Transcript Unit 8 - Phillips Scientific Methods

Unit 8 – Thermochemistry, Kinetics
& Equilibrium
I . Thermochemistry
Thermodynamics & Energy
Thermodynamics is the study of
energy.
 Energy is the ability to work or
produce heat.
 Work is a force acting over a
distance.
 Energy can be classified as either
kinetic or potential energy.

(a) Radiant energy
(c) Chemical energy
(b) Thermal energy
(d) Nuclear energy
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
(e) Electrical energy
Kinetic Energy – energy of motion
KE = ½ m v 2
A
B
mass velocity (speed)
Potential Energy – stored energy, energy of position or
composition
• Batteries (chemical potential energy)
• Spring in a watch (mechanical potential energy)
• Water trapped above a dam (gravitational potential energy)
• *Chemical bond energy = PE
Law of Conservation of Energy

Energy can be converted from one
form to another but can be neither
created nor destroyed.
 Energy of the universe is a
constant!
 1st Law of Thermodynamics
Law of Conservation of Energy
DEPICTED
Potential energy
of A is changed to
kinetic energy.
Part of this energy
is transferred to B
through work.
Some energy is
unaccounted for…
Law of Conservation of Energy
DEPICTED

Some energy is transferred to the hill
as heat.
 This transfer of energy is called
frictional heating.
 The temperature of the hill increases
very slightly.
Temperature
How is warm water different from cold
water?
 Temperature is a measure of the
random motions of the components
of a substance (avg KE)
 H2O molecules in warm water are
moving around more rapidly than
the H2O molecules in cold water.

Equal masses of hot and cold water.
What will happen in this experiment?
A Close-Up View
• Energy is transferred through the metal wall from
the hot water to the cold water.
• The two water samples eventually reach the
same temperature.
• The energy lost by the hot water EQUALS the
energy gained by the cold water.
Heat (Thermal Energy)
The flow of energy due to a
temperature difference.
 Thermal energy is transferred from a
hot object to a cold object.
 *Heat is actually a function of both temp (avg

KE) and quantity (mass). ‘Everyday’ examplewhich has more heat: a teaspoon of water or
a bathtub full of water, if both are the same
temp? Which will cause more energy flow
and why?
Energy Changes in Chemical Reactions

The universe is divided into 2 parts:
 System
- The part of the universe on which
we wish to focus attention.
 Surroundings
- Everything else in the universe!
Burning of a Match
Potential energy
System
Surroundings
(Reactants)
D(PE)
Energy released to the surrounding as heat
(Products)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293
Transferring Energy

Energy can be transferred in two distinct
ways:

Work


Heat


W=Fxd
Energy transferred from hotter objects to colder ones
No matter how energy is transferred, the First
Law of Thermodynamics must be considered:


Energy is conserved
DE = Efinal - Einitial
Internal Energy and the Sign of DE

The energy contained in a system is a very
abstract value and is nearly (if not totally)
impossible to measure; however the First Law
can be applied without knowing the final and
initial energy values:

DE = Efinal - Einitial
Ex- Cellular Respiration
Ex- Photosynthesis
Relating Changes in Energy to Heat and
Work


Because energy can be exchanged by the
transfer of both heat and work; the work done
on or by the system must be considered also:
DE = q + w
The sign of q and w are assigned by the
direction of heat or work transfer:
Sample Exercise: Relating Heat and
Work to Changes of Internal Energy

Gases A(g) and B(g) are confined in a cylinder-and-piston
arrangement like that in the Figure and react to form a solid product
C(s): A(g) + B(g) → C(s). As the reaction occurs, the system loses
1150 J of heat to the surroundings. The piston moves downward as
the gases react to form a solid. As the volume of the gas decreases
under the constant pressure of the atmosphere, the surroundings do
480 J of work on the system. What is the change in the internal
energy of the system?
Solution
Analyze The question asks us to determine ΔE, given information about q and
w.
Plan We first determine the signs of q and w (Table 5.1) and then use Equation
5.5, ΔE = q + w, to calculate ΔE.
Solve Heat is transferred from the system to the surroundings, and work is
done on the system by the surroundings, so q is negative and w is positive: q =
−1150 J and w = 480 kJ. Thus,
ΔE = q + w = (−1150 J) + (480 J) = −670 J
*The negative value of ΔE tells us that a net quantity of
670 J of energy has been transferred from the system
to the surroundings.System ‘loses’ nrg
Exp 2: Relating Heat and Work Changes
of Internal Energy
Calculate the change in the internal energy of the
system for a process in which the system absorbs 140
J of heat from the surroundings and does 85 J of work
on the surroundings.
Answer

Use the First law of thermodynamics, Q - W =
∆E where Q is heat input, W is work done by
the system and ∆E is the change in internal
energy.
Q>0 when it is heat input and W>0 when it is
work done by the system. For this exp, Q is +
and W is - , therefore:
∆E = 140 - 85 = 55J.
Exothermic & Endothermic Reactions
Exothermic: energy flows out of
system into surroundings as heat
 Endothermic: energy flows from the
surroundings into the system as heat

Conservation of Energy
in a Chemical Reaction
In this example, the
energy of the
reactants and products
Exothermic
Reaction
decreases, while the energy of the surroundings increases.
Notice
that the total energy
does+not
change.
Reactant
Product
Energy
Surroundings
Energy
Surroundings
Myers, Oldham, Tocci, Chemistry, 2004, page 41
System
Before
reaction
System
After
reaction
Conservation of Energy
in a Chemical Reaction
Endothermic
In this example, the
energy of theReaction
reactants and products
increases, while the energy of the surroundings decreases.
Notice
that the total
energy does not
change.
Reactant
+ Energy
Product
Surroundings
Energy
Surroundings
System
System
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Before
reaction
After
reaction
Exo and Endo Processes

If only the transfer of heat is
considered during the course of a
chemical reaction, then chemical
reactions can be classified as either
endothermic or exothermic
 Endothermic—heat absorbed by
the system (∆H>0)
 Exothermic—heat released to the
system ( ∆H<0)
 H= enthalpy
Enthalpy
• Enthalpy is a thermodynamic quantity that
accounts for heat flow during the course of
a chemical reaction
– Equals the energy contained by the system as
well as the pressure/volume work done on or
by the system:
H = E + PV
Or
DH = DE + P DV
Energy Diagram

Shows the change in energy during a
chemical reaction
Energy Diagram for an Exothermic Reaction

Products have
lower PE
than reactants
energy
released

∆H is –

Reactant(s)  Product(s) + Energy
Relate to Bio- where have you seen this type of rxn before?
Energy Diagram for an Endothermic Reaction
Reactants have
lower PE
than products
 ∆H is +

energy
absorbed
Reactant(s) + energy  Product(s)
Relate to Bio: where have you seen this type graph before?
Determining the Sign of ΔH
• CW- Indicate the sign of the enthalpy
change, ΔH, in the following processes
carried out under atmospheric pressure and
indicate whether each process is
endothermic or exothermic: (a) An ice cube
melts; (b) 1 g of butane (C4H10) is
combusted in sufficient oxygen to give
complete combustion to CO2 and H2O.
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Solution
•Plan We must predict whether heat is absorbed or
released by the system in each process. Processes in
which heat is absorbed are endothermic and have a
positive sign for ΔH; those in which heat is released
are exothermic and have a negative sign for ΔH.
•Solve In (a) the water that makes up the ice cube is
the system. The ice cube absorbs heat from the
surroundings as it melts, so ΔH is positive and the
process is endothermic. In (b) the system is the 1 g
of butane and the oxygen required to combust it. The
combustion of butane in oxygen gives off heat, so
ΔH is negative and the process is exothermic.
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Measuring Energy Changes

The energy required to change the
temperature of a substance depends
on:
 The amount of substance being
heated (# of grams)
 The temperature change (# of
degrees)
 The identity of the substance
Units for Energy
calorie = amount of energy needed
to raise the temperature of 1 gram of
water by one Celsius degree
 joule = SI unit for energy


1 calorie (cal) = 4.184 joules (J)
Specific Heat Capacity (c)

The amount of
energy required
to change the
temperature of
1 g of a
substance by
10C
Specific Heat Capacity Calculations

q = m x c x ΔT
 q = energy (heat) required
 m = mass of the sample in grams
 c = specific heat capacity in J/g0C
 ΔT = change in temperature in 0C
(Final temp-Initial temp)
Specific Heat Capacity Practice Problem #1
A 1.6 g sample of a metal that has the
appearance of gold requires 5.8 J of
energy to change its temperature from
230C to 410C. Is the metal pure gold?
q
q = m x c x ΔT
c
 m = 1.6 g
m  DT
 q = 5.8 J
5.8 J
 ΔT = 41 – 23 = 180C
c
(1.6 g x 18 0 C )
c = 0.20 J/g0C …
Not pure gold!
Calculating energy changes



Loss of heat by one substance = Gain of heat by another
[(m) (csub A) (DT)]
= (m) (csub B) (DT)
CW- Try the problems on the next 2
slides
Specific Heat Capacity Practice Problem #2
240 g of water (initially at 20.oC) are mixed with an unknown mass of iron (initially at 500.oC).
When thermal equilibrium is reached, the system has a temperature of 42oC.
Find the mass of the iron.
Fe
T = 500.oC
mass = ? grams
- LOSE heat = GAIN heat
T = 20.oC
mass = 240 g
DTFe = 42oC – 500.oC = -458oC
- [(m) (cFe) (DT)] = (m) (cwater) (DT)
DTwater = 42oC – 20.oC = 22oC
- [(X g) (0.45 J/goC) (-458oC)] = (240 g) (4.184 J/goC) (22oC)]
- [ (X) (0.45) (-458)] = (240 g) (4.184) (22)
206.1 X = 22092
X = 110 g Fe
MORE PRACTICE WITH HEAT
PROBLEMS
A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264
g of water at 25.0oC. If the system's final temperature is 46.0oC, what
was the initial temperature of the lead?
Pb
T = ? oC
mass = 322 g
Tf = 46oC
Ti = 25oC
mass = 264 g
Pb
- LOSE heat = GAIN heat
- [(CPb) (mass) (DT)] = (CH O) (mass) (DT)
2
- [(0.138) (322) (46.0 - Ti)] = (4.184) (264) (46.0 – 25.0)]
- [(44.44) (46.0 - Ti)] = (1104.6) (21.0)]
- 2044 + 44.44 Ti = 23197
44.44 Ti = 25241
Ti = 568oC
Practice
Classwork: Go to the kentchemistry
site (on my blog, under Unit 8) and do
practice problems AND additional
practice problems
 HW-Start Vocab WS
 HW- Heat Problems WS

Go over 3 slides from yesterday
Some how, these slides (#26-28)
were ‘turned off’ yesterday, so we
need to go over them now!
 Also, please remind me to find y’all a
chart of ‘common’ enthalpies when
we are done today

Enthalpies of Reaction
• Formation of water
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
• ΔH is proportional to amount used and will
change as amount changes
2H2(g) + O2(g) → 2H2O(l)
ΔH = - 967.2 kJ
• For reverse reactions, sign of ΔH changes
2H2O(l) → 2H2(g) + O2(g)
ΔH = + 967.2 kJ
• Treat ΔH like reactant or product
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)
Enthalpies of Reaction Practice
Consider the following rxn:
C(s) + 1/2O2(g)  CO + 458.1kJ
Is the ΔH for this reaction positive or negative?
NEGATIVE (E released as a product)
What is the ΔH for 2.00 moles of carbon, if all the carbon
is used? 2.00 mol C - 458.1 kJ
1 mol C
= - 916 kJ
What is the ΔH if 50.0g of oxygen is used?
50.0 g O2
1 mol O2
32.0 g O2
- 458.1 kJ
= -1430 kJ
0.5 mol O2
1 mol CO
28.0 g CO
458.1 kJ
= 818 kJ
1 mol CO
What is the ΔH if 50.0 g of carbon monoxide decompose,
in the reverse reaction?
50.0 g CO
Another Exp of Relating ΔH to
Quantities of Reactants and Products
using Stoichiometry
• Problem: How much heat is released when
4.50 g of methane (CH4) gas is completely
combusted to form carbon dioxide and
water in a constant-pressure system? *Given
on a chart, 890 kJ is released by the system
when 1 mol CH4 is burned at constant
pressure.
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Solution
•First, write the equation and balance.
•CH4 + 2O2 CO2 + 2H2O
•Info from a ‘standard’ chart provides us with
a stoichiometric conversion factor: (∆H of 1
mol CH4 is −890 kJ). Thus, we can convert
moles of CH4 to kJ of energy. First, however,
we must convert grams of CH4 to moles of
CH4. Thus, the conversion sequence is
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Solution con’t:
• By adding the atomic weights of C and 4 H,
we have 1 mol CH4 = 16.0 CH4. We can use
the appropriate conversion factors to
convert grams of CH4 to moles of CH4 to
kilojoules:
The neg. sign indicates the system released 250 kJ to surroundings
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Watch this video
• https://www.youtube.com/watch?v=8m_FC
e5aCqY
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Practice
• Go to the website on my blog (Day 2 under
Unit #8) and go over the practice problems.
Then do the odd numbered questions (final
ans are given)
• Do the Enthalpy Stoichiometry WS
• HW- Watch this video:
https://www.youtube.com/watch?v=Wst4q_
bns3I
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Try this one? Ask Cody if we are
doing this
• Methane gas (CH4) reacts with oxygen gas
and combusts to produce carbon dioxide
and water. The balanced rxn releases 700 J
of energy. If 100g of both CH4 and O2 are
used, how much energy will be released?
• (Hint- find the moles of limiting reactant first)
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Energy with Stoichiometry
methane
+
100 g
carbon dioxide + water + energy
Limiting
Excess
CH4
oxygen
+
2 O2
CO2
+
/ 32 g/mol
x kJ = 3.125 mol O2
6.23
700 kJ
? kJ
100 g
/ 16.05 g/mol
6.23 mol CH4
1
2 H2O +
3.125 mol O2
700 kJ
= 1100 kJ
2 mol O2
2
1.56
smaller number
is limiting reactant
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Hess’s law
• If ∆H for a particular reaction is not known,
you can use other known reactions to
calculate it.
• Doesn’t matter if you take a ‘longer route’
to get from point A to point B, you’ll still
get there all the same
• Here’s an example:
http://www.sciencegeek.net/APchemistry/P
owerpoints/6_Hess'sLaw.ppsx
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Hess’s law
• Hess’s Law states that the enthalpy of a whole
reaction is equivalent to the sum of it’s steps.
• For example: C + O2  CO2
This can occur as 2 steps
C + ½O2  CO
DH = – 110.5 kJ
CO + ½O2  CO2
DH = – 283.0 kJ
C + CO + O2
 CO + CO2
DH = – 393.5 kJ
I.e. C + O2  CO2
DH = – 393.5 kJ
• Hess’s law allows us to add equations.
• We add all reactants, products, & DH values.
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether the reaction
takes place in one step or a series of steps
Victor Hess
Why? Because enthalpy is a state function
To review:
1. If a reaction is reversed, ΔH is also reversed
2 CH4 + O2  2 CH3OH
ΔHrxn = -328 kJ
2 CH3OH  2 CH4 + O2
ΔHrxn = +328 kJ
2. If the coefficients of a reaction are multiplied by an integer,
ΔH is multiplied by that same integer
CH4 + 2 O2  CO2 + 2 H2O
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
2(CH4 + 2 O2  CO2 + 2 H2O)
ΔHrxn = -802.5 kJ
© 2015 Pearson Education, Inc.
ΔHrxn = -1605 kJ
Example: Methanol-Powered Cars
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
2 CH4(g) + O2(g)  2 CH3OH(l)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
ΔHrxn = ?
ΔHrxn = -328 kJ
ΔHrxn = -802.5 kJ
2 CH3OH(l)  2 CH4(g) + O2(g)
ΔHrxn = +328 kJ
2(CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)) ΔHrxn = -1605 kJ
2 CH3OH(l) + 2 CH4(g) + 4 O2(g)  2 CH4(g) + O2(g) + 2 CO2(g) + 4 H2O(g)
3
2 CH3OH + 3 O2  2 CO2 + 4 H2O
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
ΔHrxn = -1277 kJ
© 2015 Pearson Education, Inc.
Tips for applying Hess’s Law…
Look at the final equation that you are
trying to create first…
• Find a molecule from that eq. that is only in one of
the given equations
• Make whatever alterations are necessary to those
• Once you alter a given equation, you will not alter it again
• Continue to do this until there are no other options
• Next, alter remaining equations to get things to
cancel that do not appear in the final equation
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
1. Given the following data:
S(s) + 3/2O2(g) → SO3(g)
ΔH = -395.2 kJ
2SO2(g) + O2(g) → 2SO3(g)
ΔH = -198.2 kJ
.
Calculate ΔH for the following reaction:
S(s) + O2(g) → SO2(g)
S(s) + 3/2O2(g) → SO3(g)
2SO2(g) + O2(g) → 2SO3(g)
2SO3(g) → O2(g) + 2SO2(g)
SO3(g) → ½ O2(g) + SO2(g)
ΔH = -395.2 kJ
ΔH = -198.2 kJ
ΔH = +198.2 kJ
ΔH = +99.1 kJ
S(s) + O2(g) → SO2(g)
ΔH = -296.1kJ
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
2. Given the following data:
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJ
C(s) + O2(g) → CO2(g)
ΔH = -394 kJ
H2(g) + 1/2O2(g) → H2O(l)
ΔH = -286 kJ
Calculate ΔH for the following reaction:
2C(s) + H2(g) → C2H2(g)
C(s) + O2(g) → CO2(g)
2C(s) + 2O2(g) → 2CO2(g)
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g)
H2(g) + 1/2O2(g) → H2O(l)
2C(s) + H2(g) → C2H2(g)
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
ΔH = -394 kJ
ΔH = -788 kJ
ΔH = -1300 kJ
ΔH = +1300 kJ
ΔH = -286 kJ
ΔH = +226kJ
© 2015 Pearson Education, Inc.
3. Given the following data:
2O3(g) → 3O2(g)
O2(g) → 2O(g)
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = - 427 kJ
ΔH = + 495 kJ
ΔH = - 199 kJ
Calculate ΔH for the following reaction:
NO(g) + O(g) → NO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
O2(g) → 2O(g)
O(g) → ½ O2(g)
2O3(g) → 3O2(g)
3/2 O2(g) → O3(g)
NO(g) + O(g) → NO2(g)
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
ΔH = - 199 kJ
ΔH = + 495 kJ
ΔH = - 247.5 kJ
ΔH = - 427 kJ
ΔH = + 213.5 kJ
ΔH = +233kJ
© 2015 Pearson Education, Inc.
Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is formed
from the elements in their standard states
° = standard conditions
• Gases at 1 atm pressure
• All solutes at 1 M concentration
• Pure solids and pure liquids
f
= a formation reaction
• 1 mole of product formed
• From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
What’s the formation reaction for adrenaline, C9H12NO3(s)?
9 Cgr + 6 H2(g) + 1/2 N2(g) + 3/2 O2(g)  C9H12NO3(s)
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Hess’s Law
• See Worksheet(s) and do Problems
• Watch Video on Calorimetry (pre-lab) the
last 15 minutes of class
Chemistry: The Central Science, 13th Edition
Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus
© 2015 Pearson Education, Inc.
Unit 8 – Thermochemistry, Kinetics
& Equilibrium
Kinetics
Reaction Times

Time of reaction varies

Examples:



HCl + NaOH  NaCl + H2O Instantaneous
2 CO + O2  2CO2 Thousands of Years
It is therefore important to not only know if a
reaction will occur, but also what the reaction
rate is
Collision Theory

Collision Theory:



Atoms, ions, and molecules can react to form products when
they collide, provided that particles have enough kinetic
energy.
Concentration is very important
If the individual particles that collide do not have
enough KE, they will bounce apart when they collide.
Collision Theory

Activation energy – the minimum amount of
energy particles need in order to react

Activated complex – the temporary arrangement
of atoms at peak of activation-energy barrier

Called the transition state – can re-form reactants or
form products
An Energy Diagram
activated
complex
Ea
activation
energy
energy
reactants
ΔPE
products
course of reaction
Collision Theory
Kinetics vs. Thermodynamics

Kinetics – How FAST a reaction will occur

Thermodynamics – WHETHER a reaction will
occur in the first place
Factors Affecting Reaction Rates

Temperature

Catalysts

Concentration

Pressure

Particle Size
Factors Affecting Reaction Rate

Temperature
 high temp = fast reaction rate
 high KE
- fast-moving particles
- more likely to reach activation
energy
Concentration

Number of particles in a given area

Related to pressure (gas)

Increase collision frequency
Particle Size

Smaller ‘particles’ have larger surface area (more
‘exposure’)

Increased surface area means more reactant
exposed for reaction

Ways to increase surface area:
Dissolve solid reactants to increase surface area
 Grind into powder

Catalyst

Catalyst – substance that increases rate of
reaction without being used up itself
during the reaction

Lowers activation energy
Factors Affecting Reaction Rate

Catalyst
 substance that increases reaction rate
without being consumed in the reaction
 provides a new pathway for reaction
with a lower activation energy
 typically provides a surface where
reactants are brought together,
resulting in more successful collisions
Catalyst
Activation Energy

(Ea)
 minimum energy required for a reaction to
occur
 If a collision has an energy less than Ea,
the molecules will bounce apart
unchanged.
Activation
Energy
Activation Energy
 depends on reactants
 low Ea = fast reaction rate
Factors Affecting Reaction Rate

Enzymes
 Biological catalysts
KMT
Practice

CW: In a paragraph, and in your own
words, describe what is happening in
KMT graph. Distinguish between heat and
temperature!

HANDOUT: Reaction Energy & Rate
WS
Unit 8 – Thermochemistry, Kinetics
& Equilibrium
Equilibrium
What is equilibrium?
 Consider the following chemical reaction:
N2 + 3H2  2NH3
Reversible Reactions
 Most reactions continue until all the
reactants are used up; however some
reactions appear to stop after a certain
period of time
– This is because these reactions can proceed in
both directions:
Forward: N2 + 3H2  2NH3
Reverse: N2 + 3H2  2NH3
Chemical Equilibrium
 When the rate of the forward reaction equals
the rate of the reverse reaction, the reaction
is said to be in chemical equilibrium
– In other words, as soon as some product is
formed it reacts to produce more reactant
Equilibrium Constants
 Each reaction has a certain equilibrium
constant that tells you to what extent the
reaction proceeds to produce product
– Reactions with K values > 1 favor products
– Reactions with K values < 1 favor reactants
Chemical Equilibrium

Even though the rates of the forward
and reverse are equal, the
concentrations of components on both
sides may not be equal
 An
A
1%
equilibrium position may be shown:
B
99%
or
A
99%
B
1%
Note the emphasis of the arrow’s direction
 It depends on which side is favored; almost
all reactions are reversible to some extent

Le Châtelier's Principle
 Le Châtelier's principle essentially states
that if a system is at equilibrium and that
equilibrium is altered (placed under stress),
the system will shift so as to reestablish
equilibrium (relieve the stress)
– Most common methods of altering equilibrium:
 Adding reactant or product
 Temperature changes
Le Châtelier Translated:
When a reactant or product is added
to a system at equilibrium, the system
shifts away from the added
component.
 If a reactant or product is removed,
the system shifts toward the
removed component.

Le Châtelier’s Principle
 In order to visualize the impact of changing
the concentration of a reactant or product,
consider how adding or removing weight
from one side of a balanced teeter totter
impacts the balance.
Balanced:
At “equilibrium”
Le Châtelier’s Principle
 If you remove two blocks from the right side
of the teeter totter, what happens?
 What do you have to do to re-balance the
teeter totter (re-establish the equilibrium)?
Le Châtelier’s Principle
 To re-balance the teeter totter (re-establish
the equilibrium), you must move one of the
blocks from the left side over to the right
side.
Le Châtelier’s Principle
 What happens if you add 2 blocks to the
right side of the original teeter totter?
 What do you have to do to re-balance (reestablish the equilibrium) the teeter totter?
Le Châtelier’s Principle
 To re-balance the teeter totter (re-establish
the equilibrium), you must move one of the
blocks from the right side to the left side.
Stresses That Affect Equilibrium
Concentration
 Temperature
 Volume/Pressure

Addition of Reactant or Product at
Equilibrium
N2(g) + 3H2(g)
After addition of H2:
2NH3(g)
Stress: Change Concentration
Co(H2O)62+ + 4 Cl- ↔ CoCl42- + 6 H2O
(pink)
(blue)
Stress
Result
Ex:
Add Cl-
Forward reaction favored
Shifts right to reduce extra ClMore CoCl42- will form
Remove Cl- Reverse reaction favored
Shifts left to replace ClMore Co(H2O)62+ will form
Effect of Temperature on
Equilibrium
 Consider two different types of reactions:
– Exothermic & Endothermic
 Exothermic: Heat can be considered a product
 Endothermic: Heat can be considered a reactant
 Endothermic: Increase in T shifts reaction
to the right
 Exothermic: Increase in T shifts reaction to
the left
Stress: Change Temperature
Ex:
heat + Co(H2O)62+ + 4 Cl1- ↔ CoCl42- + 6 H2O
(pink)
(blue)
This reaction is endothermic. For Le Chatelier’s
principle, consider “heat” as a chemical.
Stress
Result
Increase T
Forward reaction favored;
shifts right to reduce extra heat
More CoCl42- will form
Decrease T
Reverse reaction favored;
shifts left to replace “lost” heat
More Co(H2O)62+ will form
Stress: Change Volume
Ex:
1 N2 (g) + 3 H2(g) ↔ 2 NH3(g)
(1 + 3 = 4 moles of gas) ↔ (2 moles of gas)
Stress Result
Decrease V Forward rxn favored;
shifts right to side with fewer moles of gas
(reduces # of mc’s packed into this
smaller volume)
Increase V
Reverse rxn favored;
shifts left to side with more moles of gas
(to fill the larger volume with more mc’s)
Catalysts & Equilibrium
MnO2
2 H2O2 (aq) ↔ 2 H2O (l) + O2 (g)
Since a catalyst increases the forward and
reverse rates equally, it will not shift the
equilibrium.
LeChatelier Example #1
A closed container of ice and water at
equilibrium. The temperature is raised.
Ice + Energy  Water
The equilibrium of the system shifts to
right
the _______
to use up the added energy.
LeChatelier Example #2
A closed container of N2O4 and NO2 at
equilibrium. NO2 is added to the container.
N2O4 (g) + Energy  2 NO2 (g)
The equilibrium of the system shifts to
left
the _______
to use up the added NO2.
LeChatelier Example #3
A closed container of water and its vapor at
equilibrium. Vapor is removed from the system.
water + Energy

vapor
The equilibrium of the system shifts to
right to replace the vapor.
the _______
LeChatelier Example #4
A closed container of N2O4 and NO2 at
equilibrium. The volume is decreased.
N2O4 (g) + Energy  2 NO2 (g)
The equilibrium of the system shifts to
left
the _______
to lower the pressure,
because there are fewer moles of gas
on that side of the equation.
Le Châtelier’s Principle SUMMARY
• Changes in Concentration
Remove
Add
Remove
Add
aA + bB
cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
Le Châtelier’s Principle SUMMARY
• Changes in Volume and Pressure
(Only a factor with gases)
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewer moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewer moles of gas
Le Chatelier’s Principle Recap
Le Chatelier’s principle:
When a system at equilibrium is disturbed, it shifts to a
new equilibrium that counteracts the disturbance.
N2(g) + 3 H2(g)
2 NH3(g)
Disturbance
Equilibrium Shift
Add more N2…………………..
“
“ H2…………………..
“
“ NH3…………………
Remove NH3…………………..
Add a catalyst………………… no shift
Decrease volume……………..
Fritz Haber
Summary of Shifts in Equilibrium
Some sulfur trioxide is sealed in a container and allowed
to equilibrate at a particular temperature. The reaction is
endothermic.
SO3(g)
SO2(g) + ½ O2(g)
In which direction will the reaction proceed
(a) if more SO3 is added to the system?
(b) if oxygen is removed from the system?
(c) if SO2 is added to the system?
(d) if the temperature is increased?
(e) if the temperature is decreased?