Chapter 6Energy Changes, Reaction Rates, and

Equilibrium

The capacity to do work◦ The ability to move or change something

Change in position Change in speed Change in state (form of matter)

Stored energy = potential energy Moving energy = kinetic energy

Energy

potential kinetic

The total energy in a system does not change. Energy cannot be created or destroyed. It can only be changed from one form into another or transferred from one object to another.

The law of conservation of energy

Stored energy◦ Exists in natural attractions and repulsions

Chemical energy◦ PE possessed by chemicals◦ Stored in chemical bonds◦ Breaking bonds requires energy◦ Forming bonds releases energy◦ PE may be released and converted to heat

Heat is a form of kinetic energy due to motion of particles

A compound with lower potential energy is more stable than a compound with higher energy

Potential energy (PE)

A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 oC.

A joule (J) is a SI unit of energy.

• Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal)

Units of energy

1 cal = 4.184 J

1,000 J = 1 kJ 1,000 cal = 1 kcal

1 kcal = 4.184 kJ

A gummy bear is 9.000 Calories (nutritional calories). How much energy is stored in a gummy bear in units of Joules?

9.000 Cal = 9.000 kcal x 1000 cal = 9000. cal

1 kcal

9000. cal x 4.184 J = 37660. J = 37.66 kJ 1 cal

Energy in food

Breaking bonds requires energy Forming bonds releases energy

Energy changes in reactions

ClCl

To cleave this bond, 58 kcal/mol must be added.

H = +58 kcal/molEndothermic To form this bond, 58

kcal/mol is released.H = −58 kcal/mol

Exothermic

H is the energy absorbed or released in a reaction; it is called the heat of reaction orthe enthalpy change.

The bond dissociation energy is the H for breaking a covalent bond by equally dividing the e− between the two atoms.

Bond dissociation energies are positive values, because bond breaking is endothermic (H > 0).

Bond formation always has negative values, because bond formation is exothermic (H < 0).

Bond dissociation energy

H H H + H H = +104 kcal/mol

H + H H H H = −104 kcal/mol

The stronger the bond, the higher its bond dissociation E.

Bond Strength

H indicates the relative strength of the bonds broken and formed in a reaction:

• H negative: Exothermic reaction: more energy is released in forming bond than is need to break bonds: products have stronger bonds.

• H positive: Endothermic reaction: more energy is needed to break bonds than is released in forming bonds: products have weaker bonds.

Bond strength

Endothermic & exothermic

For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds.

Energy diagrams

The orientation of the two molecules must be correct as well.

Energy diagrams

Ea, the energy of activation, is the difference in energy between the reactants and the transition state. It can be thought of as the energy barrier that must be overcome for the reaction to occur.

Energy diagrams

When the Ea is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow.

When the Ea is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

Energy diagrams

ΔH is negative, the reaction is exothermic:

Low Ea example

If H is positive, the reaction is endothermic

Energy diagrams

Increasing the concentration of the reactants:o Increases the number of collisionso Increases the reaction rate

Increasing the temperature of the reaction:o Increases the kinetic energy of the moleculeso Increases the reaction rate

A catalyst is a substance that speeds up the rate of a reaction and can be recovered unchanged.oCatalysts lower Activation Energy, Ea

Rates of reactions

A catalyst is a substance that speeds up the rate of reaction

A catalyst is recovered unchanged in a reaction, and does not appear in the product

A catalyst lowers Ea With a catalyst H is the same

Catalysts

The uncatalyzed reaction (higher Ea) is slower

The catalyzed reaction (lower Ea) is faster

Catalysts

Enzymes (usually protein molecules) are biological catalysts held together in a very specific three- dimensional shape

The active site binds a reactant, which then under- goes a very specific reaction with an enhanced rate.

The enzyme lactase converts the carbohydrate lactose into the two sugars glucose and galactose

People who lack adequate amounts of lactase suffer from abdominal cramping and diarrhea because they cannot digest lactose when it is ingested.

Reaction rates

A reversible reaction can occur in either direction

The system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction

The net concentrations of reactants and products do not change at equilibrium

Equilibrium

CO(g) + H2O(g) CO2(g) + H2(g)

The forward reaction proceeds to the right.

The reverse reactionproceeds to the left.

The relationship between the concentration of the products and the concentration of the reactants is the equilibrium constant, K.

Equilibrium

a A + b B c C + d D

equilibriumconstant = K =

[C]c [D]d

[A]a [B]b=

[products][reactants]

*Brackets, [ ], are used to symbolize concentration in moles per liter (mol/L).

Equilibrium

N2(g) + O2(g) 2 NO(g)

equilibriumconstant

= K =[N2] [O2]

[NO]2

*The coefficient becomes the exponent.

Magnitude of the equilibrium constant When K is much greater than 1, equilibrium

lies to the right and favors the products

When K is much less than 1, equilibrium lies to the left and favors the reactants

Equilibrium

[products][reactants]

The numerator is larger.

[products]

[reactants] The denominator is larger.

When K is around 1 (0.01 < K < 100), both reactants and products are present insimilar amounts

Equilibrium

[products][reactants]

Both are similarin magnitude.

HOW TO Calculate the Equilibrium Constant for a

Reaction

A2 + B2 2 AB

Step [1]

Write the expression for the equilibriumconstant from the balanced equation.

[AB]2

[A2][B2]K =

Step [2]

Substitute the given concentrations inthe equilibrium expression and calculate K.

[AB]2

[A2][B2]K = =

[0.50]2

[0.25][0.25] =0.25

0.0625= 4.0

If a chemical system at equilibrium is disturbed or stressed, the system will react in a direction that counteracts the disturbance or relieves the stress

Some of the possible disturbances:◦Concentration changes◦Temperature changes◦Pressure changes

Le Châtelier’s Principle

What happens if [CO(g)] is increased?◦ The concentration of O2(g) will decrease.

◦ The concentration of CO2(g) will increase.

Le Châtelier’s Principle

2 CO(g) + O2(g) 2 CO2(g)

What happens if [CO2(g)] is increased?◦ The concentration of CO(g) will increase.◦ The concentration of O2(g) will increase.

Le Châtelier’s Principle

2 CO(g) + O2(g) 2 CO2(g)

Le Châtelier’s Principle

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•What happens if a product is removed?

•The concentration of ethanol will decrease.

•The concentration of the other product (C2H4) will increase.

Le Châtelier’s Principle

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•When the temperature is increased, the reaction that absorbs heat is favored.

•An endothermic reaction absorbs heat, so increasing the temperature favors the forward reaction.

Le Châtelier’s Principle

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•Conversely, when the temperature is decreased, the reaction that adds heat is favored.

•An exothermic reaction releases heat, so increasing the temperature favors the reverse reaction.

Le Châtelier’s Principle

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•When pressure increases, equilibrium shifts in the direction that decreases the number of moles in order to decrease pressure.

Le Châtelier’s Principle

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•When pressure decreases, equilibrium shifts in the direction that increases the number of moles in order to increase pressure.

Le Châtelier’s Principle