Question

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

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

Answer 1

The correct option is B $6\times {10}^{-9}\text{s}$

The concentration of drop is given by
$\text{Conc.}=\frac{\text{mole}\times 1000}{\text{volume (in mL)}}=\frac{3\times {10}^{-6}\times 1000}{0.05}=0.06{\text{mol L}}^{-1}$

Given,
rate constant, $k=1.0\times {10}^7{\text{mol L s}}^{-1}$
Thus it is a zero order reaction.
For a zero order reaction, $t=\frac{x}{k}=\frac{0.06{\text{mol L}}^{-1}}{1.0\times {10}^7{\text{mol L s}}^{-1}}=6\times {10}^{-9}\text{s}$