The Chemistry Maths Book, Second Edition

(Grace) #1

112 Chapter 4Differentiation


and, multiplying by y,


(ii)y 1 = 1 a


x

,ln 1 y 1 = 1 x 1 ln 1 a.


Then


(iii)y 1 = 1 x


x

,ln 1 y 1 = 1 x 1 ln 1 x.


Applying the product rule,


and, therefore,


0 Exercises 69–72


EXAMPLE 4.21Logarithmic plots


For a system undergoing exponential decay (first-order kinetics), the size of the


system is given by (see Example 3.20)


x(t) 1 = 1 x


0

e


−kt

and, taking the logarithm of both sides,


ln 1 x 1 = 1 ln 1 x


0

1 − 1 kt


A plot of ln xagainst t, Figure 4.8, gives a straight line with


sloped(ln 1 x) 2 dt 1 = 1 −k, and intercept ln 1 x


0

with the axist 1 = 10.


This example is important because it demonstrates the


standard way of calculating the rate constant kfrom a linear


plot of experimental values of xand t.


0 Exercise 73


1


11


y


dy


dx


x


dy


dx


xx


x

=+ln and =( ln )+


d


dx


yx


d


dx


xx


d


dx


ln =+=+ln ln xx 1 ln


d


dx


y


y


dy


dx


a


dy


dx


aa


x

ln ==ln , ,=ln


1


and therefore


dy


dx


x


x


x


xx


=














=


+−


1


1


1


1


1


11


2

12

12 32

()()


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o


lnx


0





−k


t


lnx


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Figure 4.8

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