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To obtain the rate of the reaction $$ \begin{aligned} 3 \mathrm{I}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+& 2 \mathrm{H}^{+}(a q) \longrightarrow \\ & \mathrm{I}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ you might follow the \(\mathrm{H}^{+}\) concentration or the \(\mathrm{I}_{3}^{-}\) concentration. How are the rates in terms of these species related?

Short Answer

Expert verified
The rate of disappearance of \( \mathrm{H}^{+} \) is half the rate of formation of \( \mathrm{I}_{3}^{-} \).

Step by step solution

01

Write the Rate Expression

For any chemical reaction, the rate of the reaction is expressed as the change in concentration of reactants or products over time. For the given reaction, the rate can be expressed in terms of the change in concentration of \( \mathrm{H}^{+} \) and \( \mathrm{I}_{3}^{-} \). Denote the rate with respect to time as \( \frac{d}{dt} \).
02

Apply Stoichiometry to Rate Expression

The rate of disappearance of reactants and the rate of appearance of products are related by the stoichiometric coefficients in the balanced equation. From the balanced equation, the stoichiometric coefficient of \( \mathrm{H}^{+} \) is 2 and \( \mathrm{I}_{3}^{-} \) is 1. This means each mole of \( \mathrm{H}_{3} \mathrm{AsO}_{4} \) reacts with 2 moles of \( \mathrm{H}^{+} \) to produce \( \mathrm{I}_{3}^{-} \), and therefore the stoichiometric relationship between \( \mathrm{H}^{+} \) and \( \mathrm{I}_{3}^{-} \) dictates their rates.
03

Relate the Rates of Different Species

From the stoichiometry of the reaction, the rate of change of \( \mathrm{H}^{+} \) is related to the rate of formation of \( \mathrm{I}_{3}^{-} \) by: \[\frac{1}{2} \frac{d[\mathrm{H}^{+}]}{dt} = \frac{d[\mathrm{I}_{3}^{-}]}{dt}\]Thus, the rate of disappearance of \( \mathrm{H}^{+} \) is one-half of the rate of formation of \( \mathrm{I}_{3}^{-} \).

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Key Concepts

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

Stoichiometry
In any chemical reaction, understanding stoichiometry is key to relating different species involved. Stoichiometry originates from the Greek word for "element" and "measure," which is quite fitting, as it helps us understand the quantitative relationships in a chemical equation. When you look at a balanced chemical reaction, the stoichiometric coefficients tell you how many moles of each reactant or product are needed or produced. For example, in the reaction given:
  • 3 moles of \( \mathrm{I}^{-} \)
  • 1 mole of \( \mathrm{H}_{3} \mathrm{AsO}_{4} \)
  • 2 moles of \( \mathrm{H}^{+} \)
React to form:
  • 1 mole of \( \mathrm{I}_{3}^{-} \)
  • 1 mole of \( \mathrm{H}_{3} \mathrm{AsO}_{3} \)
  • 1 mole of water \( \mathrm{H}_{2} \mathrm{O} \)
In our case, stoichiometry helps us relate how fast a product appears or a reactant disappears based on these coefficients. Specifically, stoichiometry reveals that the rate of disappearance of \( \mathrm{H}^{+} \) is half of that of \( \mathrm{I}_{3}^{-} \), because it takes two moles of \( \mathrm{H}^{+} \) for every mole of \( \mathrm{I}_{3}^{-} \) formed.
Rate Expression
The rate of a chemical reaction tells us how quickly a reactant is being used up or a product is being formed. To define this rate, we use a mathematical expression called the "rate expression." This expression is an equation relating the rate to the concentrations of the involved chemical species. In our given reaction, the rate can be defined as the rate of change in concentration of any reactant or product over time. Here, two expressions define the rate:
  • For \( \mathrm{H}^{+} \), it's expressed as \( \frac{-1}{2} \frac{d[\mathrm{H}^{+}]}{dt} \)
  • For \( \mathrm{I}_{3}^{-} \), it's expressed as \( \frac{d[\mathrm{I}_{3}^{-}]}{dt} \)
The minus sign indicates that \( \mathrm{H}^{+} \) is decreasing in concentration, as it is a reactant. The fraction on the left side of the equation indicates the stoichiometric relationship: two moles of \( \mathrm{H}^{+} \) are used for each mole of \( \mathrm{I}_{3}^{-} \) produced. Thus, the rate of change of \( \mathrm{H}^{+} \) is half the rate of change of \( \mathrm{I}_{3}^{-} \). Understanding these rate expressions helps in connecting theoretical knowledge with practical chemical kinetics.
Reaction Mechanism
Every chemical reaction can be understood through its reaction mechanism, which is a detailed step-by-step description of how a reaction proceeds on a microscopic level. This mechanism can consist of one or several "elementary steps" leading to the overall reaction. Each step has its own rate, which contributes to the overall reaction rate.In complex reactions, the overall rate is often dictated by the slowest step, known as the "rate-determining step," which acts as a bottleneck. When analyzing a reaction mechanism, it’s crucial to consider:
  • The sequence of steps, including which substances react and form at each stage.
  • The role and conservation of reaction intermediates that are formed and consumed during the reaction.
  • How colliding molecules and the orientation of those collisions lead to bond formation or breaking.
Understanding the reaction mechanism is instrumental for predicting the effects of concentration or temperature changes on the overall reaction rate. In our specific reaction, knowing that two \( \mathrm{H}^{+} \) ions are involved in each elementary step provides insight into reaction dynamics. This not only supports the stoichiometry but also affirms the connectivity between rates and actual chemical changes during the reaction.

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Most popular questions from this chapter

Ethyl chloride, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), used to produce tetraethyllead gasoline additive, decomposes, when heated, to give ethylene and hydrogen chloride. The reaction is first order. In an experiment, the initial concentration of ethyl chloride was \(0.00100 \mathrm{M}\). After heating at \(500^{\circ} \mathrm{C}\) for \(155 \mathrm{~s}\), this was reduced to \(0.00067 \mathrm{M}\). What was the concentration of ethyl chloride after a total of \(256 \mathrm{~s}\) ?

A series of kinetics experiments are conducted on the hypothetical reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{AB}_{2} .\) Prior to conducting experiments, there are four proposed rate laws rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\) rate \(=k[\mathrm{~A}][\mathrm{B}]\) rate \(=k[\mathrm{~A}][\mathrm{B}]^{2}\) rate \(=k[\mathrm{~A}][\mathrm{B}]^{4}\) Experimental data indicates that when the concentration of \(\mathrm{A}\) is doubled while the concentration of \(\mathrm{B}\) is held constant, the rate of the reaction doubles. The data also shows that when the concentration of \(\mathrm{A}\) is held constant and the concentration of \(\mathrm{B}\) is doubled, the rate of the reaction increases by a factor of four. a. Which of the proposed rate laws is consistent with the experimental data? b. Using your answer from part a, predict the change in the initial rate for a reaction in which the concentration of \(\mathrm{A}\) is reduced by \(50 \%\) and the concentration of \(\mathrm{B}\) is increased by \(100 \%\). c. If you were to run the reaction with \([\mathrm{A}]_{0}=0.10 \mathrm{M}\) and \([\mathrm{B}]_{0}=0.10 \mathrm{M}\), sketch a concentration versus time graph depicting the changes in concentration of both \(\mathrm{A}\) and \(\mathrm{B}\).

The hypothetical reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow\) Products has the rate law Rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{3}\). If the reaction is run two separate times, holding the concentration of A constant while doubling the concentration of \(\mathrm{B}\) from one run to the next, how would the rate of the second run compare to the rate of the first run?

Consider the reaction \(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}+\mathrm{C}\). a. One rate expression for the reaction is Rate of formation of \(\mathrm{C}=+\frac{\Delta[\mathrm{C}]}{\Delta t}\) Write two other rate expressions for this reaction in this form. b. Using your two rate expressions, if you calculated the average rate of the reaction over the same time interval, would the rates be equal? c. If your answer to part b was no, write two rate expressions that would give an equal rate when calculated over the same time interval.

For the decomposition of one mole of nitrosyl chloride, \(\Delta H=38 \mathrm{~kJ}\). $$ \mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) $$ The activation energy for this reaction is \(100 \mathrm{~kJ}\) a, Is this reaction exothermic or endothermic? b. What is the activation energy for the reverse reaction? c. If a catalyst were added to the reaction, how would this affect the activation energy?

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