Question

Reaction $\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$ follows the following rate law: $\mathrm{rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{1/2}{\left[\mathrm{B}\right]}^{1/2}$​. Starting with initial conc. of $1\mathrm{M}$ of A and B each, what is the time taken for the concentration of A to become $0.25\mathrm{M}$? $(\mathrm{Given}:\mathrm{k}=2.303\times {10}^{-3}\mathrm{s})$

Answer 1

Step 1: Analyzing Given data

$$\begin{gathered} \mathrm{Rate}\mathrm{law},\mathrm{rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{1/2}{\left[\mathrm{B}\right]}^{1/2}; \\ {\left[\mathrm{A}\right]}_{\mathrm{o}}=\mathrm{initial}\mathrm{concentration}=1\mathrm{M}; \\ {\left[\mathrm{B}\right]}_{\mathrm{o}}=\mathrm{initial}\mathrm{concentration}=1\mathrm{M}; \\ {\left[\mathrm{A}\right]}_{\mathrm{t}}=\mathrm{concentration}\mathrm{at}\mathrm{time}\mathrm{t}=0.25\mathrm{M} \\ \mathrm{rate}\mathrm{constant},\mathrm{k}=2.303\times {10}^{-3}\mathrm{s} \end{gathered}$$

Step2: Understanding the rate law expression

The overall order of the reaction is first order, but the reaction is 0.5 order each with respect to A and B.

$\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$

The rate law expression for the above-mentioned reaction can be written as:

$\mathrm{rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{1/2}{\left[\mathrm{B}\right]}^{1/2}$

Step 3: Calculating half-life

It will take two half-life periods for the reactant A to reduce from 1M down to 0.25M.

The half-life of a first-order reaction is given by:

$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.303\times {10}^{-3}}=0.3\times {10}^{3}s=300s$

Step 4: Calculating the time taken for the reactant A to reduce to 0.25M of its initial value:

Since it will take two half-life periods for the reactant A to reduce from 1M down to 0.25M.

$\mathrm{T}=2\times {\mathrm{t}}_{1/2}=2\times 300\mathrm{s}=600\mathrm{s}$

Therefore, the time taken for the reactant A to reduce to 0.25M of its initial value is 600s