Transcript Rate Law

Rate Law
5-2
an expression which relates the
rate to the concentrations and a
specific rate constant
• For a general reaction between reactant
A and B at a constant temperature the
reaction can be represented by:
aA + bB -------> products
Rate law is expressed as:
Rate = k[A]m [B]n
• The rate law equation expresses the
relationship between the concentration of
the reactants and the rate of the equation.
1
2
Rate constant
(k)
Exponents
m and n
Varies with
temperature
Constant with
temperature
Constant under Can only be
constant
determined
conditions
experimentally
Exponents and Orders of reactions.
• The values of exponents (m or n) can
only be determined through
experimentation and may or may not be
the values of the coefficients in the
balanced chemical equation.
• The value of the exponents determines
the order of the reaction. If a reaction
has a single reactant and the value of
the exponent is one then it is first
order.
• If the exponent were two then it is
second order.
• If more than one reactant is present in a
reaction, the sum of the exponents (m
+ n) is called the overall reaction
order.
• 2 N2O5 (g) → 4 NO2 (g)+ O2 (g)
• Experimentally the rate was found
to be first order for N2O5 (g).
• Rate = k [N2O5 (g)] 1
• The value of exponent is not the
same as the balance for N2O5 in the
equation.
• The overall reaction order would be
first order.
• NO2 (g) + CO (g) → NO (g) + CO2
(g)
• Experimentally the rate was found
to be second order for NO2 (g) and
zero order for CO (g). The rate law
can be written:
• Rate = k [NO2] 2 [CO] 0
• or Rate = k [NO2] 2
• The value of the exponents is not the
same as the coefficients.
• The overall reaction order is second
order ( 2 + 0 ).
2 HI (g) → H2 (g) + I2 (g)
Experimentally the rate was found to
be second order for HI (g).
• Rate = k [HI] 2
Assignment #1: Rate Law
Equations
Reaction
1 A + B ----->
products
Rate
Rate law
depende
nt on
[A] 2
=k[A] 2 [B]1
[B]1
2 2NO(g) +O2 (g) [NO]1
----> NO2 (g)
[O2]1
=k[NO]1[O2]1
Reacti
on
order
3
2
3 2NO(g) + 2H2 (g)
----> N2(g) +
2 H2O(g)
4 2NO2 (g) +Cl2 (g)
------>
2NO2Cl(g)
5 2ClO2 (aq)+ 2OH
- ------> ClO 3
(aq) + ClO2- (aq)
+ H2O (l)
6 C + D ------>
products
[NO]2
[H2]1
=k[NO]2 [H2]1
3
[NO2]1
1 [Cl ]1
=k[NO]
2
[Cl2]1
2
[ClO2]0
0 [OH]1
1
=k[ClO]
[OH]
1
[C]2
[D]0
=k[C] 2 [D]0
2
• Determination of Rate Exponents
using Initial rates of reactions.
• Chemists can examine the change in
the initial rates of the reaction when
concentrations of reactants are changed
to determine the order of reactants (for
a specific equation).
Order of Reaction
Change in intial rate
when concentration
Order of reactants
is doubled
First order
second order
zero order
Doubles the rate
of reaction
Quadruples the rate
of the reaction
No change in the rate
of the reaction
• In the following questions
determine
the order of each reactant
and the overall rate
the value of the rate constant
• All the experiments were conducted
under conditions of constant
temperature.
a.. Single reactant
A -----> B + C
the following data was collected
Exp
1
Initial [A]
(mol/L)
0.01
Initial rate
(mol/L·s)
4.8 x 10-6
2
0.02
9.6 x 10-6
3
0.03
1.4 x 10-5
Finding the rate law and the
constant
• Choose the data from the table to solve for
the exponent and the constant.
• Rate law = k [ A ]m
Choose two experimental data and then
divide one by the other to find the
exponent
Calulate K by putting the experimental data
into the rate law including the value of the
expnent and solve for K.
Rate law = k [ A ]m
• Exp1: 4.8 x 10-6 = k [0.01]m
• Exp 2: 9.6 x 10-6 =k [0.02]m
• Exp1:
• Exp 2:
–
1 = [1]m
2 = [2]m
m=1
Finding k
• Chose one set of data and solve
• Exp 1
• Exp1: 4.8 x 10-6 = k [0.01]1
4.8 x 10-6 = k [0.01]1
0.01
0.01
4.8 x 10-4 = k
• Rate law = 4.8 x 10-4 [A]1
b. Two reactants
exp
Initial [A]
Initial [B]
Initial rate
1
0.01
0.03
2.4 x 10-4
2
0.03
0.03
7.2 x 10-4
3
0.01
0.06
2.4 x 10-4
• Choose the data such that the
concentration of one of the reactants, A, is
the same in both the experimental data.
• In this case exp 1 and exp 2
Rate law = k [A]m[B]n
• Exp1: 2.4 x 10-4 = k [0.01]m [0.03]n
• Exp 3: 2.4 x 10-4 =k [0.01]m[0.06]n
•
•
1
= ( 1/2 ) n
n=0
Solve for m now
• Exp1: 2.4 x 10-4 = k [0.01]m [0.03]n
• Exp 2: 7.2 x 10-4 =k [0.03]m[0.03]n
•
•
1/3
= ( 1/3 ) m
m=1
Rate law = k [A]1[B]1
• Solve for k
• Exp1: 2.4 x 10-4 = k [0.01]1 [0.03]0
•
0.01
•
•
2.4 x 10-2 = k
Rate law = 2.4 x 10-2 [A]1[B]0
0.01
2 ICl + H2 → I2 + 2 HCl
exp
[ ICl ]
[ H2 ]
Initial rate
1
0.10
0.01
0.002
2
0.20
0.01
0.004
3
0.10
0.04
0.008
Rate law = k [ICl]m[H2]n
• Exp1: 0.002 = k [0.10]m [0.01]n
• Exp 2: 0.004 = k [0.20]m[0.01]n
•
•
1/2
m
= (1/2)m
= 1
• Exp 1: 0.002 = k [0.10]m [0.01]n
• Exp 3: 0.008 = k [0.10]m[0.04]n
•
•
1/4
n
= (1/4)n
= 1
Solve for k
• Exp1: 0.002 = k [0.10] 1 [0.01] 1
• Exp1: 0.002 = k [0.10] 1 [0.01] 1
•
0.001
0.001
•
2
= k
• Rate law = 2 [ICl]
1
[H2]
1
• 1. For the general reaction:
A + B -----> C
• the following data was collected
Exp
1
[A]
(mol/L)
0.002
[B]
(mol/L)
0.05
Initial rate
(mol/L·s)
2 x 10-5
2
0.004
0.05
8 x 10-5
3
0.002
0.10
2 x 10-5
Answer
• Rate law = 5 [A]
2
[B]
0
• 2. For the reaction:
• H2O2 (aq)+2HI (aq)→ 2H2O (l)+I2
(aq)the following data was collected
exp
1
[H2O2]
(mol/L
0.05
[HI]
(mol/L)
0.05
Initial rate
(mol/L·s)
0.002
2
0.05
0.10
0.004
3
0.10
0.05
0.004
answer
• Rate law = 0.8 [H2O2
1
]
1
[HI]
3. For the reaction:
(CH3)3Br + OH- →(CH3)3COH +Br• the following data was collected
1
[(CH3)3CBr]
(mol/L)
0.03
[OH-]
(mol/L)
0.04
Initial rate
(mol/L·s)
1.2 x 10-3
2
0.03
0.08
1.2 x 10-3
3
0.06
0.04
2.4 x 10-3
exp
answer
• Rate law
1
0
• = 0.04 [(CH3)3CBr] [OH]