

● Takes advantage of H being a state function.

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!

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

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.


● 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.


● 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?

Chapter 7 Making and Breaking of Bonds

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