Collision Theory and Representing Rates of Reactions with Kinetic
Energy
and Potential Energy Curves
1. Explain in terms of the collision theory why the rate of a gaseous reaction
my increase if the total pressure of the system is increased. What are
three ways that a chemist could be able to increase the pressure of a
system?
If the pressure is increased then the particles will be closer together and
collide more, therefore more of them will have KE greater that Ea and
proper orientation and a successful collision.
Three ways that you could increase the pressure would be
- Increase concentration or number of molecules
- Increase temperature so particles speed up and collide with container
more
- Decrease volume
2. If a catalyst were added to a reacting system, what effect, if any, would
this have on the kinetic energy distribution curve and on the activation
energy for the reaction?
A catalyst will decrease the activation energy by moving it to the left on
the kinetic energy axis. The shape of the curve will be the same. As a result
area under the curve that exists to the right of the activation energy line
will increase. There are more particles that have KE > Ea and more
successful collisions.
Ea
uncatalyz
ed
Ea
catalyzed
3. A)Explain in terms of collision theory why a log burns slowly in air while if
sawdust made from the same log was scattered throughout the room and
ignited it burns explosively.
The sawdust particles have more surface area to react with the oxygen and
flame. ( more collisions with oxygen molecules per unit time).This will allow
for more potentially successful collisions leading to more product.
b) Construct a labelled kinetic energy diagram to represent both situations.
Draw curve. Ea stays in the same position.
More area will be to the right of the Ea in your graph.
Make sure axis is labelled with KE and fraction of collisions .
4. For the reaction CO + NO2  CO2 + NO , the activation energy for the
forward reaction is known to be 134 kJ.
a) Use the Standard molar enthalpies table ( pg 597) and Hess’ Law to find
the Enthalpy of this reaction.
∆Hrxn = ∑∆Hprod - ∑∆Hreact my table may be different but you get idea
= [∆H CO2 + ∆H NO] – [∆H CO + ∆H NO2]
= [40.37 – 393.5] – [ -110.5 + 33.8]
= -303.13 + 76.7
= - 226.4 kJ exothermic
b) Using both of these values, construct a labelled potential energy diagram
for the reaction.
- Are axis labelled PE and Rxn Coordinate
- Are products lower than reactants and the difference between the two
labelled as ∆H= - 226.4 kJ
- Ea f was given . Is it written on graph
- Ea r is sum of 226.4 and 134. Is it written on graph as 360.4 kJ?