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 \(\mathrm{Cl}^{-}\) is \(3.50 × 10^{-5}\, \mathrm{M}\, \mathrm{s}^{-1}\).

Step-by-step solution

Write the rate law equation

Given that the reaction is first order in each reactant, we can write the rate law equation as: \[rate = k[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}]\] where: - \(rate\) is the rate of the reaction - \(k\) is the rate constant, given as \(3.5 × 10^{-2}\, \mathrm{M}^{-1} \mathrm{s}^{-1}\) - \([\mathrm{H}_{2}\mathrm{S}]\) is the concentration of \(\mathrm{H}_{2}\mathrm{S}\) - \([\mathrm{Cl}_{2}]\) is the concentration of \(\mathrm{Cl}_{2}\)

Calculate the rate of the reaction

Now, we can plug in the concentrations of \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{Cl}_{2}\) given in the problem to find the rate of the reaction: \[rate = \left(3.5 × 10^{-2}\, \mathrm{M}^{-1} \mathrm{s}^{-1}\right)\left(2.0 × 10^{-4}\, \mathrm{M}\right)\left(0.025\, \mathrm{M}\right)\] Calculate the rate: \[rate = \left(3.5 × 10^{-2}\right)\left(2.0 × 10^{-4}\right)\left(0.025\right) = 1.75 × 10^{-5}\, \mathrm{M}\, \mathrm{s}^{-1}\] The rate of the reaction is \(1.75 × 10^{-5}\, \mathrm{M} \, \mathrm{s}^{-1}\).

Calculate the rate of formation of \(\mathrm{Cl}^{-}\)

Now that we have the rate of the reaction, we can find the rate of formation of \(\mathrm{Cl}^{-}\) using stoichiometry. The reaction equation is: \[\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\] From the equation, we can see that 1 mole of \(\mathrm{H}_{2}\mathrm{S}\) reacts with 1 mole of \(\mathrm{Cl}_{2}\) to produce 2 moles of \(\mathrm{Cl}^{-}\). Dividing the rate of the reaction by the stoichiometric coefficient of \(\mathrm{H}_{2} \mathrm{~S}\), and then multiplying by the stoichiometric coefficient of \(\mathrm{Cl}^{-}\) yields the rate of formation of \(\mathrm{Cl}^{-}\): \[rate_{\mathrm{Cl}^{-}} = \frac{2\, moles \, of \, \mathrm{Cl}^{-}}{1\, mole\, of\, \mathrm{H}_{2} \mathrm{~S}} × 1.75 × 10^{-5}\, \mathrm{M}\, \mathrm{s}^{-1}\] \[rate_{\mathrm{Cl}^{-}} = 2 × 1.75 × 10^{-5} \, \mathrm{M}\, \mathrm{s}^{-1} = 3.50 × 10^{-5}\, \mathrm{M}\, \mathrm{s}^{-1}\] The rate of formation of \(\mathrm{Cl}^{-}\) is \(3.50 × 10^{-5}\, \mathrm{M}\, \mathrm{s}^{-1}\).

Key concepts

Chemical Kinetics

Chemical kinetics is a fascinating branch of chemistry that deals with the speed or rate of chemical reactions. Rate laws, like the one used in the original exercise, describe how the concentration of reactants affects the rate of a reaction. In simple terms, chemical kinetics helps us understand how quickly a reaction progresses. To fully grasp this, think of a reaction as a sequence of events that leads to the transformation of reactants into products. The rate at which this transformation occurs can depend on several factors:

• The concentration of reactants: As demonstrated in the rate law equation, increasing the concentration of reactants generally increases the rate of reaction.
• The temperature: Typically, higher temperatures increase reaction rates as particles move faster and collide more frequently.
• The presence of a catalyst: Catalysts can significantly speed up a reaction without being consumed in the process.

Understanding these concepts is key to predicting how a reaction behaves under different conditions. It can also be crucial in applications like waste treatment, where controlling the reaction speed is necessary to efficiently remove pollutants like hydrogen sulfide.

Reaction Order

Reaction order is a fundamental concept that tells us how the rate of a reaction is affected by the concentration of its reactants. In the exercise, the reaction is `first order` in each reactant, meaning the rate of reaction depends linearly on each reactant's concentration. This can be expressed using a rate law, where the combined reaction order is the sum of the orders with respect to each reactant. For example, if the rate law is written as \[rate = k[\mathrm{H}_2\mathrm{S}][\mathrm{Cl}_2]\]the order with respect to \( \mathrm{H}_2\mathrm{S} \) is 1, the order with respect to \( \mathrm{Cl}_2 \) is 1, and therefore, the total reaction order is 2. This indicates a second-order reaction overall. Some important points to remember:

• Reaction order can be determined experimentally and is not necessarily related to the stoichiometry of the reaction.
• It provides information about the mechanism of the reaction and whether it occurs in one or multiple steps.
• The units of the rate constant \( k \) depend on the overall reaction order.

A clear understanding of reaction order helps predict how changes in conditions will affect the rate, which is particularly important in industrial processes and environmental management.

Stoichiometry

Stoichiometry is the chemistry concept that helps us understand the quantitative relationships in a chemical reaction. It allows us to predict how much of each product can be formed given certain amounts of reactants, and vice versa. In the context of the solution provided, stoichiometry is used to relate the rate of disappearance of reactants to the rate of appearance of products, like \( \mathrm{Cl}^- \) ions. From the balanced equation:\[\mathrm{H}_2\mathrm{S} + \mathrm{Cl}_2 \longrightarrow \mathrm{S} + 2\mathrm{H}^+ + 2\mathrm{Cl}^-\]We observe that 1 mole of \( \mathrm{H}_2\mathrm{S} \) reacts with 1 mole of \( \mathrm{Cl}_2 \) to produce 2 moles of \( \mathrm{Cl}^- \). This stoichiometric relationship tells us that for every unit of \( \mathrm{H}_2\mathrm{S} \) consumed, twice the amount of \( \mathrm{Cl}^- \) is formed. Key takeaways include:

• Stoichiometry is based on the law of conservation of mass, meaning the mass of the reactants equals the mass of the products.
• It ensures that chemical equations are balanced, providing exact ratios needed to predict product formation.
• Understanding stoichiometry is crucial in calculations involving reaction rates and yields.

This knowledge is vital for chemists working with reactions in the laboratory and industry, where ensuring correct proportions and yields is critical for success.