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.

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

$n=0$