Question

Following is the graph between $\log T_{50}$ and $\log a(a=$ initial concentration) for a given reaction at $27^{\circ }C$. Hence order is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

$t_{1/2}\propto {\left(\frac{1}{a}\right)}^{n-1}$ or $t_{1/2}=k(a{)}^{1-n}$
$\log t_{1/2}=\log k+(1-n)\log a(y=c+mx)$
Slope $=(1-n)=\tan 45=1$
$\therefore (1-n)=1\Rightarrow n=0$