Question

Following is the graph between $\log t_{1/2}$ and log a (a initial concentration) for a given reaction at $27^{\circ }C$

Find the order of reaction.

Answer 1

Answer: 0

$t_{1/2}\propto \frac{1}{a_0^{n-1}}$

$t_{1/2}=k\cdot \frac{1}{a_0^{n-1}}$

$\log t_{1/2}=\log k-(n-1)\log a_0$

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

$\Rightarrow n=0$