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Chapter 14: Problem 90

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 \(\mathrm{H}^{+}\) ?

Short Answer

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

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 H鈧係 and Cl鈧�, 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鈦�

Using the stoichiometry of the balanced chemical equation, we can note that as 1 mole of H鈧係 is consumed, 2 moles of H鈦� are formed. Hence, the rate of formation of H鈦� is twice the rate of disappearance of H鈧係. So, the rate of formation of H鈦� 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鈦� is \(4.0 \times 10^{-8} M s^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(\mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)\)

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(55^{\circ} \mathrm{C}\) the rate constant is \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.050 \mathrm{M} ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled to \(0.150 \mathrm{M}\) ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by \(10 \%\) to \(0.045 \mathrm{M}\) ?

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is \(3.05 \times 10^{4} \mathrm{~s},\) what is the rate constant \(k\) at this temperature? What percentage of A will not have reacted after one day?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.
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