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 the rate equation

Given that the reaction is first order in both reactants H₂S and Cl₂, we can write the rate equation as: $$ rate = k[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}] $$ where \(k\) is the rate constant, [\(\mathrm{H}_{2}\mathrm{S}\)] and [\(\mathrm{Cl}_{2}\)] represent the concentrations of H₂S and Cl₂, respectively.

02

Calculate the rate of consumption of H₂S

We are given the rate constant \(k = 3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{\; s}^{-1}\), the concentration of H₂S, [\(\mathrm{H}_{2}\mathrm{S}\)] \(= 2.0 \times 10^{-4} \mathrm{M}\), and the concentration of Cl₂, [\(\mathrm{Cl}_{2}\)] \(= 0.025 \mathrm{M}\). We can now calculate the rate of disappearance (or consumption) of H₂S using the rate equation: $$ rate = (3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{\; s}^{-1})\left(2.0 \times 10^{-4} \mathrm{M}\right)\left(0.025 \mathrm{M}\right) $$ Calculate the rate: $$ rate = 1.75 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1} $$ So, the rate of consumption of H₂S is \(1.75 \times 10^{-6}\) M/s.

03

Find the rate of formation of Cl⁻

Looking at the balanced reaction equation, we can see that the stoichiometric coefficient of H₂S is 1 and the stoichiometric coefficient of Cl⁻ is 2. This means that for each molecule of H₂S consumed, 2 molecules of Cl⁻ will be formed. Therefore, we can use stoichiometry to find the rate of formation of Cl⁻. $$ \text{Rate of formation of Cl}^{-} = 2 \times \text{Rate of consumption of H}_{2}\text{S} $$ Substitute the rate of consumption of H₂S from step 2: $$ \text{Rate of formation of Cl}^{-} = 2 \times (1.75 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1}) $$ Calculate the rate of formation of Cl⁻: $$ \text{Rate of formation of Cl}^{-} = 3.5 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1} $$ Therefore, the rate of formation of Cl⁻ is \(3.5 \times 10^{-6}\) M/s.

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