Question

Reaction $A+B\to C+D$ follows following rate law.
$\text{rate}=k\left[A{]}^{\frac{1}{2}}\right[B{]}^{\frac{1}{2}}$

Starting with initial concentration of one mole of $A$ and $B$ each, what is the time taken for amount of $A$ to become $0.25$ mole.
[Given: $k=2.31\times {10}^{–3}{\text{sec}}^{–1},ln2=0.693$]

• A. $300\text{sec}$
• B. $600\text{sec}$
• C. $900\text{sec}$
• D. None of these

Answer 1

The correct option is B $600\text{sec}$

$\begin{array}{cccccccc}& A& +& B& \to & C& +& D\\ t=0& 1& & 1& & -& & -\\ t=t& 1-x& & 1-x& & 1& & 1\end{array}$

$\text{rate}=k\left[A{]}^{\frac{1}{2}}\right[B{]}^{\frac{1}{2}}$
$-\frac{d\left[A\right]}{dt}=k(1-x{)}^{\frac{1}{2}}(1-x{)}^{\frac{1}{2}}\Rightarrow -\frac{d(1-x)}{dt}=k(1-x)\Rightarrow \frac{d(1-x)}{dt}=k(1-x)$

$\Rightarrow {\int }_0^x\frac{d(1-x)}{1-x}=k{\int }_0^{t}dt\Rightarrow [-\ln (1-x){]}_0^x=kt\Rightarrow -\ln (1-x)=kt\Rightarrow t=\frac{1}{k}\ln \frac{1}{1-x}$
When $\left[A\right]=0.25\therefore x=0.75$

$\therefore t=\frac{1}{2.31\times {10}^{-3}}\ln \frac{1}{0.25}=\frac{{10}^3}{2.31}\ln 4=\frac{2\times {10}^3\times 0.693}{2.31}=600\text{sec}$