Question

The reaction $\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$ follows the following rate law rate$=\mathrm{K}{\left[\mathrm{A}\right]}^{\frac{1}{2}}{\left[\mathrm{B}\right]}^{\frac{1}{2}}$. Starting with the initial concentration of $1\mathrm{M}$ A and B each. What is the time taken for concentration A to become $0.25\mathrm{M}$.

Answer 1

Step 1: Analyzing given data

Rate law$=\mathrm{K}{\left[\mathrm{A}\right]}^{\frac{1}{2}}{\left[\mathrm{B}\right]}^{\frac{1}{2}}$

The initial concentration of A$=1\mathrm{M}$

The concentration of A after some time$=0.25\mathrm{M}$

Step 2: Calculating Time for concentration A to become $0.25\mathrm{M}$

Order of reaction from rate law is given by the sum of the power of the reactant

Therefore, Order$=\frac{1}{2}+\frac{1}{2}\Rightarrow 1$

It will require two half times for the concentration of A to become $0.25$

Therefore, ${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\mathrm{K}}$ $\mathrm{K}=2.303\times {10}^{-3}$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{2.303\times {10}^{-3}}\times 2\Rightarrow 600\sec$

Therefore, the time required for the concentration of A to become $0.25\mathrm{M}=600\sec$