Question

Following is the graph between $log \, t_{1 / 2}$ and $loga$ ( $\text{a} =$ Concentration of reactant) for a given reaction at $27^\circ C$ . Hence, order of the reaction is



(Here, $t_{1 / 2}$ is half-life)

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$t _{\frac{1}{2}} \propto\left(\frac{1}{ a }\right)^{ n -1}$ for nth order reaction.
$ \begin{array}{l} t _{\frac{1}{2}}= k ( a )^{1- n } \\ \log t _{\frac{1}{2}}=(1- n ) \log a +\log k \end{array} $
Thus, graph between $\log t _{\frac{1}{2}}$ and $\log$ a is linear slope.
$ (1-n)=\tan 45^{\circ} $
$ n =0 $