Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the water with chlorine, in which case the following reaction occurs: H2S(aq) + Cl2(aq) → S(s) + 2 H+(aq) + 2 Cl-(aq). The rate of this reaction is first order in each reactant. The rate constant for the disappearance of H2S at 28°C is 3.5 * 10^-2 M^-1 s^-1. If at a given time the concentration of H2S is 2.0 * 10^-4 M and that of Cl2 is 0.025 M, what is the rate of formation of Cl-?

Verified step by step guidance

1. Step 1: Understand the rate law for the reaction. Since the reaction is first order in each reactant, the rate law can be expressed as: rate = k[H2S][Cl2], where k is the rate constant.
2. Step 2: Identify the given values. The rate constant k is 3.5 * 10^-2 M^-1 s^-1, the concentration of H2S is 2.0 * 10^-4 M, and the concentration of Cl2 is 0.025 M.
3. Step 3: Substitute the given values into the rate law equation to find the rate of the reaction: rate = (3.5 * 10^-2 M^-1 s^-1)(2.0 * 10^-4 M)(0.025 M).
4. Step 4: Recognize that the rate of formation of Cl- is related to the rate of the reaction. According to the balanced chemical equation, 2 moles of Cl- are produced for every mole of H2S that reacts.
5. Step 5: Calculate the rate of formation of Cl- by multiplying the rate of the reaction by 2, since 2 moles of Cl- are formed per mole of H2S consumed.