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Lecture
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How to do the famous kinetic calculations, applied to 0th, 1st and 2nd order reactions In this lecture you will learn to do the following^ •^
Determine reaction order from units • Determine reaction order from method of initial rates • Calculate “k” from rate equation • Convert how fast (differential rate equation) into how much (integrated rate equation) using calculus • Use integrated rate law to find half lives • Use integrated rate law to find extent of reaction First we review the differential rate equation and define the pieces of the rate law:
Δ^ [R ]
=^
rate =
x^ [R] A exp [-Ea/RT] =
k [R ]
x
Δ^ t
rate constant, k
RateFromSlope
rate law
Note that there are four physical parameters in the rate law that determine the rate of reaction:
1.^ [ ]
x^ concentration and order of reactants and products
2.^ Ea
activation energy
3.^ T^
temperature
the three factors making up k, the rate constant
4.^ A
pre-exponential factor
We will save learning to do calculations involving parameters 2-4 for Lecture 3 when we learn about kinetic theory andconcentrate for now on the concentration term, [R ], and how the order of reaction, x.
x[R ]
Order of Reaction In the rate equation: rate = k [R ]
x^ ,^ x is the order of reaction in the function that describes how concentration affects rate.
Example of reaction order:
2N2O
4NO
+ O 2
2
rate = k[N
]2O^5
This reaction is first order in [N
] and first order overall.2O^5
Example of reaction order:
2NO + O
2NO
rate = k[NO][O
] 2
This reaction is first order in NO and in O
and second order overall. 2
We will look at three examples of reaction order in this lecture. order
rate equation
effect of concentration on rate
rate = k[R]
0 = k^
none
rate = k[R]
1
rate increases linearly with concentration
rate = k[R]
2
rate increases as the square of the concentration
Example 1: 0
th^ order
x = 0
so Rate = k [R]
0 = k
This means that rate is independent of concentration [R] as is shown below.
Note that no matter what the concentration, the
slope does not change.
th^0 order
[ ]
t
TIME OUT FOR CALCULUS.
The rest of this lecture will follow what happens in two equations, a differential rate equation
and an integrated rate equation.
Just a reminder of what you should have learned in calculus as a scientist, if you learned
nothing else: Differential Calculus
Tells how fast something happens—it is a value found from the instantaneous slope of a function.
Integral Calculus Tells us how much we have of something and is found from area under the curve made by a function.Understand that how fast (from the differential rate equation) and how much (from the integral rate expression) are really whatwe most want to calculate in kinetics—and we are using the simple tools we learned in calculus to do it.
Can get instantaneousslopes along function
Smaller slope
Larger slope
Well^
That is pretty much it for calculus—back to kinetics.
Let’s look at kinetics calculations for reactions with different
reaction order. Look at x = 0, 1, 2 First, look at x = 1 = first order reaction
in the simple reaction below
rR^
Æ^
pP^
(assume r = 1)
We know that the rate is found from the slope of either R or P
-^ Δ^ [R] = +
Δ^ [P] = rate = k [R]
x^ y[P]
rΔ^ t^
pΔ^ t
and the rate law tells us how concentration matters—but how do we find out x and y?There are a variety of methods, and in fact I have a nice summary sheet posted on the many ways to find the order of a reaction.But experimentally, in lab you will use something called: Method of initial rates Run multiple
kinetics experiments
with different starting amounts.
For each, find the slope (rate) of the reaction right at the
start.
Note in the data below that we are being intelligent scientists by holding one concentration constant while varying the other. This allows us to isolate how the concentration effects the rate of reaction.
We want tolook here atthe initialdata pointto find rate
Can we also find the units of k? Simply cancel units in rate expression[R ] = k [R ]
cancel and 1/t = k t This means that the units for first order rate expression are reciprocal time: like sec
So k = 2 x 10
-3^ sec
Integrated Rate Equation for a first order process. Recall that earlier I suggested it might be nice to know what the function was that described the relationship betweenconcentration and time for a reaction curve. Here is our chance.The good news is that the equations we will develop to describe the kinetic reactions fall into a few simple categories which area function of the order of a reaction.
In other words, all zero order reaction have the same shape to the curve; all first order
reactions have the same shape to the curve; all second order reactions have the same shape to the curve.The bad news is that we have to use calculus to find the equation.Lets do it for just the first order integrated rate equation since it is really famous.
This is the one that describes radioactive
decay and a million other things.
Can we find the area under a curve and determine how much we have at any time:^ Now lets do some calculus to see how much occurs in first order reactions. Note that we will be using very simple functionsthat you have learned to integrate way back in baby calculus.-^ Δ^ [R] = k [R]
(^1) rΔ^ t Rearrange -
Δ^ [R] = rk
Δt [R]
Integrate
∫-^ Δ^
[R] =
∫rkΔ
t [R]f^
f
Solve
-ln [R]
= rkt^0
And solving:
-(ln [R
] – ln [Rf^
]) = rk (t 0
And rearranging:
ln [R
]^ = ln [Rf^
] – rkt 0
Å
The answer: How much is left [R
] from what starts [Rf
], for given k and t 0
values in a first order process
Applications of Integrated 1
st^ order Equation
1.^ Half life equation. How much time does it take for half of starting amount to go away in a first order reaction?So
t^ 1/
= half life when R
= 0.5 Rf
0
and substituting:
ln [R
]^ = ln [Rf^
] – rkt 0
at t
½
ln [0.5 R
] = ln [R 0
] – rkt 0
and if for example, R
-1 1 and Ro^
= 0.5f^
Calculations involving zeroth order reactions:^ Consider the simple reaction:rR^ Æ
pP^
where r = p = 1 For a zeroth order reaction the rate expression isRate = -
Δ[R] = k[R]
0 0 [P]^
= k
Δt And a plot of concentration versushas an invariant slopeFirst let’s look at a method of initial rates calculation that would involve a zeroth order reaction.
Note that no matter how the
concentrations are varied, the rate is a constant.Data for method of intitial rates calculation
[R]^
[P]^
rate
Experiment 1
.1M^
.1M^
2 x 10
Experiment 2
.2M^
.1M^
2 x 10
Experiment 3
.2M^
.2M^
2 x 10
Order of R (Experiments 1 and 2)Mathematically this is the ratio of the rate of reaction
to the concentration [R ] in experiments 1 and 2 where [P] is held
constant.
x
2 x 10
-3^ = 0.
2 x 10
-^
[R]
t
No changein slope
And solving:1 = 2
x^
so^
x = 0
Order of P (Experiments 2 and 3)Mathematically this is the ratio of the rate of reaction
to the concentration [P ] in experiments 2 and 3 where [P] is held
constant.
y
2 x 10
-3^ = 0.
2 x 10
-^
And solving:1 = 2
y^
so^
y = 0
So, Rate = k[R]
0 0 [P]
= k
What is k here?Substitute values for rate law from Experiment 1Rate = 2 x 10
3 = k[R]
00 [P]
k = 2 x 10
3 What are the units of the rate constant?
Rate = [ ]/t = k So pretty obviously,th^0 order units are k =[ ]/t
or [ ] t
-1^ or Msec
-^
and^
k = 2 x 10
3 Msec
Calculations involving second order reactions:^ Consider the simple reaction:rR^ Æ
pP^
where r = 2 2R^ Æ
P
Rate = -
Δ[R] = k[R]
x^ y[P] 2 Δt What are x and y?Use method of initial rates.
For a second order reaction the rate expression is
Rate = -
Δ[R] = k[R]
2 0 [P]^
= k
Δt First let’s look at a method of initial rates calculation that would involve a second order reaction.Data for method of intitial rates calculation
[R]^
[P]^
rate
Experiment 1
.1M^
.1M^
4 x 10
Experiment 2
.2M^
.1M^
16 x 10
Experiment 3
.2M^
.2M^
16 x 10
Order of R (Experiments 1 and 2)Mathematically this is the ratio of the rate of reaction
to the concentration [R ] in experiments 1 and 2 where [P] is held
constant.
x
16 x 10
-2^ =
4 x 10
-^
And solving:4 = 2
x^
so^
x = 2
Order of P (Experiments 2 and 3)Mathematically this is the ratio of the rate of reaction
to the concentration [P ] in experiments 2 and 3 where [P] is held
constant.
y
16 x 10
-2^ =
16 x 10
-^
And solving:1 = 2
y^
so^
y = 0
So, Rate = k[R]
2 0 [P]
= k [R]
2
What is k here?Substitute values for rate law from Experiment 1Rate = 4 x 10
-2^ = k(0.1)
(^2)
k = 4
Applications of Integrated 2nd order Equation^1
Half Life Equation. How much time does it take for half of starting amount to go away in a first order reaction?So^
t^ = half life when R1/^
= 0.5 Rf
0
and substituting:
(1/rk R
- = t1/
Notice in this case the half life is concentration dependent.
2 A garden variety integrated rate law calculation. How much do I have left if [R
] = 0.1 M and k = 4 s 0
-1-1 [ ]
after 5 seconds?
Remember r = 2 from 2R
Æ^ P
(1/[R
]) = (1/[Rf
]) + rkt 0
(1/[R
]) = (1/0.1) + (2)(4)(5)f (1/[R
]) = 50f [R] = 0.02f^ So after 5 seconds, 80% is gone
