Question

The reaction is a zero order reaction. A drop of a solution (volume $=0.05 \, mL$ ) contains $6\times 10^{- 7}$ mol of $H^{+}.$ If the rate of disappearance of $H^{+}$ is $6.0\times 10^{5}$ mol $L^{- 1}s^{- 1},$ how long will it take for the $H^{+}$ in the drop to disappear?

• A. $8.0\times 10^{- 8}s$
• B. $2.0\times 10^{- 8}s$
• C. $6.0\times 10^{- 6}s$
• D. $2.0\times 10^{- 2}s$

Answer 1

Answer: (B)

$\left[H^{+}\right]=\frac{6 \, \times 10^{- 7} m o l \, }{0.05 \, \times 10^{- 3} L}=1.2\times 10^{- 2}M$
or $r=\frac{\Delta x}{\Delta t}$ or $\Delta T=\frac{\Delta x}{r}=\frac{1.2 \, \times 10^{- 2} M}{6 \, \times 10^{5} M s^{- 1}}$
$\therefore t=2\times 10^{- 8}s.$