Many reactions which are thermodynamically
feasible are
not kinetically
feasible.
O2(g)
+ H2(g) -----> H2O(g)
DH
= -241.8 lK/mol reaction
tells us that the reaction is
allowed to go
thermodynamically.
However, if we try the experiment,
we will wait for a
VERY
long time, because the KINETICS
for this reaction
are extremely slow.
14.1
Rate
of reaction = speed of reaction = positive
quantity that tells us how [reactant] or [product]
changes with time.
Or:
how
quickly products are
made or reactants are
used up.
Consider:
aA + bB ----> cC + dD, where the lower case
letters are
the coefficients in the balanced equation
and the upper case
are the chemical species.
Rate
of reaction = -(D[A])/aDt
= -(D[B])/bDt
loss of A or loss of B with time
= (D[C])/cDt
= (D[D])/dDt
gain of C or gain of D with time
Ex.: H2(g) + 2ICl(g) -------> I2(g) + 2HCl(g)
Rate of reaction = -(D[H2])/Dt
= -(D[ICl])/2Dt
= (D[I2])/Dt
= (D[HCl])/2Dt
Data:
[H2]
[ICl]
time(sec)
rate = -(D[H2])/Dt
= -(D[ICl])/2Dt
0.750
1.500
0
0.424
0.848
1
> 0.326
Ms-1
0.276
0.552
2
> 0.148
0.186
0.372
3
> 0.090
0.124
0.248
4
> 0.062
0.078
0.156
5
> 0.046
Plot []
as
a function of time.
Instantaneous
rate of reaction at any time t is -slope
of
the
tangent
to the curve at that time t.
Rate at time 4 sec.
=
-slope of the tangent at 4 sec.
rate = (-slope) = -(rise/run) = -Dy/Dx
= -[(.06-.35)/(5-0)]
= -(-0.29/5) = 0.058 Ms-1
At t = 0, instantaneous
rate = -Dy/Dx
= -(0.600-0.750)/(0.5-0.0)
(from [H2])
= -(-0.300) = 0.300 Ms-1
OR: instantaneous
rate=-Dy/2Dx=-(1.20-1.50)/(0.5-0.0)
(from [ICl]) =
-(1/2)(-0.30/-0.5)=0.300 Ms-1
NOTE: get the same
instantaneous rate from either
reactant!!
- - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - - -
Average
rate over some time:
EX: average
rate over the 1st
3 sec.:
average
rate
= -(D[H2])/Dt
= -(0.186-0.750)/(3-0)
= -(-0.564/3) = 0.188 Ms-1
EX: average
rate for the 1st
5 sec.:
average
rate
= -(D[H2])/Dt
= -(0.078-0.750)/(5-0)
= -(-0.672/5) = 0.134 Ms-1
Ex.: Iodide ion is
oxidized
by hypochlorite ion in basic
solution:
I-1(aq) + ClO-1(aq) ---> Cl-1(aq) + IO-1(aq)
In 1.00 M NaOH at 25oC,
the [I-1] at different times was
measured to
be:
time: 2.00
s
8.00
[I-1]:
0.00169
0.00101
Calculate the average
rate of the oxidation of I-1
during
this time
interval.
average rate = -[DI-1]/DT
= -[(0.00101-0.00169)/(8.00-2.00)]
= 1.13 X 10-4 Ms-1
- - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - -
Rates of chemical reactions are affected by:
1) concentrations
of reactants: usually
the higher the
concentrations, the faster the reaction.
2) T
of the reaction: usually
the higher the T, the faster
the reaction. Therefore, bacterial spoilage of food is
slowed by lowering the T (refrigerator).
3) catalyst:
the rate of a reaction is usually
increased
by adding a catalyst, without the catalyst being used
up.
4) surface
area of solid or liquid reactants or catalysts:
the larger the surface area, the faster the reaction.
Ex.:
2 N2O5(g)
----> 4 NO2(g) +
O2(g)
14.2
The
Rate Laws
Dependence
of the Rate on Concentration:
called the rate law
rate = k[A]x[B]y[C]z
the coefficients x, y, z can ONLY be found by experiment !!!
Experimental
Determination of Rate Laws:
We will study the rate of a reaction by measuring the
instantaneous rate at t = 0
for each reactant in the
equation (initial rate of
reaction).
Ex.: 2 NO(g) + Cl2(g) ----> 2 NOCl(g)
[NO] [Cl2]
initial rate (Ms-1)
exp.1 0.10
0.10 2.53 x 10-6
exp.2 0.10
0.20 5.06 x 10-6
exp.3 0.20
0.10 10.1 x 10-6
exp.4 0.30
0.10 22.6 x 10-6
rate = k[NO]x[Cl2]y
1)
compare exps. 1 & 2:
rate 2 = 5.06 x 10-6 =
k[NO]x[Cl2]y=
k(0.10)x(0.20)Y
rate 1 2.53 x 10-6
k[NO]x[Cl2]y
k(0.10)x(0.10)Y
5.06
x 10-6 =
2 = (0.20/0.10)y
= 2y
Therefore, y
= 1
2.53
x 10-6
2)
compare exps. 1 & 3:
rate 3 = 10.1 x 10-6 =
k[NO]x[Cl2]y=
k(0.20)x(0.10)Y
rate 1 2.53 x 10-6
k[NO]x[Cl2]y
k(0.10)x(0.10)Y
10.1
x 10-6 =
4 = (0.20/0.10)x
= 2x
Therefore, x
= 2
2.53
x 10-6
Therefore, rate = k[NO]2[Cl2]1
2nd
order with respect to NO and 1st
order with respect
to Cl2 , and 3rd
order overall.
Find
k, the reaction rate constant:
Using data from Exp. 1:
2.53 x 10-6 Ms-1=
k(0.10 M)2(0.10M)1
2.53 x 10-6Ms-1/(0.10
M)2(0.10M)1
= k
2.53 x 10-3M-2s-1
= k
OR:
Using data from Exp. 4:
22.6 x 10-6 Ms-1=
k(0.30 M)2 (0.10M)1
22.6 x 10-6Ms-1/(0.30
M)2
(0.10M)1
= k
2.51 x 10-3M-2s-1
= k
THEREFORE, k = constant!!!