A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear?

Explanation

Concentration of drop `= ("Mole")/("Volume in mL") xx 1000`
`= (3 xx 10^(-6))/(0.05) xx 1000 = 0.06 mol L^(-1)`
Rate of disappearance `= ("Conc change")/("Time")`
`1 xx 10^(7) = (0.06)/("Time")`
Time `= 6 xx 10^(9) s`
Second methof
Units of `k (mol L^(-1) s^(-1))` suggest it is a zero order reaction.
`:.` For zero order `= t = (x)/(k) = ("Conc used")/("Rate constant")`
`0.05 mL` has `= 3 xx 10^(-6) mol` of `H^(o+)`
`1000 mL` has `= (3 xx 10^(-6))/(0.05) xx 10^(3)`
`t = (0.6 xx 10^(-1))/(1.0 xx 10^(-7)) = 6 xx 10^(-9) s`