Rates of Reaction and Chemical Equilibrium
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Rates of Reaction andChemical Equilibrium
ChemistryMs. Piela
Rates of Reaction
A rate is any change that occurs within an interval of time
Collision theory states that particles will react to form products when they collide This requires a certain amount of kinetic energy
As an example, imagine throwing clay balls at one another
Without enough force, they will ricochet off from one another. When they have enough kinetic energy, they will smash and fuse into something entirely new.
This is the same in chemical reactions
Activation Energy
Activation energy (Ea) is the minimum amount of energy required for a chemical reaction to occur
Activation Energy Diagram
Diagram can be used to determine in reaction is exothermic (- ∆H) or endothermic (+∆H)
Activated complex (or transition state) is the arrangement of atoms at the peak of the Ea barrier Unstable and only
appears briefly
Activation Energy Diagram
Factors Affecting Rates of Reactions
Four Major FactorsTemperature
ConcentrationParticle Size
Catalysts
Factors Affecting Rates of Reaction
Temperature Generally, an increase in temperature will increase
the rate of a reaction, and vice versa Explanation:
Increasing temperature speeds up particles, increasing kinetic energy
This allows a reaction to overcome its activation energyExample
Popcorn
Factors Affecting Rates of Reaction
Concentration Increasing the number of particles (thereby
increasing concentration) will increase the reaction rate
Explanation: More particles will increase the collision frequency More collisions will mean more chances for a
reaction to occurExample
Wooden splint in pure oxygen
Factors Affecting Rates of Reaction
Particle Size (aka Surface Area) The smaller the particle, the larger the surface area
for a given mass of particles An increase in surface area will increase the rate of
reaction Explanation:
With more exposed area to react, this increases the rate of reaction
Example Fine Dust Explosions (1902 Chicago Fire) Lycopodium dust explosion
Factors Affecting Rates of Reaction
Catalysts Any substance that
increases the rate of a reaction without being consumed
Catalysts allow a reaction to occur at lower activation energy
Enzymes are biological catalysts
Activation Energy Diagram
Factors Affecting Rates of Reaction
Review of Factors Affecting Rates
Temperature• Increase causes higher kinetic energy
Concentration• Higher concentration causes more collisions
Particle Size• More surface area leads to more exposed spaces to react
Catalysts• Substances that increase the rates of reaction without being spent
Chemical Equilibrium
Reversible reactions are reactions that simultaneously occur in both directions
Chemical equilibrium is the state in which the forward and reverse reactions take place at the same rate Sample Reaction:
2 SO2 (g) + O2 (g) 2 SO3 (g)
Chemical Equilibrium
Differences between rates and concentration One doesn’t imply the other! Although rates of forward and reverse reactions are
equal at chemical equilibrium, the concentrations are not necessarily the same
ExampleA B
1% 99% If A reacts to give B and the equilibrium mixture contains
99% B and only 1% A, the formation of B is said to be favored
Effect of Catalyst on Equilibrium
Although catalysts speed up a reaction rate, they will not affect the concentration of reactants and products at equilibrium Catalysts can only decrease the time required to reach
equilibriumAffects speed,
not concentration! Meep meep!
Le Châtelier's Principle
When a stress is applied to a system at equilibrium, the equilibrium will shift in order to relieve the stress
Factors Affecting Equilibrium
Three Major Factors
Concentration
Temperature
Pressure
Le Châtelier's Principle
Concentration Changing the amount, or concentration, of any
reactant or product in a system at equilibrium disturbs that equilibrium
Example equilibrium:2 H2CO3 (g) 2 H2O (l) + 2 CO2 (g)
Adding CO2 Shifts equilibrium to the left Concentration of H2O will decrease Concentration of H2CO3 will increase
Le Châtelier's Principle
Example equilibrium:2 H2CO3 (g) 2 H2O (l) + 2 CO2 (g)
Removing CO2
Shifts equilibrium to the right Concentration of H2O will increase Concentration of H2CO3 will decrease
Le Châtelier's Principle
Temperature Increasing the temperature causes an equilibrium
position of a reaction to shift in the direction that will absorb the heat. Consider heat as if it were a part of the reaction
2 SO2 (g) + O2 (g) 2 SO3 (g) + heat
Adding heat Shifts equilibrium to the left. Concentration of SO2 and O2 will increase. Concentration of SO3 will decrease.
Le Châtelier’s Principle
2 SO2 (g) + O2 (g) 2 SO3 (g) + heat
Removing heat Shifts equilibrium to the right. Concentration of SO2 and O2 will decrease. Concentration of SO3 will increase.
Le Châtelier's Principle
Pressure Changing the pressure only affects an
equilibrium with an uneven number of reactants and products
Increasing pressure will shift an equilibrium in the direction of the least number of moles
Be aware that a change in volume also affects pressure Reducing volume will increase the
pressure, and vice versa
Le Châtelier's Principle
Example Equilibrium:CO + 3 H2 (g) CH4 (g) + H2O (g)
Increasing pressure (reducing volume) will shift the equilibrium to the… right
Decreasing pressure (increasing volume) will shift the equilibrium to the… left
Haber Bosch Process & Le Châtelier’s Principle
Equilibrium Constant Expressions
A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached that is independent of the initial concentration of the participants
Given the symbol Keq
Can be used to determine whether products or reactants are favored in an equilibrium
Equilibrium Constant Expressions
Steps for Determining Keq
Make sure the chemical reaction is balanced and at equilibrium
Place the product concentrations in the numerator, and the reactant concentrations in the denominator
[products] [reactants]
Concentrations of any solid or liquid is left out because the concentrations never change Water is always omitted in aqueous solution. Water is
almost constant during the reaction
Keq =
Steps for Determining Keq
To complete the expression, raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation
For a general reaction: aA+ bB cC + dDThe equilibrium constant expression can be
written as
The equilibrium constant is unitless
Keq = [C]c[D]d
[A]a[B]b
Example #1
1. An aqueous solution of carbonic acid reacts to reach equilibrium described below:
H2CO3 (aq) + H2O (l) HCO3- (aq) + H3O+
(aq)
Write the equilibrium expression, Keq, for this reaction
][]][[
32
33
COHOHHCOKeq
Example #2
2. The brown gas, NO2, is in equilibrium with the colorless gas N2O5. Write the equilibrium expression for the following reaction:
N2O4 (colorless) 2 NO2 (brown)
][][42
22ON
NOeqK
Practice Problems
Practice #1
Practice #2
][][][
23
2
23
NHNHKeq
][][][
22
2
23
OSOSOKeq
Calculating using Keq
Example #1H2CO3 (aq) + H2O (l) HCO3
- (aq) + H3O+
(aq)
The solution contains the following solute concentrations: H2CO3; 3.3 x 10-2 M; HCO3
- ion, 1.19 x 10-4 M; and H3O+, 1.19 x 10-4 M. Determine the Keq expression, and the value of Keq
Example #1
][]][[
32
33
COHOHHCOKeq
]103.3[]1019.1][1019.1[
2
44
MxMxMxKeq
6101.1 x
Example #2
Keq for the equilibrium below is 1.8 x 10-5 at a temperature of 25 °C. Calculate [NH4
+] when [NH3] = 6.82 x 10-3 M and [OH-] = 3.45 x 10-3 M.
NH3 (aq) + H2O (l) NH4+
(aq) + OH- (aq)
Example #2
][]][[
3
4
NHOHNHKeq
]1082.6[]1045.3][[108.1 3
345
MxMxNHx
Mx 51056.3
Practice Problems
Problem #1
Problem #2
]102.4][108.1[][1065.1 43
23
MxMxNOx
29 ][1025.1 NOx
]1032.5[]105.1[]1098.1[
323
22
MxMxMxKeq
32752
][1054.3 5 NOMx
Magnitude of Keq
The meaning of Keq
If Keq is very large, products will be favored This implies the numerator is large in the Keq
expression compared to denominator If Keq is very small, reactants will be favored
This implies the denominator is large in the Keq expression compared to numerator
If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium
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