/*! 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 131 The data given below is for the ... [FREE SOLUTION] | 91Ó°ÊÓ

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The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at 295 $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \text { Initial rate }\left(\mathrm{mol}_{\mathbf{L}}^{-1} \mathbf{s}^{-1}\right) \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ What is the rate law? (a) \(\mathrm{r}=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) (b) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) (c) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) (d) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}\)

Short Answer

Expert verified
The rate law is (b) \( r = k[\text{Cl}_2]^1[\text{NO}]^2 \).

Step by step solution

01

Identify the Reaction Rate Formula

The general form for the rate law of a reaction involving two reactants like this one is: \( r = k[ ext{NO}]^m[ ext{Cl}_2]^n \), where \( m \) and \( n \) are the orders of reaction with respect to each reactant, and \( k \) is the rate constant.
02

Analyze Experiment 1 and 2 to Find Order of Cl2

Comparing experiments 1 and 2, where [NO] remains constant at 0.05 M while [Cl2] increases from 0.05 M to 0.15 M, the rate increases from \( 1 \times 10^{-3} \) to \( 3 \times 10^{-3} \). Since the rate triples as [Cl2] triples, \( n = 1 \) indicates a first-order reaction with respect to \( [Cl2] \).
03

Analyze Experiment 1 and 3 to Find Order of NO

Comparing experiments 1 and 3, where [Cl2] remains constant at 0.05 M while [NO] increases from 0.05 M to 0.15 M, the rate increases from \( 1 \times 10^{-3} \) to \( 9 \times 10^{-3} \). Since the rate increases ninefold as [NO] triples,\( m = 2 \) indicates a second-order reaction with respect to \( [NO] \).
04

Determine the Rate Law

Using the determined orders from Steps 2 and 3, the rate law for the reaction is \( r = k[ ext{Cl}_2]^1[ ext{NO}]^2 \), indicating it is first-order with respect to \( \text{Cl}_2 \) and second-order with respect to \( \text{NO} \).

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Key Concepts

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

Reaction Rate
The reaction rate describes how quickly or slowly a reaction takes place. It is often measured by observing the change in concentration of a reactant or product over time. Using the example from our exercise, the rate is given in terms of molarity per second (mol/L/s).

The rate can be affected by several factors:
  • Concentration of reactants: Higher concentrations typically mean that particles have a higher chance of colliding with each other to react.
  • Temperature: Temperature increases can cause particles to move faster, thus resulting in more collisions.
  • Catalysts: These substances speed up the reaction without being consumed in the process.
  • Surface area: More surface area allows for more collisions and thus a higher reaction rate.
In the exercise, a change in concentration of either \(\text{NO}\) or \(\text{Cl}_2\) results in a change in the initial reaction rate observed.
Order of Reaction
Order of reaction is key to understanding how different concentrations of reactants affect the reaction rate. It helps us pinpoint the exact relationship between reactant concentration and the rate at which it changes. Each reactant can have a different order of reaction:
  • First-order: The rate is directly proportional to the concentration of a single reactant. For example, in our exercise, the reaction is first-order with respect to \(\text{Cl}_2\).
  • Second-order: The rate of reaction is proportional to the square of the concentration of one reactant. In our example, this is the case with \(\text{NO}\), as changing its concentration triples results in rates amplified by nine.
  • Overall order: This is the sum of the orders with respect to each reactant. In our case, the overall order is\( 1 + 2 = 3 \).
Understanding reaction orders allows scientists to predict how changing conditions can affect the speed of the reaction.
Rate Law Expression
The rate law expression mathematically relates the concentrations of reactants to the reaction rate. It is expressed as:\[ r = k[ A]^m[ B]^n \]Here, \( k \) is the rate constant, and \( [ A] \) and \( [ B] \) are the concentrations of the reactants, with their respective orders \( m \) and \( n \).In our example problem, the reaction is represented by the equation:\[ r = k[ \text{Cl}_2 ]^1[ \text{NO}]^2 \]Pay close attention to the powers on the reactants:
  • The first-order with respect to \(\text{Cl}_2\) indicates that the rate increases directly with an increase in \(\text{Cl}_2\) concentration.
  • The second-order with \(\text{NO}\) suggests the rate is highly sensitive to changes in \(\text{NO}\) concentration, rising exponentially when the amount is increased.
Correct rate laws are essential for chemists to manipulate and control reactions efficiently, whether in industrial settings or laboratory experiments.

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Most popular questions from this chapter

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)

Which of the following statements are correct about half-life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half-life period (2) time required for \(75 \%\) completion of a first-order reaction is double the half-life of the reaction (3) average life \(=1.44\) times the half-life for firstorder reaction

\(3 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\) It would be a zero order reaction when (a) the rate of reaction is proportional to square of concentration of \(\mathrm{A}\) (b) the rate of reaction remains same at any concentration of \(\mathrm{A}\) (c) the rate remains unchanged at any concentration of \(\mathrm{B}\) and \(\mathrm{C}\) (d) the rate of reaction doubles if concentration of is increased to double

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{AB}\) this following rate data were obtained at \(300 \mathrm{~K}\). $$ \begin{array}{llll} \hline \text { Expt } & \text { Concentration } & \text { Rate of disappearance } \\ & {[\mathbf{A}]} & {\left[\mathrm{B}_{2}\right]} & \text { of } \mathrm{B}_{2}\left(\mathrm{~mol} \mathrm{~L} \mathbf{~ m i n}^{-1}\right) \\ \hline 1 . & 0.015 & 0.15 & 1.8 \times 10^{-3} \\ 2 . & 0.09 & 0.15 & 1.08 \times 10^{-2} \\ 3 . & 0.015 & 0.45 & 5.4 \times 10^{-3} \\ \hline \end{array} $$ What is the rate law? (a) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]\) (b) \(r=[\mathrm{A}]^{2}\left[\mathrm{~B}_{2}\right]^{1}\) (c) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]^{2}\) (d) \(\mathrm{r}=k\left[\mathrm{~B}_{2}\right]\)

A gaseous compound decomposes on heating as per the following equation: \(\mathrm{A}(\mathrm{g}) \longrightarrow B(\mathrm{~g})+2 \mathrm{C}(\mathrm{g})\). After 5 minutes and 20 seconds, the pressure increases by \(96 \mathrm{~mm} \mathrm{Hg}\). If the rate constant for this first order reaction is \(5.2 \times 10^{-4} \mathrm{~s}^{-1}\), the initial pressure of \(\mathrm{A}\) is (a) \(226 \mathrm{~mm} \mathrm{Hg}\) (b) \(37.6 \mathrm{~mm} \mathrm{Hg}\) (c) \(616 \mathrm{~mm} \mathrm{Hg}\) (d) \(313 \mathrm{~mm} \mathrm{Hg}\)

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