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Chapter 14: Problem 12

Consider the zeroth-order reaction: \(\mathrm{A} \longrightarrow\) product. (a) Write the rate law for the reaction. (b) What are the units for the rate constant? (c) Plot the rate of the reaction versus [A].

Short Answer

Expert verified
(a) Rate = k; (b) Units: \(\text{mol L}^{-1} \text{s}^{-1}\); (c) Plot: Horizontal line.

Step by step solution

01

Identify the Order of the Reaction

The problem states that this is a zeroth-order reaction. In a zeroth-order reaction, the rate is independent of the concentration of the reactant.
02

Write the Rate Law

For a zeroth-order reaction, the rate is constant and given by the rate law: \[\text{Rate} = k\]where \(k\) is the rate constant.
03

Determine the Units for the Rate Constant

The rate of a reaction is typically expressed in units of concentration per unit time, such as \(\text{mol L}^{-1} \text{s}^{-1}\). For a zeroth-order reaction, the rate constant \(k\) must have the same units as the rate to ensure dimensional consistency. Therefore, the units for \(k\) are \(\text{mol L}^{-1} \text{s}^{-1}\).
04

Plot the Rate of the Reaction versus [A]

In a zeroth-order reaction, the rate is constant and does not depend on the concentration of \([A]\). Thus, when plotting the rate against the concentration \([A]\), the graph is a horizontal line, indicating that the rate remains constant as \([A]\) varies.

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Key Concepts

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

Rate Law for Zeroth-Order Reactions
In chemical kinetics, the rate law is an equation that describes the rate of a reaction in terms of the concentration of reactants. For a zeroth-order reaction like \(\mathrm{A} \longrightarrow \text{product}\), the rate is independent of the concentration of the reactant. This means the rate doesn't change even if the concentration of \([\text{A}]\) increases or decreases.

The rate law for a zeroth-order reaction is:
  • \(\text{Rate} = k\)
Here, \(k\) is the rate constant. In zeroth-order reactions, since the rate doesn't change with \([\text{A}]\), it remains constant throughout the reaction.
Understanding the Rate Constant
The rate constant, symbolized by \(k\), is a crucial concept in understanding reaction rates. For a zeroth-order reaction, the units for \(k\) are the same as that of the rate itself because the rate is constant. Typically, the units are expressed as concentration over time.

For example, the units for \(k\) in a zeroth-order reaction are:
  • \(\text{mol L}^{-1} \text{s}^{-1}\)
These units indicate that the reaction rate reflects how the concentration of the reactant changes per second or other time unit. It's essential that \(k\)'s units match the rate's units to maintain dimensional consistency across the equation.
Concentration vs. Time Plot
Plotting a concentration vs. time graph helps visualize the progress of a reaction. For a zeroth-order reaction, the concentration of \([\text{A}]\) decreases linearly with time. This is because the rate is constant regardless of the reactant's concentration.

If you were to plot the rate of the reaction against \([\text{A}]\), you would see a horizontal line. Here's why:
  • The rate is constant, shown by a straight line parallel to the x-axis.
  • The slope of the line is zero since changes in concentration do not affect the rate.
This graph clearly shows that the reaction progresses at a uniform speed, supporting the concept of a zeroth-order reaction where the reaction rate doesn't change with varying concentrations of \([\text{A}]\).

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Most popular questions from this chapter

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

The reaction of \(\mathrm{G}_{2}\) with \(\mathrm{E}_{2}\) to form \(2 \mathrm{EG}\) is exothermic, and the reaction of \(\mathrm{G}_{2}\) with \(\mathrm{X}_{2}\) to form \(2 \mathrm{XG}\) is endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic reaction. Sketch the potential-energy profile diagrams for these two reactions on the same graph.

Determine the molecularity, and write the rate law for each of the following elementary steps: (a) \(\mathrm{X} \longrightarrow\) products (b) \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products (c) \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products (d) \(\mathrm{X}+\mathrm{X} \longrightarrow\) products (e) \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products

Chlorine oxide \((\mathrm{ClO}),\) which plays an important role in the depletion of ozone, decays rapidly at room temperature according to the equation: $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \end{array} $$

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of rate determining.
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