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 $30{ }^{\circ} \mathrm{C}\( is \)4.0 \times 10^{-2} M^{-1} \mathrm{~s}^{-1}$. If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is $2.5 \times 10^{-4} \mathrm{M}\( and that of \)\mathrm{Cl}_{2}\( is \)2.0 \times 10^{-2} \mathrm{M},$ what is the rate of formation of \(\mathrm{H}^{+}\) ?

Step by step solution

01

- Write down the balanced chemical equation

The balanced chemical equation for the reaction is given by: \( \mathrm{H}_{2} \mathrm{S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \)

02

- Write down the rate constant \(k\) and reactants' concentrations

We are given the following information: Rate constant (\(k\)): \(4.0 \times 10^{-2} M^{-1} s^{-1}\) Concentration of \(\mathrm{H}_{2} \mathrm{S}\): \(2.5 \times 10^{-4} M\) Concentration of \(\mathrm{Cl}_{2}\): \(2.0 \times 10^{-2} M\)

03

- Calculate the reaction rate

Since the reaction is first order with respect to H₂S and Cl₂, the reaction rate is given by the equation: \( \text{rate} = k \times [\mathrm{H}_{2} \mathrm{S}] \times [\mathrm{Cl}_{2}] \) Now plug in the given values: \( \text{rate} = (4.0 \times 10^{-2} M^{-1} s^{-1}) \times (2.5 \times 10^{-4} M) \times (2.0 \times 10^{-2} M) \) Calculate the result: \( \text{rate} = 2.0 \times 10^{-8} M s^{-1} \)

04

- Determine the rate of formation of H⁺

Using the stoichiometry of the balanced chemical equation, we can note that as 1 mole of H₂S is consumed, 2 moles of H⁺ are formed. Hence, the rate of formation of H⁺ is twice the rate of disappearance of H₂S. So, the rate of formation of H⁺ is: \( \text{rate}_{\mathrm{formation \ of} \ H^{+}} = 2 \times \text{rate} \) Plug in the previously calculated rate: \( \text{rate}_{\mathrm{formation \ of} \ H^{+}} = 2 \times (2.0 \times 10^{-8} M s^{-1}) \) Calculate the result: \( \text{rate}_{\mathrm{formation \ of} \ H^{+}} = 4.0 \times 10^{-8} M s^{-1} \) The rate of formation of H⁺ is \(4.0 \times 10^{-8} M s^{-1}\).

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