Question

A substance '$A$' undergoes conversion by elementary step to '$B$' and '$C$' in aqueous phase as shown:
$A_{\left(aq\right)}+H_{2}O\to B_{\left(aq\right)}+C_{\left(aq\right)}$
If concentration of $A_{\left(aq\right)}$ initially and after $6.93$ minutes is $1\text{M}$ & $\frac{1}{4}\text{M}$ respectively then calculate rate constant in terms of $M^{-1}{\text{min}}^{-1}$.
(Given: $\log 2=0.3010$)

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $3.6\times {10}^{-3}$
• D. $1.8\times {10}^{-3}$

Answer 1

The correct option is C $3.6\times {10}^{-3}$

$A_{\left(aq\right)}+H_{2}O\to B_{\left(aq\right)}+C_{\left(aq\right)}$
Rate is $k\left[A\right]\left[H_{2}O\right]$
Reaction is pseudo first oreder reaction with respect to $A$.
$\therefore$ Rate $=k^{\prime }\left[A\right]$
Where $k^{\prime }=k\left[H_{2}O\right]$
$k^{\prime }=\frac{2.303}{6.93}\log \frac{1}{1/4}=\frac{1}{5}{\text{min}}^{-1}$
Also, $\left[H_{2}O\right]=55.5\text{M}$
$\therefore k=\frac{k^{\prime }}{\left[H_{2}O\right]}=\frac{1}{5\times 55.5}=3.6\times {10}^{-3}{\text{M}}^{-1}{\text{min}}^{-1}$