New example:
Given: A + B + C ----> Prods.
Find the rate law and k (specific rate constant).
[A]
[B]
[C] initial rate (Ms-1)
exp.1 0.20
0.10 0.90 1.8 x 10-2
exp.2 0.10
0.20 0.30 3.0 x 10-3
exp.3 0.20
0.20 0.30 1.2 x 10-2
exp.4 0.20
0.20 0.10 4.0 x 10-3
rate = k[A]x[B]y[C]z
1)
Compare exps. 4 & 3:
rate
4 =4.0 x 10-3 =
k[A]x[B]y [C]z=k(0.20)x(0.20)y(0.10)z
rate
3
1.2 x 10-2 k[A]x[B]y
[C]z k(0.20)x(0.20)y(0.30)z
4.0
x 10-3 =
4.0 = 0.333 = (0.10)z
= (0.333)z
1.2
x 10-2
12.0
(0.30)z
Therefore, z = 1
2)
Compare exps. 2 & 3:
rate
3 =1.2 x 10-2 =k[A]x[B]y
[C]z=k(0.20)x(0.20)Y(0.30)z
rate
2
3.0 x 10-3
k[A]x[B]y [C]z
k(0.10)x(0.20)Y(0.30)z
1.2
x 10-2=
12.0 = 4.0
= (0.20)x
= 2x
Therefore, x = 2
3.0
x 10-3
3.0
(0.10)x
3)
Compare exps. 1 & 3:
rate
1 =1.8 x 10-2 =k[A]x[B]y
[C]z=k(0.20)x(0.10)Y(0.90)z
rate
3
3.0 x 10-3
k[A]x[B]y [C]z
k(0.20)x(0.20)Y(0.30)z
1.8
x
10-2 =
1.8 = 1.5 = (0.10)y(0.90)z
1.2 x 10-2
1.2
(0.20)y(0.30)z
=(0.5)y(3)z =(0.5)y(3)1
1.5 = (0.5)y(3)
1.5 = (0.5)y;
0.5 = (0.5)y
Therefore, y = 1
3
We now
have
the values for x, y , z, and therefore have
the rate law
rate = k[A]x[B]y[C]z
= k[A]2[B]1[C]1
We
can now calculate k, the specific rate constant:
1.8
x 10-2 Ms-1 =
k(0.20M)2(0.10M)1(0.90M)1
k = 1.8 x 10-2Ms-1
= 5.0 M-3s-1
(0.20M)2(0.10M)(0.90M)
- - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - -
14.3
Relation
Between Reactant Concentration and Time
Concentration vs. Time:
Integrated
rate equation
We can
use
this to find out how much of a reactant remains
after a
specified amount of time. We can also find out what
the
half-life,
t1/2,
the time it takes for 1/2 of that reactant
to be
used
up. These are different for different orders of
reactions.
We will be looking at only one reactant going to
product.
0th-Order Reactions: for me: check this all out!!!!!!!!!
aA -----> Products: if the reaction is 0th order in A
and 0th order overall, then:
Rate = k[A]0 = k = -(1/a)(D[A]/Dt)
which means that the rate is independent of the
concentration of A.
The integrated equation is: [A] = [A]o - akt
And
the
half-life is: t1/2
= [A]o/2ak
(THIS
IS ONLY
FOR 0th ORDER!!)
Therefore,
a linear plot of [A] vs. time yields a straight
line function, where y = mx + b, and
m = slope = -k and b
= the
y-intercept
at time 0
UNITS of k are: Mtime-1
=
Ms-1 =
Mmin-1 =
Mhr-1
*************************************************************************
1st-Order Reactions:
aA -----> Products: if the reaction is 1st order in A
and 1st order overall, then:
Rate = k[A]1
which
means
that the rate is directly dependent on the
concentration of A.
The integrated equation is: ln([A]/[A]o) = -akt
where: [A]o = initial
concentration of A
[A] = concentration of A at some time t
a = coefficient in the balanced equation
k = specific rate constant
t
= time after reaction begins
And
the
half-life
is: t1/2
= 0.693/ak
(THIS IS ONLY
FOR 1ST-ORDER!!)
***************************************************************************
Derivation of 1st order
integrated
rate equation:
rate = -D[A]/aDt = k[A]1
(D[A])/[A] = - akDt
Integrate left-hand side
from
Ao to A and the right side
from t = 0 to t = t:
ln([A]/[Ao]) = - akt
ln[A] - ln[Ao] = - akt
****************************************************************************
Therefore,
a linear plot of [A] vs. time yields an exponential
decaying curve; a plot of ln[A] vs. time yields a straight
line function, where y = mx + b, and
m = slope = -ak
and
b = the
y-intercept
at time 0 = ln [A]o
UNITS of k are: Rate = k[A]1
Rate/[A]1 = k ; therefore, Ms-1/M = time-1
Ex.:
The decomposition of N2O5
to NO2 and O2
is 1st order,
with a rate constant of 4.80 x 10-4/s
at
45oC. If the initial
concentration of N2O5 is
1.65 x 10-2 mol/L, what is the
concentration after 825 s?
Balance the equation!!! 2 N2O5(g) ----> 4 NO2(g) + O2(g)
ln([A]o/[A]) = akt
ln[A]o- ln[A] = akt
ln (1.65 x 10-2M) - ln[A] = akt = 2(4.80
x 10-4/s)(825s)
-4.104 - ln[A] = 0.792
- ln[A] = 0.792 + 4.104 = 4.896
ln[A] = -4.896
[A] = 0.00747 M = 7.47 x 10-3 M
How long
would it take for the concentration of N2O5
to
decrease to 0.950 x 10-2 mol/L from
the
initial value
given above?
ln([A]o/[A])
= akt
ln (1.65 x 10-2 M/0.950 x 10-2
M) = 2(4.80 x 10-4/s)(t)
ln(1.737) = (9.60 x 10-4/s)(t)
0.552/(9.60 x 10-4/s) = t
575 s = t
- - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - - -
DERIVATION OF 1/2-LIFE:
ln([A]o/[A]) = akt
ln([A]o/([A]o/2)) = akt1/2
ln(1/2) = akt1/2
ln2 = akt1/2
0.693 = akt1/2
- - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - - - - -
At 25oC,
the specific rate constant for a reaction A ---> B
is 0.0450
s-1.....Therefore, 1st order!!
What is the
half-life of A?
t1/2
= 0.693/ak
t1/2 = 0.693/1(0.0450
s-1) = 15.4 s
This means that 1/2 of
A is gone after 15.4 s, another 1/2
of the 1/2
is gone after another 15.4 s (therefore only
1/4 is left!!), etc.
*****************************************************************************
2nd Order Reactions:
aA -----> Products: if the reaction is 2nd order in A
and 2nd order overall, then:
Rate = k[A]2
The integrated equation is: (1/[A]) - (1/[A]o) = akt
where: [A]o = initial
concentration
of A
[A] = concentration of A at some time t
a = coefficient in the balanced equation
k = specific rate constant
t = time after reaction begins
And
the
half-life is: t1/2
= 1/ak[A]o
(THIS IS ONLY
FOR 2nd ORDER!!)
Therefore,
a linear plot of [A] vs. time yields an exponential
decaying curve; a plot of ln[A] vs. time also yields an
exponential decaying curve; a plot of 1/[A] vs. time
yields a straight line function, where y = mx + b,
and m = slope = ak
and
b = the
y-intercept
at time 0 = 1/[A]o
UNITS of k are:
Rate = k[A]2
Rate/[A]2 = k; therefore, Ms-1/M2 = M-1time-1
1) Plot [NO2]
vs. time and get an exponential decay;
therefore, NOT 0th order !!!
2) Plot ln[NO2]
vs. time and get another exponential
decay; therefore, NOT 1st order !!!
3) Plot 1/[NO2]
vs. time and get a straight line;
therefore, the reaction IS 2nd
order
!!
slope = ak = Dy/Dx
= (333-100)/(300-0) = 0.777
therefore, k = slope/a = 0.777/2 = 0.388 M-1s-1