Question

For a certain reaction of order $n$, the time for half change, $t_{\frac{1}{2}}$, is given by $t_{\frac{1}{2}}=\frac{(2-\sqrt{2})}{k}\times C_o^{\frac{1}{2}}$ where $k$ is constant and $C_0$ is the initial concentration. What is the value of $n$?

• A. $1$
• B. $2$
• C. $0$
• D. $0.5$

Answer 1

The correct option is D $0.5$

Given,
$t_{\frac{1}{2}}=\left(\frac{2-\sqrt{2}}{k}\right)C_o^{\frac{1}{2}}$
We know,
$t_{\frac{1}{2}}\propto (C_0{)}^{1-\text{order}}$
For $n^{\text{th}}$ order reaction,
$t_{\frac{1}{2}}\propto (C_0{)}^{1-\text{n}}$
$\Rightarrow 1-n=\frac{1}{2}$
So, $n=0.5$.