Question

Match List I with List II and select the correct answer using the code given below the lists:

	List I		List II
(a)	Zero order	(p)	$k=\frac{1}{2t}[\frac{1}{(a_0-x{)}^2}-\frac{1}{a_0^2}]$
(b)	1st order	(q)	$k=\frac{1}{t}[\frac{1}{(a_0-x)}-\frac{1}{a_0}]$
(c)	Second order	(r)	$k=\frac{x}{t}$
(d)	Third order	(s)	$k=\frac{1}{t}ln\left(\frac{a_0}{(a_0-x)}\right)$

• A. (a) - r, (b) - s, (c) - q, (d) - p
• B. (a) - s, (b) - p, (c) - q, (d) - r
• C. (a) - r, (b) - s, (c) - p, (d) - q
• D. (a) - p, (b) - s, (c) - q, (d) - r

Answer 1

The correct option is A (a) - r, (b) - s, (c) - q, (d) - p

We know that for zero order reaction,

$K=\frac{x}{t}$

where, $K$=rate constant, $t$=time,

$x$=amount of reactant reacted

For first order reaction,

$K=\frac{1}{t}ln\left(\frac{a_0}{a_0-x}\right)$

where, $a_0$= Initial concentration

For second order reaction we have

$K=\frac{1}{t}[\frac{1}{(a_0-x)}-\frac{1}{a_0}]$

For third order reaction, we have:-

$\frac{dx}{dt}=K(a-x{)}^3$

$\Rightarrow \frac{dx}{(a-x{)}^3}=Kdt$

On integrating both sides,

$\Rightarrow \int \frac{dx}{(a-x{)}^3}=\int Kdt$

$\frac{1}{2(a-x{)}^2}=Kt+C$

Now, at $t=0,x=0$

$\Rightarrow C=\frac{1}{2a^2}$

$\Rightarrow Kt=\frac{1}{2}[\frac{1}{(a-x{)}^2}-\frac{1}{a^2}]$