A drop of solution (Volume 0.05 mL) contains 3.0 × 10–6 mole of H+ ions. If the rate constant of disappearance of H+ is 1.0 × 107 mol L–1 s–1, how long will it take for H+ ions to disappear?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(6 × 10^{-9} s\)
The unit of the rate constant, mol L−1 s−1, shows that the disappearance of \( \ce{H^{+}} \) is a zero order reaction. For a zero order reaction the rate is constant (rate = k), so the time for complete disappearance is
\( t = \displaystyle\frac{[R]_0}{k} \)
The initial concentration of \( \ce{H^{+}} \) in the drop:
\( V = 0.05\ \text{mL} = 0.05 \times 10^{-3}\ \text{L} = 5 \times 10^{-5}\ \text{L} \)
\( [H^+]_0 = \displaystyle\frac{3.0 \times 10^{-6}\ \text{mol}}{5 \times 10^{-5}\ \text{L}} = 6.0 \times 10^{-2}\ \text{mol L}^{-1} \)
Time to disappear:
\( t = \displaystyle\frac{6.0 \times 10^{-2}\ \text{mol L}^{-1}}{1.0 \times 10^{7}\ \text{mol L}^{-1}\ \text{s}^{-1}} = 6 \times 10^{-9}\ \text{s} \)
The \( \ce{H^{+}} \) ions disappear in 6 × 10−9 s.