Chem 1310: Introduction to physical chemistry

Part 2c: integrated rate laws

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t½

The integrated rate law

If we know how the rate depends on all concentrations,we can predict how it will change over time.

This is called the "integrated rate law"At each point we know concentrations, hence the rate,hence we can predict concentrations at the next moment:

rate = [Z]/t[Z] = rate*tnew [Z] = old [Z] + [Z] = old [Z] + rate*t

The integrated rate law

A rigorous derivation of integrated rate laws requires calculus, but we can understand and use the results without using calculus.

We will discuss only three cases:• First-order (most important by far!)• Second-order• Zero-order

The integrated rate lawfor a first-order reaction

For a first-order rate lawrate = k [A]

the integrated rate law is[A]t = [A]0 e-k t ("exponential decay")

or equivalentlyln [A]t = ln [A]0 - k t

So, plotting ln [A]t vs t should give a straight line with slope -k and intercept ln [A]0.

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Checking for thecrystal violet reaction

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[CV+] vs t ln [CV+] vs t

ln plot is nicely linear...

First-order integrated rate laws

First-order rate laws are very common, not only in chemistry.

They express a situation where the chance of something happening to each of a set of objects is constant in time and the same for each object, independent of what happens to the other objects.

Exponential decay rate law

Reasonable examples:• People not winning the lottery, if they play consistently for

the same amount of money and the lottery company doesn't "change the rules".

• People never losing at blackjack, provided they and the dealer keep the same strategy (and there are infinitely many cards).

• Decay of radioactive nuclei.• Houses not being hit by a meteorite (assuming all houses

have the same roof area etc).

Exponential decay rate law

Keep in mind:• Molecules are perfect at playing the game of chance.• They never react "all at the same time".• Each molecule has at any time a certain probability to

undergo the reaction.• If the probability is high, most molecules will react

quickly.

Exponential decay and half-life

A simple exponential-decay curve.

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Exponential decay and half-life

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t½

At time t½, the concentration is halfof its original value.

Exponential decay and half-life

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t½

We take the remainder of the curve (past t½)...

Exponential decay and half-life

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t½

Stretch *2

Expand it vertically by a factor of 2...

Exponential decay and half-life

Perfect fit! So every part of the curve has the same "shape", and every time we wait a time of t½ the concentration halves.

For other decays, the half-life is not constant.

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t½

Move back And move it back to t = 0...

Relation between rate constantand half-life

½½

½

½

½

0½

0½

0.6930.6930.6932ln

2ln½ln½

]A[½]A[]A[

tkor

kt

tktk

tke

tket

Using exponential decay(ex 43, p 661)

SO2Cl2 decomposes into SO2 and Cl2 in a first-order reaction:

SO2Cl2 SO2 + Cl2

The half-life at 600K is 1.47*104 s (4.08 hr).• If you begin with 1.6*10-3 mol in a 2-L flask, how

long will it take till only 1.2*10-3 mol is left?• And if we do the same experiment in a 1-L flask?• How much is left after 5 hr?

Decomposition of SO2Cl2

The rate constant isk = 0.693/t½ = 0.0000472 s-1

and we have

Reducing the amount from 1.6*10-3 mol to1.2*10-3 mol reduces the concentration from 0.8*10-3 mol/L to 0.6*10-3 mol/L,so we want [A]t/[A]0 = 0.6/0.8 = 0.75.

tkttkettket 00

0 ]A[

]A[ln

]A[

]A[]A[]A[

Decomposition of SO2Cl2

We now calculate t from:

If we do the reaction in a smaller flask, begin and end concentrations will increase (*2), but their ratio stays the same same answer.

hrsst

tkt

7.110*1.60000472.0/288.0

288.075.0ln]A[

]A[ln

4

0

Decomposition of SO2Cl2

After 5 hr (t = 18000 s):[A]t = [A]0 e-0.000047*18000 = [A]0 *0.429

= 0.000343 mol/LAmount remaining: [A]t * 2 L = 0.69*10-3 mol.

For a second-order rate lawrate = k [A]2

the integrated rate law is

So, plotting 1/[A]t vs t should give a straight line with slope k and intercept 1/[A]0.

The integrated rate lawfor a second-order reaction

tkortk t

t

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]A[1

]A[1]A[]A[ 0.00E+00

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The integrated rate lawfor a zero-order reaction

For a zero-order rate lawrate = k (constant!)

the integrated rate law is[A]t = [A]0 - k t

(only valid as long as k t <= [A]0).

Plot [A]t vs t: slope -k, intercept [A]0.

Zero-order reactions are rare!

Using graphs to analyze rate laws

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[A] vs t ln[A] vs t

1/[A] vs tNot zero-order Not first-order

Looks like second-order