South Pasadena • AP Chemistry
Name
2 ▪ Chemical Kinetics
Period
1 ·
Date
KINETIC ENERGY DIAGRAMS
Draw how the KE diagram would change if:
the temperature of
the sample is increased.
a catalyst is added
to the reaction mixture.
more chemical is
added at the same temperature.
What is the name of the vertical line intersecting the graph? Threshold energy (activation energy)
Key Idea: In terms of collisions, why does an increase in temperature result in a higher rate of reaction?
Molecules are moving with more energy, and collide with enough energy to form the activated
complex/transition state.
2 ▪ Chemical Kinetics
2 ·
POTENTIAL
The energy of the reactants is 30 kJ/mol.
H = –20 kJ/mol
Ea = +40 kJ/mol
Draw the PE curve.
The energy of the products is 10 kJ/mol
The activation energy for the
reverse reaction is 60 kJ/mol
Draw the same curve with a catalyst included. (- - - - - - -)
Key Idea: In terms of collisions, why does adding a catalyst result
in an increase in the rate of reaction?
A catalyst allows collisions to take place more effectively to
form the products.
ENERGY DIAGRAMS
2 ▪ Chemical Kinetics
3 · RATES FROM
The vertical axis is [R] in mol·L
GRAPHS
-1
Show your work for these two problems.
The average rate of the reaction for the first
2.0 hours is _________________.
10 M – 40 M
= –15 M/hr
2.0 hr
The instantaneous rate of the reaction at 2.0
hours is ________________.
4.5 M – 15.6 M
= –5.6 M/hr
3.0 hr – 1.0 hr
If this reaction is either zero, first, or secondorder with respect to [R], what is the order?
Justify your answer.
This is not first order because the graph is
not linear. This is not first order because
the half lives are not the same. This must
be second order.
2 ▪ Chemical Kinetics
4 · CALCULATING RATES OF
Consider the combustion of propane, C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
The rate of disappearance of O2(g) is 60. mol·L-1·s-1




What is the rate of disappearance of C3H8(g)?
∆[C3H8] 1 ∆[O2] 1
=
= (60 M/s) = 12 M/s
∆t
5 ∆t
5
What is the rate of appearance of CO2(g)?
∆[CO2] 3 ∆[O2] 3
=
= (60 M/s) = 36 M/s
∆t
5 ∆t
5
What is the rate of appearance of H2O(g)?
∆[H2O] 4 ∆[O2] 4
=
= (60 M/s) = 48 M/s
∆t
5 ∆t
5
What is the rate of the reaction?
∆[C3H8]
Rate of Reaction =
= 12 M/s
∆t
REACTION
2 ▪ Chemical Kinetics
5 · RATE LAWS: METHOD OF
INITIAL RATES
Here is some initial rate data for the reaction, A + B  2C.
[A]
[B] Rate (mol·L-1·s-1)
0.400 0.200
1.44 × 104
0.200 0.100
1.80 × 103
0.200 0.500
4.50 × 104
a) Determine the orders of reactants A 1st order and B 2nd order
b) Write the rate law for this reaction: rate = k[A]1[B]2
c) Calculate the value of the rate constant, k, with units. 9.00 × 105 M–2 s–1
2 ▪ Chemical Kinetics
5 · INTEGRATED RATE
Which graph corresponds to the change in concentration of
a reactant that is first order? (D)
LAW GRAPHS
This graph is a chemical that is second order. The
horizontal axis is time. What is the vertical axis? 1/[X]
2 ▪ Chemical Kinetics
7 · REACTION
MECHANISMS
Consider this reaction mechanism:
HCOOH + H2SO4  HCOOH2+ + HSO4
HCOOH2+  COH+ + H2O
COH+ + HSO4  CO + H2SO4
a) What is the overall reaction?
HCOOH → CO + H2O
b) List any “intermediates.” HCOOH2+, HSO4–, COH+
c) List any catalysts. H2SO4_ Justify your answer.
H2SO4 is present at the beginning and regenerated at the end, so it is not used up.
d) If the first step is the slow step, what is the rate law? Rate = k[HCOOH]1[H2SO4]1
2 ▪ Chemical Kinetics
8 ·
HALF
A radioactive substance has a half-life of 8.00 minutes.
How long will it take for 93.75% of this chemical to decay?
ln 2
0.693
k=
=
= 0.0866 min–1
t½
8.00 min
ln(0.0625) – ln (1.00) = – (0.0866 min–1)t
t = 32.0 min
The reaction X  Y follows first-order kinetics with a half-life of 4.00 minutes. What is the value of k?
If the initial concentration of X is 3.6 M, what is the concentration after 15.0 minutes?
ln 2
0.693
k=
=
= 0.173 min–1
t½
4.00 min
ln [X] – ln(3.6) = –(0.173 min–1)(15.0 min)
[X] = 0.268 M
LIFE
2 ▪ Chemical Kinetics
9 · ARRHENIUS
Calculate the activation energy, Ea, for N2O5(g)  2 NO2(g) + ½ O2(g)
given k (at 25C) = 3.46 × 10-5 s-1 and k (at 50C) = 1.10 × 10-3 s-1.
k1
Ea 1 1
ln  = –  – 
R T1 T2
k2
3.46 × 10–5 s–1
Ea
 1 – 1 
ln
–3 –1 = –
8.314 J/mol-K298 K 323 K
1.10 × 10 s 
Ea = 111 kJ/mol
EQUATION