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Question: A drop of solution (volume 0.05 mL) contains \[3 \times {10^{ - 6}}\] mole of \[{H^ + }\]. If the ra...

A drop of solution (volume 0.05 mL) contains 3×10−63 \times {10^{ - 6}} mole of H+{H^ + }. If the rate constant of disappearance of H+{H^ + } is 107mol litre−1sec−1{10^7}mol{\text{ }}litr{e^{ - 1}}se{c^{ - 1}} , how long would it take for H+{H^ + } in the drop to disappear?

Explanation

Solution

We can approach this question by the very simple formula of the rate equation. We know that the rate of disappearance of the reactant is equal to the change in concentration at a given time divided by the time. So we need to first find the concentration of the drop and thus substitute in the formula and get the time.

Complete answer: We are provided with the rate at which the concentration of H+{H^ + } in terms of liters and thus we need to find out the concentration of the given drop in moles per liter.
We know that concentration of any substances can be found out by the given equation:
⇒Concentration of drop  =molevolume in mL×1000\Rightarrow Concentration{\text{ }}of{\text{ }}drop\; = \dfrac{{mole}}{{volume{\text{ }}in{\text{ }}mL}} \times 1000
Thus we can substitute the values given to us in this equation.
We are given 3×10−63 \times {10^{ - 6}} moles of H+{H^ + } and 0.05 mL of solution.
⇒Concentration of drop  =3×10−60.05 mL×1000\Rightarrow Concentration{\text{ }}of{\text{ }}drop\; = \dfrac{{3 \times {{10}^{ - 6}}}}{{0.05{\text{ }}mL}} \times 1000
⇒Concentration of drop  =0.06mol litre−1\Rightarrow Concentration{\text{ }}of{\text{ }}drop\; = 0.06mol{\text{ litr}}{{\text{e}}^{ - 1}}
Now we can substitute this value in the rate equation which is given by:
⇒Rate of disappearance  =conc. changetime\Rightarrow Rate{\text{ }}of{\text{ }}disappearance\; = \dfrac{{conc.{\text{ }}change}}{{time}}
The rate of disappearance is given as 107mol litre−1sec−1{10^7}mol{\text{ }}litr{e^{ - 1}}se{c^{ - 1}}
⇒1×107=0.06time\Rightarrow 1 \times {10^7} = \dfrac{{0.06}}{{time}}
⇒Time=0.061×107\Rightarrow Time = \dfrac{{0.06}}{{1 \times {{10}^7}}}
⇒Time=6×10−9sec⁡\Rightarrow Time = 6 \times {10^{ - 9}}\sec
Thus we can say that the correct answer for the given question is 6×10−96 \times {10^{ - 9}} seconds.

Note:
Another method to solve this question is by using the unit of rate constant to predict the reaction.
mol litre−1sec−1mol{\text{ }}litr{e^{ - 1}}se{c^{ - 1}} is the unit for zero-order reaction rate constant and thus we can say that the time can be found out by using the equation for zero-order reaction.
⇒Time=Conc. usedRate constan⁡t\Rightarrow Time = \dfrac{{Conc.{\text{ }}used}}{{Rate{\text{ }}cons\tan t}}
⇒Time=0.061×107\Rightarrow Time = \dfrac{{0.06}}{{1 \times {{10}^7}}}
⇒Time=6×10−9sec⁡\Rightarrow Time = 6 \times {10^{ - 9}}\sec
Thus we can find out the answer.