Problem 90 - Chemistry: The Central Science in SI Units, Global Edition

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 \(30{ }^{\circ} \mathrm{C}\) 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

Answer

The rate of formation of H⁺ is \(4.0 \times 10^{-8} M s^{-1}\).

Step 1 - Write down the balanced chemical equation

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

Step 2 - 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 \(\mathrm{Cl}_{2}\): \(2.0 \times 10^{-2} M\)

Step 3 - 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: \( \text{rate} = 2.0 \times 10^{-8} M s^{-1} \)

Step 4 - 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 \(4.0 \times 10^{-8} M s^{-1}\).

Key Concepts

Rate Law

The rate law of a chemical reaction expresses the relationship between the reaction rate and the concentrations of its reactants. In our exercise, the rate law is particularly simple as we are dealing with a reaction that is first order in both reactants, Hydrogen sulfide (\(\text{H}_2 \text{S}\)) and chlorine (\(\text{Cl}_2\)). This implies that the rate of reaction depends linearly on the concentration of each reactant.

For the given reaction, the rate law can be expressed as:\[\text{rate} = k \cdot [\text{H}_2 \text{S}] \cdot [\text{Cl}_2]\]where \(k\) is the rate constant, \([\text{H}_2 \text{S}]\) indicates the molar concentration of \(\text{H}_2 \text{S}\), and \([\text{Cl}_2]\) indicates the molar concentration of \(\text{Cl}_2\).

Understanding the rate law is crucial because it allows us to calculate the speed of the reaction under different concentrations of reactants, providing insights into how quickly products form. In this particular exercise, given concentrations help us determine the rate of formation for the \(\text{H}^{+}\) ions based on this relationship.

Reaction Order

A reaction order is an important aspect of the rate law. It indicates how the reaction rate is affected by the concentration of each reactant. When the reaction is first order in a substance, it means that if the concentration of that substance is doubled, the reaction rate also doubles. In our problem, the reaction is first order with respect to both \(\text{H}_2 \text{S}\) and \(\text{Cl}_2\).

In mathematical terms, for a reaction \(A + B \rightarrow C\) with rate law:\[\text{rate} = k \cdot [A]^m \cdot [B]^n\]the overall reaction order is \(m + n\). In this specific exercise, \(m = 1\) and \(n = 1\), giving us an overall reaction order of 2. This tells us that the reaction depends on the product of the concentrations of \(\text{H}_2 \text{S}\) and \(\text{Cl}_2\), confirming that the reaction rate changes as the concentrations of these reactants vary.

Understanding reaction orders helps in predicting how different conditions will affect the speed of a reaction, which is essential in industrial applications where reactions need to be controlled precisely.

Stoichiometry

Stoichiometry is the quantitative relation between reactants and products in a chemical reaction. In our balanced chemical equation:\[\text{H}_2 \text{S} + \text{Cl}_2 \rightarrow \text{S} + 2\text{H}^{+} + 2\text{Cl}^{-}\]we see the stoichiometric coefficients are crucial. They show us that each mole of \(\text{H}_2 \text{S}\) and \(\text{Cl}_2\) consumed produces 2 moles of \(\text{H}^{+}\) and \(\text{Cl}^{-}\) each.

In the context of reaction kinetics, stoichiometry aids in linking the rate of appearance of products with the rate of disappearance of reactants. As shown in the original problem, the rate of formation of \(\text{H}^{+}\) is double that of the rate of disappearance of \(\text{H}_2 \text{S}\) due to its coefficient of 2 in the balanced equation.

Understanding stoichiometry gives insight into how amounts of substances consumed and produced are interrelated. This is fundamental for ensuring the correct ratios of reactants for optimal product yields in chemical processes.