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Chapter 16: Problem 15

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ \mathrm{D}(g) \rightarrow \frac{3}{2} \mathrm{E}(g)+\frac{5}{2} \mathrm{~F}(g) $$ When [E] is increasing at \(0.25 \mathrm{~mol} / \mathrm{L}\) -s, how fast is [F] increasing?

Short Answer

Expert verified
[F] is increasing at 0.4175 mol/L-s.

Step by step solution

01

Identify the given data

The reaction is \( \mathrm{D}(g) \rightarrow \frac{3}{2} \mathrm{E}(g) + \frac{5}{2} \mathrm{~F}(g) \), and the rate of change in concentration of [E] is given as 0.25 mol/L-s.
02

Write the rate expression

The rate of the reaction can be expressed in terms of the change in concentration of reactants and products: \[ \text{Rate} = -\frac{1}{\Delta t} \frac{d[\text{D}]}{dt} = \frac{1}{\left(3/2\right)} \frac{d[\text{E}]}{dt} = \frac{1}{\left(5/2\right)} \frac{d[\text{F}]}{dt} \].
03

Calculate the rate of the reaction

Using [E]'s rate of increase, the rate is \[ \text{Rate} = \frac{d[\text{E}]}{dt} \cdot \frac{2}{3} \]. Given \( \frac{d[\text{E}]}{dt} = 0.25 \text{ mol/L-s} \), so \[ \text{Rate} = 0.25 \cdot \frac{2}{3} = 0.167 \text{ mol/L-s} \].
04

Relate [F]'s rate of increase to the reaction rate

Using the fact that \[ \frac{d[\text{F}]}{dt} = \text{Rate} \cdot \frac{5}{2} \], substitute \( \text{Rate} = 0.167 \text{ mol/L-s} \), so \[ \frac{d[\text{F}]}{dt} = 0.167 \cdot \frac{5}{2} = 0.4175 \text{ mol/L-s} \].

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Key Concepts

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

Rate of Reaction
The rate of reaction is a measure of how quickly reactants turn into products. It's important because it tells us the speed of a chemical reaction. Generally, the rate can be determined by observing changes in concentration over time. For this exercise, the reaction rate can be written as: \[ \text{Rate} = -\frac{1}{\text{stoich. coef. of D}} \frac{d[\text{D}]}{dt} = \frac{1}{\text{stoich. coef. of E}} \frac{d[\text{E}]}{dt} = \frac{1}{\text{stoich. coef. of F}} \frac{d[\text{F}]}{dt} \]. These expressions show how changes in the concentrations of reactants and products relate to the overall reaction rate. Each component's change is tied to its stoichiometric coefficient. This coefficient helps balance the reaction, ensuring the rate expression is consistently maintained across all substances involved.
Concentration Change
As the reaction proceeds, the concentrations of reactants and products change. For example, the concentration of [D] decreases as it gets converted into products [E] and [F]. In our current scenario, we've been given the rate of change for [E] at 0.25 mol/L-s. This value is crucial as it helps calculate the reaction rate and the rate of change for [F]. To determine how fast [F] is increasing, we need to adjust this given rate according to the stoichiometric coefficients of [E] and [F] in the reaction. Thus, the step-by-step approach links these rates using their stoichiometric relationships. This approach ensures consistency while describing how fast concentrations change.
Stoichiometry
Stoichiometry is the relationship between the quantities of reactants and products in a chemical reaction. Understanding this is key to calculating rates properly. In our given reaction: \( \text{D} \rightarrow \frac{3}{2} \text{E} + \frac{5}{2} \text{F} \). This equation tells us that one mole of D produces \( \frac{3}{2} \) moles of E and \( \frac{5}{2} \) moles of F. This ratio allows us to determine how the concentration changes for one component will affect the others.For example, from our solution step where we used \( \frac{d[\text{E}]}{dt} = 0.25 \text{ mol/L-s} \) and applied it to the rate of [F]'s increase by adjusting the rate using stoichiometry: \[ \frac{d[\text{F}]}{dt} = 0.167 \times \frac{5}{2} = 0.4175 \text{ mol/L-s} \]. Thus, understanding the stoichiometric coefficients directly influences the correct computation of rates and ensures balanced reactions.

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Most popular questions from this chapter

Nitrification is a biological process in which \(\mathrm{NH}_{3}\) in wastewater is converted to \(\mathrm{NH}_{4}^{+}\) and then removed according to the following reaction: \(\mathrm{NH}_{4}^{+}+2 \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) The first-order rate constant is given as \(k_{1}=0.47 e^{0.095\left(T-15^{\circ} \mathrm{C}\right)}\) where \(k_{1}\) is in day \(^{-1}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). (a) If the initial concentration of \(\mathrm{NH}_{3}\) is \(3.0 \mathrm{~mol} / \mathrm{m}^{3},\) how long will it take to reduce the concentration to \(0.35 \mathrm{~mol} / \mathrm{m}^{3}\) in the spring \(\left(T=20^{\circ} \mathrm{C}\right) ?\) (b) In the winter \(\left(T=10^{\circ} \mathrm{C}\right) ?\) (c) Using your answer to part (a), what is the rate of \(\mathrm{O}_{2}\) consumption?

Chlorine is commonly used to disinfect drinking water, and inactivation of pathogens by chlorine follows first-order kinetics. The following data are for \(E\). coli inactivation: $$ \begin{array}{cc} \text { Contact Time (min) } & \text { Percent (\%) Inactivation } \\ \hline 0.00 & 0.0 \\ 0.50 & 68.3 \\ 1.00 & 90.0 \\ 1.50 & 96.8 \\ 2.00 & 99.0 \\ 2.50 & 99.7 \\ 3.00 & 99.9 \end{array} $$ (a) Determine the first-order inactivation constant, \(k\). [Hint: \% inactivation \(\left.=100 \times\left(1-[\mathrm{A}] /[\mathrm{A}]_{0}\right) .\right]\) (b) How much contact time is required for \(95 \%\) inactivation?

The overall equation and rate law for the gas-phase decomposition of dinitrogen pentoxide are \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad\) rate \(=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Which of the following can be considered valid mechanisms for the reaction? I One-step collision II \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}_{3}(g)+2 \mathrm{NO}_{2}(g) \quad[\) slow \(]\) \(2 \mathrm{NO}_{3}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+2 \mathrm{O}(g)\) [fast] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) [fast] III \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)\) [fast] \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 3 \mathrm{NO}_{2}(g)+\mathrm{O}(g) \quad\) [slow] \(\mathrm{NO}_{3}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{IV} 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{3}(g)+3 \mathrm{O}(g) \quad[\) fast \(]\) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) [slow] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{V} \quad 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{N}_{4} \mathrm{O}_{10}(g)\) [slow] \(\mathrm{N}_{4} \mathrm{O}_{10}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) [fast]

Define the half-life of a reaction. Explain on the molecular level why the half-life of a first-order reaction is constant.

By what factor does the rate change in each of the following cases (assuming constant temperature)? (a) A reaction is first order in reactant \(\mathrm{A}\), and \([\mathrm{A}]\) is doubled. (b) A reaction is second order in reactant \(\mathrm{B},\) and \([\mathrm{B}]\) is halved. (c) A reaction is second order in reactant \(\mathrm{C},\) and \([\mathrm{C}]\) is tripled.
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