Reaction Rate
How Fast Does the Reaction Go?
Collision Theory
 In
order to react molecules and atoms
must touch each other.
 They must hit each other hard enough to
react.
– Must break bonds
 Anything that increases how often and
how hard will make the reaction faster.
Energy
Reactants
Products
Reaction coordinate
Energy
Activation Energy Minimum energy to
make the reaction
happen – how hard
Reactants
Products
Reaction coordinate
Energy
Activated
Complex or
Transition State
Reactants
Products
Reaction coordinate
Activation Energy
 Must
be supplied to start the reaction
 Low activation energy
– Lots of collision are hard enough
– fast reaction
 High Activation energy
– Few collisions hard enough
– Slow reaction
Activation energy
 If
reaction is endothermic you must keep
supplying heat
 If it is exothermic it releases energy
 That energy can be used to supply the
activation energy to those that follow
Energy
Reactants
Overall energy
change
Products
Reaction coordinate
Things that Affect Rate
 Temperature
– Higher
temperature faster particles.
– More and harder collisions.
– Faster Reactions.
 Concentration
– More concentrated molecules closer
together
– Collide more often.
– Faster reaction.
Things that Affect Rate
 Particle
size
– Molecules can only collide at the
surface.
– Smaller particles bigger surface area.
– Smaller particles faster reaction.
– Smallest possible is molecules or ions.
– Dissolving speeds up reactions.
– Getting two solids to react with each
other is slow.
Things that Affect Rate
 Catalysts-
substances that speed up a
reaction without being used up.(enzyme).
 Speeds up reaction by giving the reaction a
new path.
 The new path has a lower activation
energy.
 More molecules have this energy.
 The reaction goes faster.
 Inhibitor- a substance that blocks a
catalyst.
Energy
Reactants
Products
Reaction coordinate
Catalysts
H H
 Hydrogen
bonds to
surface of metal.
 Break H-H bonds
H
H
Pt surface
H H
H H
Catalysts
H
H
H
C
C
H
H H
H
H
Pt surface
Catalysts
 The
double bond breaks and bonds to the
catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H
Catalysts
 The
hydrogen atoms bond with the carbon
H
H
H
C
H
C
H
H
Pt surface
H H
Catalysts
H
H
H
H
C
C
H
H
H
Pt surface
H
Reversible Reactions
 2H2(g)
+ O2(g)  2H2O(g) + energy
+ energy  2H2(g) + O2(g)
 2H2(g) + O2(g)
2H2O(g) + energy
 2H2O(g)
Equilibrium
 When
you first put reactants together the
forward reaction starts.
 Since there are no products there is no
reverse reaction.
 As the forward reaction proceeds the
reactants are used up so the forward
reaction slows.
 The products build up, and the reverse
reaction speeds up.
Equilibrium
 Eventually
you reach a point where the
reverse reaction is going as fast as the
forward reaction.
 This is dynamic equilibrium.
 The rate of the forward reaction is equal to
the rate of the reverse reaction.
 The concentration of products and
reactants stays the same, but the reactions
are still running.
Equilibrium
 Equilibrium
position- how much product
and reactant there are at equilibrium.
 Catalysts speed up both the forward and
reverse reactions so do not affect
equilibrium position.
 Just get you there faster
Measuring equilibrium
 At
equilibrium the concentrations of
products and reactants are constant.
 We can write a constant that will tell us
where the equilibrium position is.
 Keq equilibrium constant
coefficients
 Keq = [Products]
[Reactants]coefficients
 Square brackets [ ] means concentration
in molarity (moles/liter)
Writing Equilibrium Expressions
 General
equation
aA + bB
 Keq
cC + dD
= [C]c [D]d
[A]a [B]b
 Write
the equilibrium expression for this
reactions.
 3H2(g) + N2(g)
2NH3(g)
Writing Equilibrium
Expressions
3H2(g) + N2(g)
Keq = [NH3]2
[H2]3 [N2]
2NH3(g)
Calculating Equilibrium
 Keq
is the equilibrium constant, it is only
effected by temperature.
 Calculate
the equilibrium constant for the
following reaction.
3H2(g) + N2(g)
2NH3(g) if at
25ºC there 0.15 mol of N2 , 0.25 mol of
NH3 , and 0.10 mol of H2 in a 2.0 L
container.
Calculating Equilibrium
Calculate molarity
[H2] = 0.10 mol of H2
2.0 L
[N2] = 0.15 mol of N2
2.0 L
[NH3] = 0.25 mol of NH3
2.0 L
Calculating Equilibrium
Keq = [NH3]2
[H2]3 [N2]
Keq =
[.125 M]2
[.05 M]3 [.075 M]
Keq = 1667
What it tells us
 If
Keq > 1 Products are favored
– More products than reactants at
equilibrium
 If Keq < 1 Reactants are favored
LeChâtelier’s Principle
Regaining Equilibrium
LeChâtelier’s Principle
 If
something is changed in a system at
equilibrium, the system will respond to
relieve the stress.
 Three types of stress are applied.
– Changing concentration
– Changing temperature
– Changing pressure
Changing Concentration
 If
you add reactants (or increase their
concentration).
 The forward reaction will speed up.
 More product will form.
 Equilibrium “Shifts to the right”
 Reactants  products
Changing Concentration
 If
you add products (or increase their
concentration).
 The reverse reaction will speed up.
 More reactant will form.
 Equilibrium “Shifts to the left”
 Reactants  products
Changing Concentration
 If
you remove products (or decrease their
concentration).
 The reverse reaction will slow down.
 More product will form.
 Equilibrium reverse“Shifts to the right”
 Reactants  products
Changing Concentration
 If
you remove reactants (or decrease their
concentration).
 The forward reaction will slow down.
 More reactant will form.
 Equilibrium “Shifts to the left”.
 Reactants  products
 Used to control how much yield you get
from a chemical reaction.
Changing Temperature
 Reactions
either require or release heat.
 Endothermic reactions go faster at higher
temperature.
 Exothermic go faster at lower
temperatures.
 All reversible reactions will be exothermic
one way and endothermic the other.
Changing Temperature
 As
you raise the temperature the reaction
proceeds in the endothermic direction.
 As you lower the temperature the reaction
proceeds in the exothermic direction.
 Reactants + heat  Products at high T
 Reactants + heat  Products at low T
 H2O (l)
H2O(s) + heat
Changes in Pressure
 As
the pressure increases the reaction
will shift in the direction of the least
gases.
 At high pressure
2H2(g) + O2(g)  2 H2O(g)
 At low pressure
2H2(g) + O2(g)  2 H2O(g)
 Low pressure to the side with the most
gases.
Two Questions
 How
Fast?
– Depends on collisions and activation
energy
– Affected by
• Temperature
• Concentration
• Particle size
• Catalyst
 Reaction Mechanism – steps
Keq =
 How
far?
– Equilibrium
• Forward and reverse rates are equal
• Concentration is constant
– Equilibrium Constant
• One for each temperature
– LeChâtelier’s Principle
CO2 (g) + C (g) + heat ↔ 2 CO (g)
Inc. CO2
Dec. CO
Inc. heat
Inc. Pressure
Decrease [C]
Hess’s Law
 Enthalpy
is a state function.
 It is independent of the path.
 We can add equations to to come up with
the desired final product, and add the DH
 Two rules
 If the reaction is reversed the sign of DH
is changed
 If the reaction is multiplied, so is DH
Standard Enthalpy
 The
enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1 M
solutions)
 Symbol DHº
 When using Hess’s Law, work by adding
the equations up to make it look like the
answer.
 The other parts will cancel out.
Calculate ΔH for the reaction
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g),
from the following data.
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) Δ H = -483.6 kJ
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) Δ H = -483.6 kJ
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) Δ H = -483.6 kJ
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) Δ H = -483.6 kJ
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) Δ H = -483.6 kJ
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
2 N2 (g) + 2 O2 (g) → 4 NO (g)
ΔH = 2 (-180.5 kJ)
4 NH3 (g) → 2 N2 (g) + 6 H2 (g)
ΔH = 2 (+91.8 kJ)
6 H2 (g) + 3 O2 (g) → 6 H2O (g)
ΔH = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
ΔH = -1628. kJ
Standard Enthalpies of Formation
 Hess’s
Law is much more useful if you
know lots of reactions.
 There are tables of standard heats of
formation. The amount of heat needed to
for 1 mole of a compound from its
elements in their standard states.
 Standard states are 1 atm, 1 M and 25ºC
 For an element it is 0
Standard Enthalpies of Formation
We can use heats of formation to
figure out the heat of reaction.
( DH of products) - ( DH of reactants) = DH o
Standard Enthalpies of Formation
Use standard enthalpies of formation to
determine the change in enthalpy for this
reaction.
ΔHf NaOH(s) = -426.7 (kJ/mol)
ΔHf HCl(g) = -92.3 (kJ/mol)
ΔHf NaCl(s) = -411.0 (kJ/mol)
ΔHf H2O(g) = -241.8 (kJ/mol)
NaOH(s) + HCl(g) → NaCl(s) + H2O(g)
Standard Enthalpies of Formation
o
o
o
( DH f products) - ( DH f reactants) = DH
∆H = [-411.0 kJ + -241.8 kJ]
– [-426.7 kJ + -92.3 kJ]
= -133.8 kJ