$\begin{array}{1 1}(a)\;k=\large\frac{2}{t}\normalsize(A_o^{-1/2}-A^{-1/2})\\(b)\;\;K=\large\frac{2}{t}\normalsize(A_o^{1/2}-A^{1/2})\\(c)\;t_{1/2}=\large\frac{k}{2[A_o]^{1/2}}\\(d)\;t_{1/2}=\large\frac{ln\;2}{k}\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer: $ t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg)$

By differential rate law, $\large\frac{d[A]}{dt}=$$-k[A]^{3/2}$

Integrating, $\large \int_{[A_0]}^{[A_t]}$$\large\frac{d[A]}{[A]^3/2}$$=-\large\int_0^t $$kdt$

$\Rightarrow 2([A_o]^{-1/2}-[A_t]^{1/2})=kt$

$k=\large\frac{2}{t}$$((A_o)^{-1/2}-(A_t)^{-1/2})$

Simplifying for $t = t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg)$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...