Chemical kinetics practical solutions

1. kinetic vs. potential energy diagrams
• Recall the Maxwell-Boltzman distribution (i.e.
kinetic energy diagram)
Kinetic energy →
Fraction
molecule
Ea
• The Ea is a critical point. To examine it more
closely we can use a potential energy graph
Potential
Energy(Ep)
Path of reaction →
• The axes are not the same, thus the Ep graph
is not a blow up of the Ek graph; however it
does correspond to the part of the Ek graph
that is circled

2. potential energy graph: a closer look
Collision
begins
molecules
speed up
Ep ↓, Ek ↑
Activated complex /
transition state
Reactants
Products
Ep(Potentialenergystored
inchemicalbonds)
Path of reaction
Collision ends
molecules
slow down
Ep ↑, Ek ↓
A2 & B2
rush
together
2AB
molecules
float apart
Overall Ep(reactants)>Ep(products)
Ek(reactants)<Ek(products)
Ep + Ek =constant
throughout
Ea
∆H

3. Ep graph: Important points
• Forward and reverse reactions are possible
• Ea is the difference between Ep at transition
state and initial or final Ep
∀∆H is the difference between initial and final
Ep. It is -ve for exothermic,+ve for endothermic
Ep
Exothermic Endothermic
Ea
forward Ea
reverse
• The graph depicts an exothermic reaction.
Endothermic reactions are also possible
∆ H is
positive

4. The collision theory
• Related to the Ep graph is the “collision theory”
- the idea that for molecules to react they must
meet with sufficient force
• Factors that affect reaction rate can be
explained via the collision theory:
• Increased temperature causes molecules to
move faster (increased number of collisions
per unit time and greater kinetic energy)
• Increased concentration means more collisions
• Homogenous reactions occur faster because
reacting molecules collide more frequently
• Catalysts decrease Ea, decreasing the amount
of kinetic energy needed to overcome Ea

5. Catalysts
• Recall, catalysts speed a reaction
• This can be explained by the Ek or Ep graphs
• In both, the catalyst works by lowering the Ea:
• Catalysts speed forward and reverse reactions
• However, most reactions favour the side that has the
lowest potential energy (most stable)
• Catalysts are heterogenous or homogenous
• They provide a substrate (p. 768) for a reaction or they
can bond temporarily to a molecule, increasing the odds of
a favourable meeting
Fractionof
molecules
Kinetic energy →
potential
energy
Path of reaction

6. Transition state lab: purpose
Purpose: 1) to visualize an activated complex, 2)
to observe the influence of a catalyst
We will be examining the following reaction:
NaKC4H4O6(aq) + H2O2(aq) → CO2(g) + …
Procedure:
1. Turn hot plates immediately to medium heat
2. Get a 10 mL graduated cylinder, a 100 mL
beaker, a test tube, and a rubber stopper.
3. Weigh 1.7 g NaKC4H4O6. Add to beaker along
with 10 mL distilled H2O. Swirl to dissolve.
4. Add 4.5 mL of 10% H2O2 to beaker. Heat.
5. Get 5 mL of CoCl2 but don’t add it yet.

7. Transition state lab: procedure
Procedure:
6. As soon as tiny bubbles start to form and
rise, remove the beaker from the hot plate.
Add the CoCl2 at this point.
7. Record your observations (in order to answer
the questions). Clean up – wash everything
down the drain, wipe off your lab bench.
Questions: answer on a separate sheet of paper
1. Look at the chemical equation that
represents the reaction. What physical sign
will there be when a reaction is occurring?
2. The products of the reaction are colourless.
What colour are the reactants?

8. Transition state lab: conclusions
Questions: read 18.11 (pg. 767 – 769)
3. What was the catalyst in the lab? What colour
was it? Is it homogenous or heterogeneous?
4. At the beginning of step 5, both reactants
were present; why was there no reaction?
(Illustrate with a Ek diagram).
5. Why is the reaction still slow after heat is
added? (illustrate using the Ek diagram)
6. Was the catalyst a different colour at the end
of the experiment than at the beginning?
7. What colour was the activated complex?
8. Illustrate the affect the catalyst had on the
reaction (using both Ek and Ep diagrams)

9. Answers
1. The production of CO2 (bubbling) is a
physical sign that the reaction is occurring
2. The reactants are colourless
3. CoCl2 was the catalyst in the lab (pink,
homogenous)
4. There was no reaction because the Ea was
not reached (Illustrate with a Ek diagram).
5. The reaction still slow after heat is added
because very few molecules exceed Ea.
Fractionof
molecules
Kinetic energy →

10. Answers
6. The catalyst was the same colour at the end
of the experiment (catalysts don’t change).
7. The activated complex was green
8. Illustrate the affect the catalyst had on the
reaction (using both Ek and Ep diagrams)
Fractionof
molecules
Kinetic energy →
potential
energy
Path of reaction
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