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 \(\mathrm{Cl}^{-} ?\)

Step by step solution

01

Write down the rate law for the reaction

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

02

Calculate the reaction rate using given values

We are given that the rate constant \(k = 3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1}\), the concentration of H₂S at that time \([\mathrm{H}_{2}\mathrm{S}] = 2.0 \times 10^{-4} \mathrm{M}\), and the concentration of Cl₂ at that time \([\mathrm{Cl}_{2}] = 0.025 \mathrm{M}\). Plug these values into the rate law: \(Rate = (3.5\times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1})(2.0 \times 10^{-4} \mathrm{M})(0.025 \mathrm{M})\) Now calculate the rate: \(Rate = 1.75 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\)

03

Determine the stoichiometric relationship between the reactants and products

From the balanced reaction, we can see that for every 1 mole of H₂S that reacts, 2 moles of Cl⁻ are produced: \(\mathrm{H}_{2}\mathrm{S}(aq) + \mathrm{Cl}_{2}(aq) \longrightarrow \mathrm{S}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{Cl}^{-}(aq)\) So, the rate of production of Cl⁻ is twice the rate of disappearance of H₂S.

04

Calculate the rate of formation of Cl⁻

Using the stoichiometric relationship and the calculated rate of reaction, we can find the rate of formation of Cl⁻: \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 2 \times Rate_{disappearance}\, of\, \mathrm{H}_{2}\mathrm{S}\) \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 2 \times 1.75 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\) Now calculate the rate: \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 3.5 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\) So, the rate of formation of Cl⁻ is \(3.5 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\).

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