Transcript The Rate Law

The Rate Law
And Reaction Orders
Rate Law
• Expresses the relationship of rate of
reaction to the rate constant and the
concentrations of the reactants
raised to some powers
• For general reaction:
aA + bB  cC + dD
• The raw law is:
Where x and y are
determined
rate= k[A]x[B]y numbers
experimentally
Rate Law
• Reaction Order: is the number that
expresses how the rate changes
with concentration, expressed as a
power
• Overall reaction order: is the sum of
the powers to which all the reactant
concentrations appearing in the rate
law are raised(reaction orders)
Reaction Order
0) -A reaction order of 0 with respect
to A means that the rate is
independent of the concentration of
A.
- The rate law is given by:
Rate= k[A]0
Rate=k
Reaction Orders
1) -A reaction order of 1 with respect to A
means that when the concentration of
A is doubled, the rate doubles(21),
when the concentration of A is tripled,
the rate triples(31)
-Means that the concentration of A is
directly proportional to the rate
-The rate law is given by
rate=k[A]
Reaction Orders
2) –A reaction order of 2 with respect
to A means that when the
concentration of A is doubled the
reaction rate quadruples(22). When
the concentration of A is tripled, the
rate becomes 9 times greater(32)
-The rate law is given by
rate=k[A]2 or rate= k[A][B]
Overall Reaction Order
• Determines the units of rate
constant(k)
0  M/s because M/s=M/s
1st s-1 because M/s=(M-1s-1)(M2)
2nd M-1s-1 because M/s=(M-2s-1)(M3)
Finding the rate constant
• We can calculate the rate constant because
we are given concentrations and the rate
rate
1.2 x103 M / s
k

 1.2M * s
[ F2 ][ClO2 ] (0.10M )(0.010M )
First-Order Reactions
Relating concentration to time
• In a first order reaction of the type
Aproducts
• Rate= -Δ[A]/Δt
• From the rate law we also know
rate=k[A]
• Combining the two equations we get

[ A]  k [ A]
t
• Using Calculus we can show for the above
equation that
[ A]t
ln
 kt
[ A]0
First-Order Reactions
• We can rearrange the equation to:
ln[A]t  kt  ln[A]0
Example
The conversion of cyclopropane to propene in the
gas phase is a 1st order reaction with a rate
constant of 6.7x10-4s-1 at 500°C.
a) If the initial concentration of cyclopropane was
0.25M, what is the concentration after 8.8min?
b) How long will it take for the concentration of
cyclopropane to decrease from 0.25M to
0.15M?
c) How long will it take to convert 74% of the
starting material?
Example
a) First we convert 8.8mins to sec because the
rate constant is given in units s-1
8.8 min X
60 sec
 528 sec
1 min
We are given [A]0=0.25M and asked for [A]t after
8.8mins so we use the equation below and
substitute the values to find [A]t
ln[A]t  kt ln[A]0
ln[A]t  (6.7 x10 4 s 1 )(528s)  ln(0.25)
ln[A]t  1.74
[ A]t  e 1.74
[ A]t  0.18M
Example
b) We are given concentrations and we need to
solve for t, use the same equation as a) but
rearrange for t
0.15M
 (6.7 x10 4 s 1 )t
0.25M
1 min
t  7.6 x102 sX
60s
t  13min
ln
Example
c) We do not have any concentrations in this part
but we know there was 100% to start and 74%
reacted so 26% is left over, we can use that ratio,
because it would be equal to the actual
concentrations [A]t/[A]0, or 0.26/1.00. Using
equation we get
0.26
 4 1
ln
 (6.7 x10 s )t
1.00
1 min
3
t  2.0 x10 sX
60s
t  33min
Half-Life of First Order
• Half-life-time required for the concentration of a
reactant to decrease to half of its initial
concentration.
• The expression for t½ for a first order reaction is
below
• When t= t½, [A]t=[A]0/2, so
[ A]0
1
t 1  ln
or
k [ A]0 / 2
2
1
t 1  ln 2
k
2
0.693
t1 
k
2
Second-Order Reactions
Relating concentration to time
• In a second-order reaction of the type
Aproducts
• Rate= -Δ[A]/Δt
• From the rate law we also know
rate=k[A]2
• Another type of second-order reaction is
A + Bproducts
• From the rate law we know
rate=k[A][B]
Second-Order Reactions
• Using calculus we obtain expression below for
Aproducts second-order:
1
1
 kt 
[ A]t
[ A]0
• We can also find an equation for half-life by
setting [A]t=[A]0/2 to get
1
1
 kt 1 
[ A]0 / 2
[ A]0
2
1
t1 
k[ A]0
2
Refer to table 13.3 on page 554 for
equations