Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Kinetics and Equilibrium
Text Chapter 13: ALL ODD QUESTIONS! (Show all work when applicable)
Kinetics: the branch of chemistry that deals with how (mechanism) and how fast or slow (rate)
reactions occur.
collision theory - in order for a reaction to occur the reactants must collide with sufficient
energy and proper orientation (an effective collision).
more collisions = faster rate (less time)
fewer collisions = slower rate (more time)
-You will be able to explain why a reaction is fast or slow. Then you can explain how to alter the speed
of a reaction (make a reaction speed up or slow down).
Reaction Mechanism (steps of a reaction):
In order to understand rate, we must know that all reactions occur in a series of steps known as
the reaction mechanism. The slowest step of the mechanism is known as the rate-determining step, and
predicts the over- all rate of the reaction. (Think of a relay race-the team is only as fast as the slowest
runner).
Example:
The Haber Process (the synthesis of ammonia from its elements):
Hydrogen + Nitrogen  Ammonia
3H2 + N2  2NH3
In order for this reaction to occur there is a series of steps that must take place:
1st) the 3 moles of Hydrogen and the 1 mole of Nitrogen must split up -- Endothermic
Breaking Bonds is Endothermic (BARF)
2nd) the free atoms must rearrange
3rd) 3 Hydrogen atoms arrange and bond themselves to one Nitrogen atom -- Exothermic
Forming Bonds is Exothermic (BARF)
** In order for the reacting particles to react there must be both frequent and effective collisions
between the particles (COLLISION THEORY)!
_________________________________
Endothermic/Exothermic
All reactions have both endothermic and exothermic components.
-When we classify a reaction as Exothermic, we are saying that the exothermic component is
larger than the endothermic component.
-When we classify a reaction as Endothermic, we are saying that the endothermic component is
larger than the exothermic component.
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Heat of Reaction = Enthalpy of Reaction = ∆H
The net change in heat determines whether the reaction is endothermic or exothermic. This net change
in heat is called the heat of reaction=enthalpy of reaction=∆H. ∆H varies with temperature and
pressure, so 25 ˚C and 1 atm have been considered standard state conditions for reporting heat data.
∆H0 = ∆H under standard conditions.
∆Hreaction = ∆Hproduct-∆Hreactant
∆H = PEproduct – PEreactant
∆Hreaction will always be positive for an endothermic reaction.
∆Hreaction will always be negative for an exothermic reaction.
Heat of Formation
∆H0ƒ = the standard enthalpy (heat) of
formation = molar heat of formation =
the change in energy (enthalpy) that
happens when one mole of that
compound is formed from its elements
(under standard conditions).
You do not need to memorize this
table. These tables can always be
found in reference books.
Remember these are for ONE MOLE of
that substance.
1)
How much heat is released when 20.0 grams of H2(g) reacts with excess O2(g) resulting in water
vapor? The reaction equation is:
H2(g) + ½ O2(g)  H2O(g)
∆H0ƒ = -241.8kJ/mol
Answer: 2418 kJ released
2)
How much heat is released when 75.0 grams of oxygen react with excess calcium to form
calcium oxide?
Answer: 2979 kJ released
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Heat of Combustion
∆Hc = heat of combustion =the heat released by the complete combustion of one mole of a substance.
This is in terms of one mole of reactant (∆Hc = one mole of product)
Hess’s Law of Heat Summation
When a reaction can be expressed as the sum of two or more other reactions, the heat of reaction is the
sum of the heats of those individual reactions.
Examples:
1) Determine the heat of reaction of carbon monoxide from the standard heat of formation of
carbon dioxide.
C(s) + O2(g CO2(g)
CO(g) + ½ O2(g)  CO2(g)
∆H0ƒ=-393.5 kJ/mol
∆H0c=-283.0 kJ/mol
The equation we want is: C(s) + ½ O2(g)  CO(g)
**When you reverse an equation, you must reverse the sign for ∆H.
Answer: -110.5 kJ/mol
2)
Calculate the enthalpy change for the formation of methane from solid carbon (as graphite) and
hydrogen gas. The enthalpy change for this reaction cannot be measured in the laboratory
because the reaction is very slow. Below are the heats of combustion for carbon, hydrogen, and
methane.
C (s) + O2 (g)  CO2 (g)
H2 (g) + ½ O2 (g)  H2O (l)
CH4 (g) + 2O2 (g)  2H2O (l) + CO2(g)
∆H0 = -393.5 kJ/mol
∆H0 = -285.8 kJ/mol
∆H0 = -890.3 kJ/mol
Answer: -74.8 kJ/mol
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Honors Chemistry
3)
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Calculate ∆H for the decomposition (for one mole) of sodium chlorate.
∆H NaCl(s) = -410.5 kJ/mol
∆H NaClO3(s) = -358.2 kJ/mol
∆H O2(g) = 0 (all elements = 0)
Answer: -52.3 kJ/mol
4)
Calculate the heat of reaction for the following reaction:
(∆H NO(g) = +90.4 kJ/mol)
4NH3 (g) + 5O2 (g)  6H2O (g) + 4NO (g)
Answer: -904.4 kJ/mol
Bond Energy
The amount of energy required to break bonds or
amount of energy released when bonds are formed
differs depending on the number and type of bonds
and is known as the bond energies. The bond
energies are tabulated, and can be looked up in
references books (our course will not go this far
however AP Chemistry will)
Single bond energy
Double bond energy
Triple bond energy
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Factors that affect rates of Reactions
How can you speed-up a slow reactions/How can you slow down a fast reaction
All based on the COLLISION THEORY: the more frequent and effective collisions the faster the
reaction will take place
1. Temperature (a measure of a.k.e.)
Incr. Temp  incr. KE (more forceful collisions) particles move faster Incr. # of collisions 
incr.rate (faster)
2. Concentration ([x] = concentration of x in molarity)
Amount of Reactants
Decreasing Volume
If incr.[ ]  incr. # of particles in a given volume (ex: per L)  incr. # of collisions  faster rate
3. Pressure (gases only)
Incr. P  Incr. concentration  incr. # particles per liter  incr. # collisions  incr. rate
(faster)
4. Surface Area (solid substances)
incr. SA, incr. exposure of reactants, incr. # collisions, incr. rate (faster)
5. Stir or Agitate
6. Nature of the Reactants (have no control of this just understand)
-some reactants are faster to react than others.
Ionic = fast
Covalent = slow
(organic = has C = covalent = slow)
-Reactions between ionic compounds in solutions react faster than reactions between molecules
(covalent bonds)
-Reactions that produce precipitates or gases tend to go to completions faster than reactions that do
not.
7. Catalyst (works by providing an alternative path for the reaction, a path that requires lower activation
energy, lower energy to start the reaction)
-A catalyst speeds up a reaction, without being part of (or consumed by) the reaction, by lowering the
activation energy.
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Potential Energy Diagrams
Reminders: Reactants are ALWAYS written on the left and products are ALWAYS written on the right.
Reactants ⇒ Products
-Every reaction has its own change in energy
-Energy is either absorbed or released
Endothermic = energy is absorbed
Exothermic = energy is released
Potential energy diagrams show the change in potential energy as a reaction occurs.
There are only 2 types of PE diagrams that you are responsible for: Endothermic PE Diagrams and
Exothermic PE Diagrams
Which is the general shape of an endothermic potential energy diagram (shows energy being
absorbed/taken in)? (hint: if you start with 75 pens and end up with 225 pens, did you take pens in or
give pens away?)
Which is the general shape of an exothermic potential energy diagram (shows energy being
released/given off)? (hint: if you start with 30 pens and end up with 10 pens, did you take in pens or
give pens away?)
Reaction coordinate/reaction pathway represents progress of reaction (unmeasured time).
Activated complex: the intermediate step of the reaction. It will always have a higher PE than both the
reactants and the products
ex: 2A + B --> C
1) A + B --> AB
2) AB + A --> C
AB = activated complex (intermediate)
Activation energy: the amount of energy needed to start the reaction.
Activation energy = PEactivated complex - PEreactants
-Some reactions can proceed in both directions. This means the reactants form products AND the
products form reactants. Whenever you see "reverse reaction" simply read from right to left.
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Endothermic Potential Energy Diagrams
-Reactants will always have less PE than the products.
-Products will always have more PE than the reactants.
-Energy is absorbed/taken in during the reaction
-Products will always be higher than the reactants
-Reactants will always be lower than the products
______________________________________
*Arrows could be drawn ANYWHERE on the graph!
1. Potential energy of the reactants
2. Potential energy of the products
3. Potential energy of the activated complex
4. Activation energy
5. Heat of reaction = ΔH = PEproducts - PEreactants
ΔH is always positive (+) in an endothermic reaction
6. Activation Energy of the reverse reaction
__________________________________
State what each arrow represents.
1
2
3
4
5
6
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Exothermic Potential Energy Diagrams
-Reactants will always have more PE than the products.
-Products will always have less PE than the reactants.
-Energy is released during the reaction
-Products will always be lower than the reactants
-Reactants will always be higher than the products
*Arrows could be drawn ANYWHERE on the graph!
1. Potential energy of the reactants
2. Potential energy of the products
3. Potential energy of the activated complex
4. Activation energy
5. Heat of reaction = ΔH = PEproducts - PEreactants
ΔH is always negative (-) in an exothermic
reaction
6. Activation Energy of the reverse reaction
____________________________
State what each arrow represents.
1
2
3
4
5
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Identify each PE diagram as endo or exo and explain why IN TERMS OF ENERGY:
Reminder: A catalyst speeds up a reaction by lowering the activation energy/providing an alternate
pathway (lowers the bump).
Show how each of the PE diagrams below would be changed if a catalyst was introduced:
Using the diagram below, draw a line representing each of the following values and write what the
actual values are for each:
1.
2.
3.
4.
5.
6.
PE of the reactants
PE of the products
PE of the activated complex
Activation Energy
Heat of reaction
Activation Energy of the reverse reaction
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
1) A + B  C + D
∆ H=+30 kJ
If the PE of the Product is 80 kJ
If the PE Activated Complex is 140 kJ
Sketch the PE Diagram and Label all 6 points
What is the PE of the Reactant____________
What is the Enthalpy of the Reaction_______
What is the Activation Energy Forward Rxn______
What is the Activation Energy Reverse Rxn______
2) A + B  C + D + 60 kJ
If the PE of the Reactant is 100 kJ
Activation Energy Reverse Rxn is 85 kJ
Sketch and Label all 6 points on the PE Diagram
What is the PE Product______
What is the PE of the Activated Complex_______
What is the Activation Energy of Forward Rxn_____
What is the Heat of the Reaction_______
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Entropy and Enthalpy
Entropy = randomness = tendency for disorder.
-(g) has more entropy that (l).
-(l) has more entropy than (s).
-(aq) has more entropy than (s).
-more (g) has more entropy than less (g).
-Everything in nature favors conditions of higher entropy. (Would you rather be stuck in this desk with
no breaks for 3 hours straight or out on recess for 3 hours straight?)
-a positive ∆S represents a move toward higher entropy
Enthalpy: ΔH=Heat of Reaction=PEproducts - PEreactants
-Everything in nature favors low enthalpy. Would you rather be running a marathon (exerting a lot of
energy) for 10 hours straight or sitting on the couch (exerting little energy) for 10 hours straight?
-Which type of reaction (endo or exo) moves towards a lower energy?
-Many, many, many reactions are exothermic because everything in nature favors lower energy.
-A negative ΔH represents a move toward lower energy.
Table I:
-tells you how much energy is absorbed or released for specific reactions.
-State of matter (s, l, g) does matter - BE CAREFUL!
-Note that these values of ΔH for these reactions are under conditions of 101.3 kPa and 298K.
-If ΔH is positive (+): endothermic: energy is written as a reactant (left of the arrow)
-If ΔH is negative (-): exothermic: energy is written as a product (right of the arrow).
-NEVER SUBTRACT IN A CHEMICAL EQUATION. (Anything "lost" or released as part of a reaction is
simply written as a product).
So, if ΔH is (-) the numerical value will be written as an addition to the product side (but
ΔH is still negative).
-Pay close attention to coefficients. Remember coefficients represent moles in balanced equations.
ExampleCH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)
has a ΔH of -890.4 kJ
to show energy as part of the equation:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) + 890.4 kJ
This means when 1 mole of CH4(g) reacts with 2 moles of O2(g), 1 mole of CO2(g) 2 moles of H2O(l) and
890.4 kJ of heat are released.
How many kJ of heat will be released when 2 moles of CH4 react?
1)
a.
b.
Write a chemical equation including energy for the formation of gaseous nitrogen monoxide
from its gaseous elements.
How many moles of nitrogen monoxide will be produced when 250. kJ are absorbed in this
reaction?
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Honors Chemistry
Kinetics and Equilibrium
HW: chapter 13 all odd questions
Reminders about endothermic and exothermic reactions:
-When a process is endothermic, heat is absorbed in order for the process to occur. This heat that is
absorbed must come from somewhere! This heat that is absorbed comes from the surroundings. So,
the temperature of the surrounding area decreases (gets cooler/goes down) when a process is
endothermic.
-When a process is exothermic, heat is released in order for the process to occur. This heat that is
released must go somewhere! This heat that is released goes to the surroundings. So, the temperature
of the surrounding area increases (gets warmer/goes up) when a process is exothermic.
Spontaneous Reactions
There are 2 factors that we must consider:
1. Energy/Heat
2. Entropy
Energy/Heat: (Enthalpy)Reactions that result in lower energy are more stable; more spontaneous -EXOTHERMIC RXN (negative H)
Entropy (∆S): measure of disorder, chaotic-ness. Simply put, it is how spread-out the particles are.
slg=entropy is increasing ∆S=+
Factors that Favor a Spontaneous Reaction
Low Enthalpy(energy/heat), Exothermic Reaction, Rxn that Release Energy
and
High Entropy, Rxn that end up with the products being more spread-out than the reactants
Factors that Favor a NonSpontaneous Reaction
High Enthalpy(energy/heat), Endothermic Reaction, Rxn that absorb Energy
and
Low Entropy, Rxn that end up with the products being less spread-out than the reactants
Putting together both the Enthalpy and Entropy: Gibbs Free Energy (∆G)
∆G = ∆H - T∆S
**Temperature must be KELVIN
Enthalpy: low (favored)
Entropy: high (favored)
∆G=NEGATIVE VALUE = spontaneous
Enthalpy: high (not favored)
Entropy: low (not favored)
∆G=POSITIVE VALUE – not spontaneous
What happens when ∆H and ∆S disagree?
∆G = ∆H - T∆S
Depends on the Temperature!!
Enthalpy: low (favored)
Entropy: low (not favored)
Rx will occur at low temperatures
Enthalpy: high (not favored)
Entropy: high (favored)
Rx will occur at high temperatures
At high temperatures – Entropy (∆S)predicts
At low temperatures – Enthalpy (∆H) predicts
∆G = 0 The reaction is in equilibrium
∆H = T∆S
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