energy changes, reaction rates, and equilibrium. the capacity to do work ◦ the ability to move or...
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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