Transcript Topic C: Reaction Mechanisms

TOPIC C:
REACTION MECHANISMS
• Mechanism - the sequence of elementary steps that make
up a chemical reaction
• Each step will be relatively fast or relatively slow
• The overall rate of the chemical reaction is dependent
upon the slowest elementary step.
• For this reason, the slowest step is known as the rate
determining step or RDS.
• To study reaction rates we need to express changing
concentrations of reactants in some quantitative form
• We use a rate equation or rate law.
•
General form, Rate = k [A]x [B]y [C]z
•
k is the rate constant
•
x, y and z are the orders with respect to the
concentrations of reactants A, B and C.
• The order with respect to a given reactant is the power to
which the concentration of that reactant is raised to in the
rate equation.
• The overall order of the chemical reaction is the sum of
the individual orders.
• The reactants in the rate-determining step are ones that
affect the rate
• Only these reactants appear in the overall rate equation
• Often the RDS rate law will match the overall rate law
Important Points:
• 1. You cannot determine the order of a reaction from the
balanced equation
•
Orders must be determined experimentally
•
It is possible to determine the overall order of a
reaction from the slowest elementary step
•
• [ the coefficient in the slow step is the power that the concentration
of that substance is raised to in the rate equation ]
• 2. Units and magnitude of the rate constant are important.
They vary a great deal (since reaction rates vary
widely), and have often been the subject of AP
questions.
3. A reactant that has no effect on the rate has an order
equal to zero.
It has no effect on the rate
Any number raised to the power of zero is equal to
1,
so it can be omitted from the rate equation.
• 4. Orders can be fractional, e.g., ½.
• 5. In valid mechanisms the sum of the elementary steps
must add up to the overall chemical reaction.
• 6. Intermediates are formed in one elementary step during
the overall reaction, but are then used up in a subsequent
elementary step.
• 7. If a substance is present at the beginning of a reaction
and present in the same form at the end of the reaction, it
can be identified as a catalyst.
• The overall reaction is shown.
• This allows the cancellation of I-, since it is present both at the
beginning and the end of the reaction (considered a catalyst).
• (Note, IO- also cancels out, but not because it is a catalyst, rather it is
an intermediate).
• Step 1: H2O2 + I-  H2O + IO• Step 2: IO- + H2O2  H2O + I- + O2
_______________________________
• Overall: 2H2O2  2H2O + O2
• Practice:
• Identify the catalyst and the intermediate in the
mechanism below.
•
Step 1: SO2 + V2O5  SO3 + V2O4
•
Step 2: V2O4 + ½O2  V2O5
• Determining rate law by inspection
• Example #1 – see handout
• Example #2 – see handout
• Example #3 – see handout
• Practice Problems (handout)
Reaction Rate (additional background)
• Rate = [A] at time t2 - [A] at time t1
•
t2 – t1
• Rate = D[A]
Dt
• Reaction Rate – is a change in concentration of
a reactant or product per unit time
• For this reaction:
• 2NO2 (g) → 2NO (g) + O2 (g)
Reaction Rates:
2NO2(g)  2NO(g) + O2(g)
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically
Reaction Rates:
2NO2(g)  2NO(g) + O2(g)
D[NO2]
Dt
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
D[ N O2 ]
Dt
 constant
Defining Rate of a Reaction
• We can define rate in terms of the disappearance of the
reactant or in terms of the rate of appearance of the
product.
• In our example
N2 + 3H2
2NH3
• -D[N2]
Dt
or -3D[H2] or 2D[NH3]
Dt
Dt
Rate Laws
• Reactions are reversible.
• As products accumulate they can begin to turn back into
reactants.
• Early on the rate will depend on only the amount of
reactants present.
• We want to measure the reactants as soon as they are
mixed.
• This is called the Initial rate method.
Rate Laws
• Two key points
1.
The concentration of the products do not appear in the rate law
because this is an initial rate.
2.
The order must be determined experimentally, it cannot be
obtained from the equation.
2 NO2(g) → 2 NO(g) + O2(g)
• Rate will only depend on the concentration of the
reactants. Products have not built up yet. To define rate
in terms of NO2:
• Rate = k[NO2]
n
• This is called a rate law expression.
• k is called the rate constant.
• n is the order of the reactant - usually a positive integer. (must be
determined by experiment)
Types of Rate Laws
• Differential Rate law - describes how rate depends on
concentration. (Typically just called Rate Law)
• Integrated Rate Law - Describes how concentration
depends on time.
• For each type of differential rate law there is an integrated rate
law and vice versa.
• Rate laws can help us better understand reaction mechanisms.
Determining Rate Laws
• The first step is to determine the form of the rate law
(especially its order).
• Must be determined from experimental data.
• For this reaction in CCl4 soln.
2 N2O5 (aq) → 4NO2 (aq) + O2 (g)
Oxygen escapes the solution and therefore there is no
reverse reaction
Now graph the data
[N2O5] (mol/L)
1.00
0.88
0.78
0.69
0.61
0.54
0.48
0.43
0.38
0.34
0.30
Time (s)
0
200
400
600
800
1000
1200
1400
1600
1800
2000
1.2
1
0.8
0.6
0.4
• To find rate we have to find the slope at two points
0.2• We will use the tangent method.
0
2000
1800
1600
1400
1200
1000
800
600
400
200
0
1.2
1
0.8
0.6
0.4
At .90 M the rate is - 5.4 x 10 -4
0.2
0
2000
1800
1600
1400
1200
1000
800
600
400
200
0
1.2
1
0.8
0.6
0.4
0.2
At .45 M the rate is - 2.7 x 10 -4
0
2000
1800
1600
1400
1200
1000
800
600
400
200
0
.90 = -5.4 x 10-4
.45
-2.7 x 10-4
• As the [
] doubles the rate of reaction doubles.
(Twice as fast)
∆[
2x
]
Rate
2
Order of Reaction
Rate Doubles
1st Order (x = 1)
1 =
Quadruples
2ndk[N
Order
(x = 2)
2•x Rate =4 -D[N2ORate
]
=
k[N
O
]
5
2 5
2O 5]
3rd Order (x = 3)
2x Dt
8
• We say this reaction is first order in N2O5
• Mathematically (Initial rates method)
• It is also possible to use the initial rates method to find the
orders
• We use the ratio of the rate equations.
The method of Initial Rates
• This method requires that a reaction be run several times.
• The initial concentrations of the reactants are varied.
• The reaction rate is measured just after the reactants are
mixed.
• Eliminates the effect of the reverse reaction.
An example
-
-
+
• For the reaction BrO3 + 5 Br + 6H
3Br2 + 3 H2O
• The general form of the Rate Law is
-]n [Br-]m [H+]p
3
• We use experimental data to determine the values of
n,m,and p
• Rate = k[BrO
Initial concentrations (M)
Rate (M/s)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Now we have to see how the rate changes
with concentration