Question

The stoichiometric equation for the oxidation of bromide ions by hydrogen peroxide in acid solution is
$2Br\left(aq\right)+H_2{O}_2\left(aq\right)+{2H}^{+}\left(aq\right)⟶{Br}_2\left(\ell \right)+2H_{2}O\left(\ell \right)$.
Since the reaction does not occur in one stage the rate equation does not correspond to this stochiometric equation but is rate = $k\left[H_2{O}_2\right]\left[H^{+}\right]\left[{Br}^{-}\right]$

(a) If the concentration of $H_2{O}_2$ is increased by a a factor of 3, by what factor is the rate of consumption of $\left[{Br}^{-}\right]$ ions increased.

(b) If, under certain conditions, the rate of consumption of $\left[{Br}^{-}\right]$ ions is $7.2\times {10}^{-3}mole{dm}^{-3}{s}^{-1}$, what is the rate of consumption of hydrogen peroxide. What is the rate od production of bromine.

(c) What is the effect on the rate constant k of increasing the concentration of bromide ions.

(d) If by the addition of water to the reaction mixture the total volume were doubled, what would be the effect on the rate of change of the concentration of ${Br}^{-}$. What would be the effect on the rate constant k.

Answer 1

$2Br+H_2{O}_2+{2H}^{+}\to {Br}_2+{2H}_{2}Orate=k\left[H_2{O}_2\right]\left[H^{+}\right]\left[{Br}^{-}\right]$

(a) When $H_2{O}_2$ concentration is increased by a factor of 3, rate of consumption of ${Br}^{-}$ is increased by a factor of 3.

This is because they're both following first order.

(b) $-\frac{1}{2}\frac{d\left[Br\right]}{dt}=-\frac{d\left[H_2{O}_2\right]}{dt}=\frac{7.2}{2}\times {10}^{-3}=3.6\times {10}^{-3}=\frac{d\left[{Br}_2\right]}{dt}=\frac{7.2}{2}\times {10}^{-3}=3.6\times {10}^{-3}$

(c) The rate constant is proportional to the concentration of [Br] ions. It will increase proportionally when conc. of [Br] ions increases.

(d) On dilution the concentration of [Br] ions decreases which will result in a proportional decrease in the rate constant.