Question

A drop of solution (volume 0.05 mL) contains $3.0\times 10^{-6}$ moles of $H^{+.}$ If the rate constant of disappearance of $H^{+}$ is $1.0 \times 10^7 mole \,litre^{-1} sec^{-1.}$ How long would it take for $H^{+}$ in drop to disappear:

• A. $6 \times 10^{-8}$ sec
• B. $6 \times 10^{-7}$ sec
• C. $6 \times 10^{-9}$ sec
• D. $6 \times 10^{-10}$sec

Answer 1

Answer: (C)

time Total for drop to disappears$\left(a_{0}-a_{t}\right)=k_{t} a_{t}=0$
$\frac{3.0\times10^{-6}}{\left(0.05\times10^{-3}\right)\times1.0\times10^{7}}=t_{100\%}$
$\Rightarrow t_{100\%}=6\times10^{-9}$ sec
Alternative method Units rate constant $=molL^{-1}sec^{-1}\Rightarrow $ Zero order reaction
$\left[A_{0}\right]-\left[A\right]=kT \left[A_{0}\right]=\frac{3\times10^{-6}\times100}{0.05\times10^{-3}}=\frac{3}{50}$
$\frac{3}{50}-0=kT$
$\Rightarrow t=\frac{3}{50}\times\frac{10^{-7}}{1}$
$=6\times10^{-9}$ sec