Question

Following is the graph between $log \, \text{t}_{\frac{1}{2}}$ and $log \text{a}$ ( $\text{a} = \, $ concentration of reactant) for a given reaction at $27^{o}C$ . Hence, order of the reaction is



(Here, $\text{t}_{1 / 2}$ is half-life)

• A. $\text{0}$
• B. $\text{1}$
• C. $\text{2}$
• D. $\text{3}$

Answer 1

Answer: (A)

$t _{\frac{1}{2}} \propto\left(\frac{1}{ a }\right)^{ n -1}$ for nth order reaction.

$t _{\frac{1}{2}}= k ( a )^{1- n }$

$\log \mathrm{t}_{\frac{1}{2}}=(1-\mathrm{n}) \log \mathrm{a}+\log \mathrm{k}$

Thus, graph between $log \text{t}_{\frac{1}{2}}$ and $log \text{a}$ is linear slope.

$(1-n)=\tan 45^{\circ}$

$n=0$