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For each of the rate laws below, determine the order of the reaction with respect to the hypothetical substances \(\mathrm{X}, \mathrm{Y},\) and \(\mathrm{Z}\). What is the overall order? (a) Rate \(=k[\mathrm{X}][\mathrm{Y}][\mathrm{Z}]\) (b) Rate \(=k[\mathrm{X}]^{2}[\mathrm{Y}]^{1 / 2}[\mathrm{Z}]\) (c) Rate \(=k[\mathrm{X}]^{1.5}[\mathrm{Y}]^{-1}\) (d) Rate \(=k[\mathrm{X}] /[\mathrm{Y}]^{2}\)

Short Answer

Expert verified
(a) 3, (b) 3.5, (c) 0.5, (d) -1.

Step by step solution

01

Analyze Rate Law (a)

For rate law \( \text{Rate} = k[\mathrm{X}][\mathrm{Y}][\mathrm{Z}] \), identify the reaction order with respect to each component. \( [\mathrm{X}] \) has a power of 1, \( [\mathrm{Y}] \) also has a power of 1, and \( [\mathrm{Z}] \) has a power of 1. Add these to get the overall order: \( 1 + 1 + 1 = 3 \).
02

Analyze Rate Law (b)

For rate law \( \text{Rate} = k[\mathrm{X}]^{2}[\mathrm{Y}]^{1 / 2}[\mathrm{Z}] \), determine the reaction order for each component. \( [\mathrm{X}] \) has a power of 2, \( [\mathrm{Y}] \) has a power of 0.5, and \( [\mathrm{Z}] \) has a power of 1. Sum these to find the overall order: \( 2 + 0.5 + 1 = 3.5 \).
03

Analyze Rate Law (c)

For rate law \( \text{Rate} = k[\mathrm{X}]^{1.5}[\mathrm{Y}]^{-1} \), find the reaction order for each component. \( [\mathrm{X}] \) has a power of 1.5, and \( [\mathrm{Y}] \) has a power of -1. There is no \( [\mathrm{Z}] \), so its order is 0. Combine these to find the overall order: \( 1.5 - 1 + 0 = 0.5 \).
04

Analyze Rate Law (d)

For rate law \( \text{Rate} = k[\mathrm{X}] / [\mathrm{Y}]^{2} \), determine the reaction order for each component from the expression. \( [\mathrm{X}] \) has a power of 1, and \( [\mathrm{Y}] \) has a power of -2, since it is in the denominator. No \( [\mathrm{Z}] \) means its order is 0. Sum these to get the overall order: \( 1 - 2 + 0 = -1 \).

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Key Concepts

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

Rate Law
The rate law of a chemical reaction is a mathematical expression that describes the relationship between the rate of a reaction and the concentrations of its reactants. Understanding the rate law helps you determine how the concentration of each reactant affects the speed of the reaction. In a rate law, such as \( \text{Rate} = k[\text{X}]^m[\text{Y}]^n[\text{Z}]^p \), the exponents \( m \), \( n \), and \( p \) represent the orders of the reaction with respect to each reactant. These exponents can be zero, positive, or even negative.

Determining the order involves looking at these exponents directly from the rate law. For instance:
  • If the exponent is zero, the reaction rate is independent of that reactant's concentration.
  • A positive exponent means the reaction rate increases with an increase in that reactant's concentration.
  • A negative exponent indicates the reaction rate actually decreases as the concentration of that reactant increases.
The sum of these exponents gives the overall reaction order, impacting the analysis of reaction kinetics.
Reaction Kinetics
Reaction kinetics refers to the study of rates of chemical processes and the factors affecting them. It explores how changes in conditions like concentration, temperature, and surface area alter the speed of reactions. Within reaction kinetics, the rate law serves as a crucial tool for predicting and interpreting the behavior of a reaction over time.

A key factor in reaction kinetics is the rate constant \( k \), which is specific to a given reaction at a particular temperature. The value of \( k \) provides insight into the intrinsic properties of the reaction. However, it does not depend on the concentrations of the reactants. Instead, it can change if the temperature or the presence of a catalyst is altered.
  • Concentration: The rate typically increases with higher concentrations, depending on the reaction order.
  • Temperature: Higher temperatures usually increase the reaction rate, often modeled by the Arrhenius equation.
  • Catalysts: Presence of a catalyst allows a reaction to proceed at a faster rate or via a different pathway with lower activation energy.
Understanding these aspects helps chemists control and optimize reactions in various fields, from pharmaceutical to industrial settings.
Chemical Kinetics
Chemical kinetics not only focuses on the rate at which reactions occur, but also provides insight into the mechanistic pathways of the reactions. By observing the rate laws and the orders of reactions, chemists can deduce likely steps of the reaction mechanism, which might include multiple elementary steps or stages.

The overall reaction order, derived from individual reactant orders, offers a glimpse into these elementary steps as it is often equal to the sum of all individual step orders in the proposed mechanism. The study of chemical kinetics helps with:
  • Determining reaction mechanisms by matching observed rates to predicted rates from potential pathways.
  • Evaluating the effect of conditions (like pressure and temperature) on the reaction pathway.
  • Investigating the formation and consumption of intermediates that are not visible in the overall balanced equation.
By gaining a comprehensive understanding of chemical kinetics, scientists can improve efficiency and yields in chemical manufacturing, contributing significantly to advancements in materials science and energy production.

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Most popular questions from this chapter

The transfer of an oxygen atom from \(\mathrm{NO}_{2}\) to CO has been studied at \(540 \mathrm{~K}\) : \(\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})\) These data were collected: \begin{tabular}{ccc} \hline Initial Rate & \multicolumn{2}{c} { Initial Concentration (mol/L) } \\ \cline { 2 - 3 } (mol \(\mathrm{L}^{-1} \mathrm{~h}^{-1}\) ) & {\(\left[\mathrm{NO}_{2}\right]\)} \\ \hline \(5.1 \times 10^{-4}\) & \(0.35 \times 10^{-4}\) & \(3.4 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.70 \times 10^{-4}\) & \(1.7 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.18 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.0 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.5 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(10.2 \times 10^{-8}\) \\ \hline \end{tabular} (a) Write the rate law. (b) Determine the reaction order with respect to each reactant (c) Calculate the rate constant and express it in appropriate units.

For the reaction $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) $$ make qualitatively correct plots of the concentrations of \(\mathrm{NO}_{2}(\mathrm{~g}), \mathrm{NO}(\mathrm{g}),\) and \(\mathrm{O}_{2}(\mathrm{~g})\) versus time. Draw all three graphs on the same axes; assume that you start with \(\mathrm{NO}_{2}(\mathrm{~g})\) at a concentration of \(1.0 \mathrm{~mol} / \mathrm{L}\). Explain how you would determine, from these plots, (a) the initial rate of the reaction. (b) the final rate (that is, the rate as time approaches infinity).

Assuming that each reaction is elementary, predict the rate law. (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\) (d) \(2 \mathrm{HI}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\)

If a reaction has the experimental rate law, Rate \(=k[\mathrm{~A}]^{2}\), explain what happens to the rate when (a) the concentration of A is tripled. (b) the concentration of A is halved.

Nitrosyl bromide, NOBr, is formed from \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\). $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g}) $$ Experiment shows that the reaction is first-order in \(\mathrm{Br}_{2}\) and second-order in NO. (a) Write the rate law for the reaction. (b) If the concentration of \(\mathrm{Br}_{2}\) is tripled, determine how the reaction rate changes. (c) Determine what happens to the reaction rate when the concentration of \(\mathrm{NO}\) is doubled.

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