Download 1 Solid State - Unique Solutions

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338
Nature of graph : When log10[A]t is plotted against time(t), we get a straight line
–k
Slope : The slope of the line is equal to
2.303
Equation :
k=
[A] 0
2.303
log10
. The above integrated rate
t
[A] t
law expression can be rearranged to give,
log10
[A] 0
k
=
t + 0. The above equation
[A] t
2.303
is of the form, y = mx + C
[A] 0
Nature of graph : When log 10 [A] is plotted
t
against time(t), we get a straight line
Slope : The slope of the line is equal to
6.
k
2.303
Half life of first order reaction (t 1/ 2 ) :
The half life of a reaction is defined as the time needed for the reactant concentration
to fall to one half of its initial value.
For the first order reaction the integrated rate law is given by the equation
k=
[A] 0
2.303
log10
t
[A] t
where [A]0 is the initial concentration of the reactant at t = 0. Its concentration falls to
[A]t at time t from the start of the reaction. The concentration of the reactant falls to
[A]0/2 at time t 1/2 , the half life period. Hence, at t = t 1/2 , [A]t = [A]0/2. With this condition,
the equation can be written as
k
=
=
=
[A] 0
2.303
log 10
[A] 0
t 1/2
2
2.303
log10 2
t 1/2
2.303
× 0.301 = 0.693
t 1/2
t 1/2
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I
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