Question

Section (C) : Integrated Rate law : Second order & Pseudo first order reaction
The rate constant for a certain second order reaction is $8$ $\times$ ${10}^{-5}{M}^{-1}mi{n}^{-1}$ How long will it take at 1 M sollution to be reduced to $0.5M$ in reactant ? How long will it take from that point until the solution $0.25M$ in reactant.

Answer 1

Rate law equation for second order is

$\frac{1}{\left[A\right]}=\frac{1}{\left[A_o\right]}+kt$

Given $\left[A_o\right]=1M$

$\Rightarrow \left[A\right]=0.5M$

$\Rightarrow K=8\times {10}^{-5}$

$\Rightarrow \frac{1}{0.5}=\frac{1}{1}+8\times {10}^{-5}\times t$

$\Rightarrow 1M^{-1}=8\times {10}^{-5}t{M}^{-1}mi{n}^{-1}$

$\Rightarrow t=\frac{1}{8\times {10}^{-5}}min=12500$ minute

Thus to reduce the concentration from $1M$ to $0.5M$ it will take $12500$ minute.

$\Rightarrow$ Now, to reduce from $0.5M$ to $0.25M$

$\Rightarrow \frac{1}{\left[0.25\right]M}=\frac{1}{\left[0.5\right]M}+8\times {10}^{-5}{M}^{-1}mi{n}^{-1}\times t$

$\Rightarrow 2M^{-1}=8\times {10}^{-5}{M}^{-1}mi{n}^{-1}\times t$

$\Rightarrow t=\frac{2}{8\times {10}^{-5}}min=25000min$

Thus $25000$ minute will be taken to reduce concentration from $0.5M$ to $0.25M$.