Ncert Exercises & Solutions

Match the graph given in Column I with the order of reaction given in Column II.
More than one item in Column I may be linked to the same item in Column II:

Column I	Column II
(i)	(a)	1st order
(ii)	(b)	Zero order
(iii)
(iv)

	(i)	(ii)	(iii)	(iv)
1.	(a)	(b)	(a)	(b)
2.	(a)	(b)	(b)	(a)
3.	(a)	(a)	(b)	(b)
4.	(b)	(b)	(a)	(a)

HINT:

k = \frac{[R_{0}] - [R]}{t}

for zero order reaction.

Explanation:

For zero-order reaction rate equation may be written as

A = -kt + [R]0 ...(i)

Which denotes a straight line equation similar to y = mx + C

$On transforming (i)$
$\begin{array}{l} \frac{[R] - [R_{0}]}{l} = - k \\ k = \frac{[R_{0}] - [R]}{t} \end{array}$

For a first-order reaction $\frac{dx}{dt} \propto$ [concentration]

Therefore the graph between rate and concentration may be drawn as

$\begin{array}{l} k = \frac{2 .303}{t} \log \frac{[R]_{0}}{[R]} \\ \frac{kt}{2 .303} = \log [\frac{R^{\circ}}{R}] \\ \frac{kt}{2 .303} = \log [R]_{0} - \log [R] \\ \log [R] = (\frac{- k}{2 .303}) t + \log [A]_{0} \\ slope intercept \end{array}$

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