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91Ó°ÊÓ

For a reaction involving the decomposition of a hypothetical substance \(\mathrm{Y}\), these data are obtained: Rate \([\mathrm{Y}]\) \(\begin{array}{lllll}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}\right) & 0.288 & 0.245 & 0.202 & 0.158 \\ & 0.200 & 0.170 & 0.140 & 0.110\end{array}\) (a) Determine the order of the reaction. (b) Write the rate law for the decomposition of \(Y\). (c) Calculate \(k\) for the experiment above.

Short Answer

Expert verified
First-order reaction with rate law: \\( ext{Rate} = k[Y] \\). Rate constant \\( k = 0.694 \: ext{min}^{-1} \\).

Step by step solution

01

Understand the Data Relationships

We are given rate data dependent on concentrations \([Y]\). The task is to determine the relationship between these rates and the concentrations. To determine the order of the reaction, observe how the rate changes as \([Y]\) changes.
02

Determine the Reaction Order

The reaction order can be determined by examining how changes in concentration affect the rate. Assume \text{Rate} = k[Y]^n\. Calculate \(n\) using different pairs: let the first pair represent \( [Y]_1 = 0.288 \: ext{mol L}^{-1} \, ext{Rate}_1 = 0.200 \: ext{mol L}^{-1}\text{min}^{-1}\) and the second pair \( [Y]_2 = 0.245 \: ext{mol L}^{-1} \, ext{Rate}_2 = 0.170 \: ext{mol L}^{-1}\text{min}^{-1}\).

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Key Concepts

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

Reaction Order
In reaction kinetics, the reaction order tells us how the rate of a chemical reaction depends on the concentration of reactants. Knowing the reaction order is crucial to understanding how different concentration levels will affect the speed of the reaction. We define the order of reaction as an exponent value in the rate law equation, which indicates the power to which the concentration of a reactant is raised.

To determine the reaction order, observe how changes in the concentration of reactants influence the rate. For example, if doubling the concentration leads to doubling the reaction rate, the reaction order with respect to that reactant is likely first order (n = 1). If doubling the concentration leads to a fourfold increase, it's second order (n = 2). The reaction can be zero, fractional, or even negative order, depending on how the concentration change affects the rate. Analyzing experimental data is critical to determining these values.
Rate Law
The rate law provides a mathematical relationship that shows the role of reactant concentrations in influencing the rate of reaction. It is usually expressed as:

\[ ext{Rate} = k[Y]^n\]

Where \( ext{Rate} \) is the rate of reaction, \( [Y] \) is the concentration of reactant Y, \( n \) is the reaction order, and \( k \) is the rate constant. The rate law is specific to each chemical reaction and must be determined experimentally. It tells us how sensitive the reaction rate is to changes in the concentration of the reactants. The form of the rate law suggests mechanisms through which reactants are turned into products.

By analyzing concentration and rate data, one can derive the order of the reaction with respect to each participating species and construct the rate law to model how the reaction proceeds under various conditions.
Rate Constant
The rate constant, denoted as \( k \), is a crucial part of the rate law that relates the rate of reaction to the concentrations of reactants and the reaction order. This constant is determined experimentally and is unique to each reaction. It varies with temperature and can also be affected by the presence of catalysts.

The units of the rate constant depend on the overall order of the reaction and are derived to satisfy the rate equation. For a first-order reaction, for example, the units of \( k \) are \( ext{min}^{-1} \), showing that it's reliant on time. Being able to calculate \( k \) gives an opportunity to predict reaction rates at different concentrations, which aids in designing chemical processes and understanding reaction mechanisms.

Ultimately, the rate constant is an essential factor that allows chemists to predict and control the progress of reactions in both laboratory and industrial settings.

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Most popular questions from this chapter

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. $$ \text { Step 1: } \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {fast, endothermic } $$ $$ \text { Step } 2: \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} $$ slow (a) Write an equation for the overall reaction. (b) Draw a reaction energy diagram for this reaction. (c) Show that the rate law for this reaction is $$ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right] $$

Indicate whether each of these statements is true or false. Change the wording of each false statement to make it true. (a) It is possible to change the rate constant for a reaction by changing the temperature. (b) The reaction rate remains constant as a first-order reaction proceeds at a constant temperature. (c) The rate constant for a reaction is independent of reactant concentrations. (d) As a second-order reaction proceeds at a constant temperature, the rate constant changes.

The reaction of \(\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g})\) is second-order in \(\mathrm{NO}_{2}\) and zeroth-order in \(\mathrm{CO}\) at temperatures less than \(500 \mathrm{~K}\). (a) Write the rate law for the reaction. (b) Determine how the reaction rate changes if the \(\mathrm{NO}_{2}\) concentration is halved. (c) Determine how the reaction rate changes if the concentration of CO is doubled.

Make an Arrhenius plot and calculate the activation energy for the gas-phase reaction \(2 \mathrm{NOCl}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & Rate Constant \(\left(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline \(400 .\) & \(6.95 \times 10^{-4}\) \\ \(450 .\) & \(1.98 \times 10^{-2}\) \\ \(500 .\) & \(2.92 \times 10^{-1}\) \\ \(550 .\) & 2.60 \\ \(600 .\) & 16.3 \\ \hline \end{tabular}

For the reaction $$ \mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g}) $$ make qualitatively correct plots of the concentrations of \(\mathrm{O}_{3}(\mathrm{~g}), \mathrm{O}(\mathrm{g}),\) and \(\mathrm{O}_{2}(\mathrm{~g})\) versus time. Draw all three graphs on the same axes, assume that you start with \(\mathrm{O}_{3}(\mathrm{~g})\) and \(\mathrm{O}(\mathrm{g})\), each at a concentration of \(1.0 \mu \mathrm{mol} / \mathrm{L}\). 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).

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