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In acid solution, methyl formate forms methyl alcohol and formic acid. \(\mathrm{HCO}_{2} \mathrm{CH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{HCOOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq})\) methyl formate \(\begin{array}{ll}\text { formic acid } & \text { methyl alcohol }\end{array}\) The rate law is Rate \(=k\left[\mathrm{HCO}_{2} \mathrm{CH}_{3}\right]\left[\mathrm{H}^{+}\right] .\) Why does \(\mathrm{H}^{+}\) appear in the rate law but not in the overall equation for the reaction?

Short Answer

Expert verified
\(\text{H}^+\) is a catalyst, affecting rate but not in the reaction equation.

Step by step solution

01

Understand the Reaction Components

Identify the reactants and products in the chemical reaction. The reactants are methyl formate \(\text{HCO}_2\text{CH}_3\) and water \(\text{H}_2\text{O}\). The products are formic acid \(\text{HCOOH}\) and methyl alcohol \(\text{CH}_3\text{OH}\).
02

Identify the Rate Law Components

Analyze the given rate law: Rate \(= k[\text{HCO}_2\text{CH}_3][\text{H}^+]\). It shows that the reaction rate depends on the concentrations of methyl formate \([\text{HCO}_2\text{CH}_3]\) and the hydrogen ion \([\text{H}^+]\).
03

Understand the Role of \(\text{H}^+\)

In the rate law, \(\text{H}^+\) appears because it acts as a catalyst for the reaction. Catalysts can influence the rate of a reaction by lowering the activation energy but do not appear in the overall balanced equation, as they are not consumed in the process.

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Key Concepts

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

Catalysis
Catalysis is a key concept in chemical reactions, particularly in the way it affects reaction rates. In the given reaction, hydrogen ions (\(\mathrm{H}^+\)) act as catalysts. Catalysts are substances that increase the rate of a chemical reaction without undergoing permanent changes themselves. Instead of being consumed, they offer a pathway with a lower activation energy required for the reaction to proceed. This makes it easier and faster for reactants to convert into products.

A catalyst might not appear in the overall stoichiometric equation of the reaction but plays a crucial role in determining the speed of the reaction. Some important points about catalysts include:
  • They do not change the equilibrium position of a reaction.
  • They are not consumed or altered during the reaction.
  • They can be used repeatedly to catalyze many reactions.
For this reaction, the presence of \(\mathrm{H}^+\) is essential in lowering the activation energy barrier while not being shown in the final equation.
Reaction Mechanism
The reaction mechanism provides insight into the step-by-step sequence of elementary reactions that leads to the formation of products. Understanding these steps allows chemists to see how reactants transform into products, including which molecules and intermediates are involved.

In the case of the given reaction, the mechanism would involve steps that possibly incorporate \(\mathrm{H}^+\) as a temporarily involved entity that helps facilitate each elementary step. Reaction mechanisms help in explaining why certain species, like \(\mathrm{H}^+\), are present in the rate law but absent from the overall balanced equation.

Considering the mechanism is essential because:
  • It reveals intermediates and transition states which aren't visible in the overall reaction.
  • It explains why certain species affect the reaction rate, even if they're catalysts.
  • It helps in identifying potential rate-determining steps, where the reaction is slowest.
Through analyzing the reaction mechanism, the role of \(\mathrm{H}^+\) as a catalyst becomes clear as it participates in one or more of these elementary steps.
Chemical Kinetics
Chemical kinetics involves the study of reaction rates and how different conditions affect these rates. A fundamental aspect of kinetics is the rate law, which provides an expression that relates the concentration of reactants to the rate of a chemical reaction.

For the methyl formate reaction, the rate law is given as \[\text{Rate} = k[\text{HCO}_2\text{CH}_3][\text{H}^+]\]. This indicates that the rate is dependent on the concentrations of methyl formate and hydrogen ions. The rate constant, \(k\), is specific to the reaction and can change with temperature. Understanding the rate law helps predict how variations in reactant concentrations affect the overall speed of the reaction.

Some key points to remember about chemical kinetics include:
  • The rate law is determined experimentally and cannot be inferred solely from the balanced chemical equation.
  • The order of reaction pertains to the power of the concentration terms in the rate law, not the stoichiometry.
  • Temperature is a crucial factor; it influences the rate constant, \(k\), often described by the Arrhenius equation.
By grasping these kinetics concepts, one can understand why the hydrogen ion concentration is included in the rate law, demonstrating its catalytic role.

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Most popular questions from this chapter

For the reaction of iodine atoms with hydrogen molecules in the gas phase, these rate constants were obtained experimentally. \(2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) 2 \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & \(10^{-5} \mathrm{k}\left(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\right)\) \\ \hline 417.9 & 1.12 \\ 480.7 & 2.60 \\ 520.1 & 3.96 \\ 633.2 & 9.38 \\ 666.8 & 11.50 \\ 710.3 & 16.10 \\ 737.9 & 18.54 \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant of the reaction at \(400.0 \mathrm{~K}\).

These data were obtained for the rate constant for reaction of an unknown compound with water: \begin{tabular}{cc} \hline\(T\left({ }^{\circ} \mathrm{C}\right)\) & \(\mathrm{k}\left(\mathrm{s}^{-1}\right)\) \\ \hline 56.2 & \(1.04 \times 10^{-5}\) \\ 78.2 & \(1.45 \times 10^{-4}\) \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant of the reaction at a temperature of \(100.0^{\circ} \mathrm{C}\).

Draw an energy versus reaction progress diagram (similar to the one in Question 60 ) for each of the reactions whose activation energy and enthalpy change are given below. Draw arrows to represent the activation energies of the forward and reverse reaction and \(\Delta_{\mathrm{r}} H^{\circ} .\) (a) \(\Delta_{r} H^{\circ}=105 \mathrm{~kJ} \mathrm{~mol}^{-1} ; E_{\mathrm{a}}=175 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(\Delta_{r} H^{\circ}=-43 \mathrm{~kJ} \mathrm{~mol}^{-1} ; E_{\mathrm{a}}=95 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(\Delta_{r} H^{\circ}=15 \mathrm{~kJ} \mathrm{~mol}^{-1} ; E_{\mathrm{a}}=55 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Using data given in the table for the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}\) calculate the average rate of reaction during each of these intervals: \(\begin{array}{ll}0.50 \mathrm{~h} . & \text { (b) } 0.50 \text { to } 1.0 \mathrm{~h} .\end{array}\) 0.00 to (a) 1.0 to (c) \(2.0 \mathrm{~h}\) (d) 2.0 to \(3.0 \mathrm{~h}\). \(4.0 \mathrm{~h}\) (f) 4.0 to \(5.0 \mathrm{~h}\). (e) 3.0 to \begin{tabular}{cccc} \hline Time (h) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} & Time \((\mathrm{h})\) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} \\ \hline 0.00 & 0.849 & 3.00 & 0.352 \\ 0.50 & 0.733 & 4.00 & 0.262 \\ 1.00 & 0.633 & 5.00 & 0.196 \\ 2.00 & 0.472 & & \\ \hline \end{tabular}

Radioactive gold-198 is used in the diagnosis of liver problems. \({ }^{198}\) Au decays in a first-order process, emitting a \(\beta\) particle (electron). The half-life of this isotope is 2.7 days. You begin with a 5.6 -mg sample of the isotope. Calculate how much gold-198 remains after 1.0 day.

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