/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 Hydrogen sulfide \(\left(\mathrm... [FREE SOLUTION] | 91影视

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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

Expert verified
The rate of formation of Cl鈦 is \(3.5 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\).

Step by step solution

01

Write down the rate law for the reaction

Since the reaction is first order with respect to both reactants, the rate law can be written as: \(Rate = k[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}]\) where \(Rate\) is the reaction rate, \(k\) is the rate constant, and \([\mathrm{H}_{2}\mathrm{S}]\) and \([\mathrm{Cl}_{2}]\) are the concentrations of H鈧係 and Cl鈧, respectively.
02

Calculate the reaction rate using given values

We are given that the rate constant \(k = 3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1}\), the concentration of H鈧係 at that time \([\mathrm{H}_{2}\mathrm{S}] = 2.0 \times 10^{-4} \mathrm{M}\), and the concentration of Cl鈧 at that time \([\mathrm{Cl}_{2}] = 0.025 \mathrm{M}\). Plug these values into the rate law: \(Rate = (3.5\times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1})(2.0 \times 10^{-4} \mathrm{M})(0.025 \mathrm{M})\) Now calculate the rate: \(Rate = 1.75 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\)
03

Determine the stoichiometric relationship between the reactants and products

From the balanced reaction, we can see that for every 1 mole of H鈧係 that reacts, 2 moles of Cl鈦 are produced: \(\mathrm{H}_{2}\mathrm{S}(aq) + \mathrm{Cl}_{2}(aq) \longrightarrow \mathrm{S}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{Cl}^{-}(aq)\) So, the rate of production of Cl鈦 is twice the rate of disappearance of H鈧係.
04

Calculate the rate of formation of Cl鈦

Using the stoichiometric relationship and the calculated rate of reaction, we can find the rate of formation of Cl鈦: \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 2 \times Rate_{disappearance}\, of\, \mathrm{H}_{2}\mathrm{S}\) \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 2 \times 1.75 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\) Now calculate the rate: \(Rate_{formation}\, of\, \mathrm{Cl}^{-} = 3.5 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\) So, the rate of formation of Cl鈦 is \(3.5 \times 10^{-6} \mathrm{M}\cdot\mathrm{s}^{-1}\).

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Key Concepts

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

Rate Laws
In chemistry, rate laws describe how the rate of a reaction changes with the concentration of its reactants. This is captured by the rate law equation. For our given reaction, the rate law is expressed as: \( \text{Rate} = k[\text{H}_2\text{S}][\text{Cl}_2] \) This means the rate at which the reaction proceeds is directly proportional to the concentration of both hydrogen sulfide \( \text{[H}_2\text{S}] \) and chlorine \( \text{[Cl}_2] \). - **Rate**: Indicates how quickly the reactants are consumed, or products are formed.- **k**: Represents the rate constant, which remains constant for a reaction at a given temperature.- **[H\(_2\)S] and [Cl\(_2\)]**: The concentrations of each reactant in the reaction.The rate law is essential for determining reaction speed and is derived from the experimental data.
Order of Reaction
The order of reaction refers to the power to which the concentration of a reactant is raised in the rate law equation. In our case, the reaction with hydrogen sulfide and chlorine is first-order in each of its reactants. - **First-order**: If a reaction is first order in a particular reactant, the rate of reaction is directly proportional to the concentration of that reactant. For example, when the reaction is first-order with respect to both \( \text{H}_2\text{S} \) and \( \text{Cl}_2 \), it means if the concentration of either one of these reactants doubles, the reaction rate would also double. Understanding the order of the reaction allows scientists to predict the effects of concentration changes on reaction rates. It's vital for safe scaling of reactions in industrial settings.
Stoichiometry
Stoichiometry helps us understand the quantitative relationships between reactants and products in a chemical reaction. In our reaction: \( \text{H}_2\text{S}( ext{aq}) + \text{Cl}_2( ext{aq}) \rightarrow \text{S}(s) + 2\text{H}^+(\text{aq}) + 2\text{Cl}^-(\text{aq}) \) Stoichiometry tells us the ratio of reactants to products. Here, every mole of \( \text{H}_2\text{S} \) consumed by the reaction produces two moles of \( \text{Cl}^- \). This is termed a **stoichiometric coefficient**. - **Reactant ratios**: Determine how much product will form from the given amounts of reactants. - **Product formation**: Allows calculation of how much product can be expected, such as the rate of \( \text{Cl}^- \) production being twice that of \( \text{H}_2\text{S} \) disappearance.Stoichiometry is essential when calculating yields in reactions and figuring out how much reactant is needed to fully react.
Rate Constant
The rate constant \( k \) is a crucial component of the rate law, affecting the speed of the reaction. It's a proportionality factor that links the reaction rate to reactant concentration. For the given reaction at \( 28^{\circ} \text{C} \), the rate constant \( k \) is \( 3.5 \times 10^{-2} \text{M}^{-1}\text{s}^{-1} \). - **Temperature Dependence**: The rate constant is specific to a given temperature; increasing temperature generally increases \( k \), thus speeding up the reaction rate. - **Units**: The unit of \( k \) changes based on the order of the reaction. For a second-order reaction like ours (first-order in each reactant), its unit is \( \text{M}^{-1}\text{s}^{-1} \).Understanding the rate constant helps predict how quickly a reaction will occur, crucial for designing processes in laboratories and industry.

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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: \(\begin{array}{l}{\text { (a) } 2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\\ {\text { (c) } 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)}\end{array}\)

\(\begin{array}{l}{\text { (a) What is meant by the term elementary reaction? }} \\ {\text { (b) What is the difference between a unimolecular }} \\\ {\text { and a bimolecular elementary reaction? (c) What is a }}\end{array}\) \(\begin{array}{l}{\text {reaction mechanism?}(\mathbf{d}) \text { What is meant by the term rate- }} \\ {\text { determining step? }}\end{array}\)

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c),\) has an activation energy of 86.8 \(\mathrm{kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) A g \(\mathrm{KOH}\) in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. At 300 \(\mathrm{K}\) the rate constant equals \(7.0 \times 10^{-4} \mathrm{s}^{-1}\) . Calculate the half-life at this temperature. (b) If the activation energy for this reaction is \(75 \mathrm{kJ} / \mathrm{mol},\) at what temperature would the reaction rate be doubled?

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