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Q.

For a first order reaction the graph log [A] vs t is given below
‘x’ is equal to: ${[A]}_{0}$ = initial conc. and [A] = Conc. after time ‘t’]

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a

$\frac{kt}{2 .303}$

b

$\frac{0 .693}{k}$

c

$\log {[A]}_{0}$

d

$- \frac{k}{2 .303}$

answer is C.

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Detailed Solution

$\log \frac{[A]_{0}}{[A]_{t}} = \frac{kt}{2 .303}$
$\log [A]_{0} - \log [A]_{t} = \frac{kt}{2 .303}$
$\log [A]_{t} = - \frac{kt}{2 .303} + \log [A]_{0}$
In the graph drawn between log[A] and t, slope = -k / 2.303

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