/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 The following kinetic data are c... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 94

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{X}]_{0}(M)} & {[\mathrm{Z}]_{0}(M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.25 & 0.25 & 4.0 \times 10^{1} \\ 2 & 0.50 & 0.50 & 3.2 \times 10^{2} \\ 3 & 0.50 & 0.75 & 7.2 \times 10^{2} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(X\) is \(0.75 M\) and that of \(Z\) is \(1.25 M ?\)

Short Answer

Expert verified
(a) The rate law for this reaction is: Rate = \( k [\mathrm{X}]^2 [\mathrm{Z}] \) (b) The value of the rate constant is approximately \( 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \) (c) The reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M is approximately \( 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \).

Step by step solution

01

Determine the order of the reaction with respect to reactant X

Compare Experiment 1 and Experiment 2, as both have a change in concentration for X but not for Z. The rate and concentration ratio can help us determine the order of the reaction with respect to X. For the change between Experiment 1 and Experiment 2: Change in concentration of X: \( \cfrac{[\mathrm{X}]_{0}\text{ (Experiment 2)}}{[\mathrm{X}]_{0}\text{ (Experiment 1)}} = \cfrac{0.50}{0.25} = 2 \) Change in rate: \( \cfrac{\mathrm{Rate} \cdot \mathrm{(Experiment 2)}}{\mathrm{Rate} \cdot \mathrm{(Experiment 1)}} = \cfrac{3.2 \times 10^2}{4.0 \times 10^1} = 8 \) Since the rate changes in 8 times while the concentration of X changes in 2 times, we can conclude that the reaction is second order with respect to X.
02

Determine the order of the reaction with respect to reactant Z

Compare Experiment 2 and Experiment 3, as both have a change in concentration for Z but not for X. The rate and concentration ratio can help us determine the order of the reaction with respect to Z. For the change between Experiment 2 and Experiment 3: Change in concentration of Z: \( \cfrac{[\mathrm{Z}]_{0}\text{ (Experiment 3)}}{[\mathrm{Z}]_{0}\text{ (Experiment 2)}} = \cfrac{0.75}{0.50} = 1.5 \) Change in rate: \( \cfrac{\mathrm{Rate} \cdot \mathrm{(Experiment 3)}}{\mathrm{Rate} \cdot \mathrm{(Experiment 2)}} = \cfrac{7.2 \times 10^2}{3.2 \times 10^2} = 2.25 \) Since the rate changes in 2.25 times while the concentration of Z changes in 1.5 times, we can conclude that the reaction is first order with respect to Z.
03

Determine the rate law and calculate the rate constant

From Steps 1 and 2, we know that the reaction is second order with respect to X and first order with respect to Z. Therefore, the rate law can be written as: Rate \( = k[\mathrm{X}]^2[\mathrm{Z}] \) Now let's use the data from Experiment 1 to calculate the rate constant k: \( 4.0 \times 10^1 = k (0.25)^2(0.25) \) \( k = \cfrac{4.0 \times 10^1}{(0.25)^2(0.25)} \) Calculate k: \( k \approx 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \)
04

Calculate the reaction rate for the given initial concentrations

Now that we have the rate law and the rate constant, we can calculate the reaction rate for the given initial concentrations of X = 0.75 M and Z = 1.25 M: Rate \( = k [\mathrm{X}]^2 [\mathrm{Z}] \) Rate \( = (2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}}) (0.75 \, \mathrm{M} )^2 (1.25 \, \mathrm{M}) \) Calculate the reaction rate: Rate \( \approx 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \) #Summary# (a) The rate law for this reaction is: Rate = \( k [\mathrm{X}]^2 [\mathrm{Z}] \) (b) The value of the rate constant is approximately \( 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \) (c) The reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M is approximately \( 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \).

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Most popular questions from this chapter

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{~nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a 1 -cm cuvette in a spectrometer and obtain the following data: From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\). and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100{ }^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the halflife for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

A first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) has the rate constant \(k=3.2 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(2.5 \times 10^{-2} M\), what is the rate of the reaction at \(t=660 \mathrm{~s}\) ?

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?
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