Question

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-} ?\)

Short answer

The rate of formation of Cl⁻ ions is \(3.5 \times 10^{-6} \mathrm{M} \mathrm{s}^{-1}\).

Step-by-step solution

Write the rate law for the reaction

Since the rate of the reaction is first order with respect to both reactants, the rate law for the reaction can be written as: $$ \text{Rate} = k[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}] $$ where, 'Rate' is the rate of the reaction, 'k' is the rate constant, and the concentrations of the reactants are represented by [H2S] and [Cl2].

Substitute the given rate constant and concentrations

We are given the rate constant for the disappearance of H2S (\(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1}\)), and concentrations of H2S (\(2.0 \times 10^{-4} \mathrm{M}\)) and Cl2 (\(0.025 \mathrm{M}\)). Substitute these values into the rate law: $$ \text{Rate} = (3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1})(2.0 \times 10^{-4} \mathrm{M})(0.025 \mathrm{M}) $$ Calculate the rate of the reaction: $$ \text{Rate} = 1.75 \times 10^{-6} \mathrm{M} \mathrm{s}^{-1} $$

Use stoichiometry to determine the rate of formation of Cl-

According to the balanced chemical equation provided, for every mole of H2S that reacts, two moles of Cl- are formed: $$ \mathrm{H}_{2} \mathrm{S}(aq)+\mathrm{Cl}_{2}(aq) \longrightarrow S(s)+2H^{+}(aq)+2Cl^{-}(aq) $$ Thus, the rate of formation of Cl- is twice the rate of disappearance of H2S. To find the rate of Cl- formation, we multiply the rate of the reaction by 2: $$ \text{Rate of Cl- formation} = 2 \times (1.75 \times 10^{-6} \mathrm{M} \mathrm{s}^{-1}) = 3.5 \times 10^{-6} \mathrm{M} \mathrm{s}^{-1} $$ The rate of formation of Cl- ions is found to be \(3.5 \times 10^{-6} \mathrm{M} \mathrm{s}^{-1}\).