Ncert Exercises & Solutions

The correct graphical representation of first-order reaction is:

(a)		(b)
(c)		(d)

1.	(a) and (b)	2.	(b) and (c)
3.	(c) and (d)	4.	(a) and (d)

HINT: $k = \frac{2.303}{t} log \frac{[R_{o}]}{[R]}$ $k = \frac{2.303}{t} \log \frac{[R_{o}]}{[R]}$

Explanation:

Step 1:

For the first-order reaction

$k = \frac{2.303}{t} log \frac{[R_{o}]}{[R]}$ $k = \frac{2.303}{t} \log \frac{[R_{o}]}{[R]}$
$log \frac{[R_{o}]}{[R]} = \frac{k}{2.303} t$ $\log \frac{[R_{o}]}{[R]} = \frac{k}{2.303} t$ $. . . . . . (1)$ $. . . . . . (1)$

The first equation is the same as y=mx+c. Thus, straight-line graph is obtained. The slope is $\frac{k}{2.303}$ $\frac{k}{2.303}$ .

The correct plot between log $\frac{{[R]}_{0}}{[R}}$ $\frac{{[R]}_{0}}{[R}}$ vs t can be represented by (d)

Step 2:

The time taken for any fraction of the reaction to complete is independent of the initial concentration Let, us consider it for half of the reaction to complete.

$t$ $t$ $=$ $=$ $\frac{2.303}{k} log$ $\frac{2.303}{k} \log$ $\frac{a}{a - x}$ $\frac{a}{a - x}$
$For$ $For$ $half$ $half$ $life$ $life$
$t$ $t$ $=$ $=$ $t_{1 / 2};$ $t_{1 / 2};$ $x = \frac{a}{2}$ $x = \frac{a}{2}$
$t_{1 / 2} = \frac{2 .303}{k} log$ $t_{1 / 2} = \frac{2 .303}{k} \log$ $\frac{a}{a - \frac{a}{2}}$ $\frac{a}{a - \frac{a}{2}}$
$t_{1 / 2} = \frac{2 .303}{k} log$ $t_{1 / 2} = \frac{2 .303}{k} \log$ $2$ $2$
$t_{1 / 2} = \frac{0 .693}{k}$ $t_{1 / 2} = \frac{0 .693}{k}$

t_1/2 is independent of initial concentration. Hence, the correct plot of t_1/2, and [R_o] can be represented by (a)

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