Hydrogen peroxide `H_(2)O_(2)` (aq) decomposes to `H_(2)O`(l) and `O_(2)`(g) in a reaction that is first order in `H_(2)O_(2)` and has a rate constant `k=1.06xx10^(3) min^(-1)`
(i) How long will it takes for 15% of a sample of `H_(2)O_(2)` to decompose?
(ii) How long will it take for 875% of the sample to decompose?

Explanation

The decomposition of `H_(2)O_(2)` in aqueous solution (catalysed by finely divided pt) takes place as follows:
`H_(2)O_(2)(l) to H_(2)O(l) + 1/2O_(2)(g)`
The kinetics of the reaction is based on the fact that `H_(2)O_(2)` solution can be oxidised by `KMnO_(4)` and thus, can be titrated against this reagent. The volume of `KMnO_(4)` used corresponds to the volume of `H_(2)O_(2)` present.
Volume of `KMnO_(4)` used at time `t(V_(t) propto "Initial concentration of " H_(2)O_(2))` i.e. (a)
Substituting the values in the reaction,
`k=(2.303)/t log(a/(a-x))=(2.303/t)logV_(0)/V_(t)`
a) According to first order reaction,
`t = 2.303/k log (a)/((a-x))`
`t=(2.303)/(1.06 xx 10^(-3)min^(-1)) log 100/80 = (2.303 xx 0.0706)/(1.06 xx 10^(-3)min^(-1))=154.4 min`
b) `t=(2.303)/k loga/(a-x)=2.303/(1.06 xx 10^(-3)min^(-1)) log 100/15 = (2.303)/(1.06 xx 10^(-3)min^(-1))log(100/15) = (2.303 xx 0.8239)/(1.06 xx 10^(-3) min^(-1))` = 1790.0 min