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Chapter 10: Problem 139

For a zero order reaction, the plot of concentration versus time is linear with (a) positive slope with zero intercept (b) positive slope with non-zero intercept (c) negative slope with non-zero intercept (d) parallel to time axis.

Short Answer

Expert verified
(c) negative slope with non-zero intercept.

Step by step solution

01

Understand Zero Order Reactions

In a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactants. This means that the rate at which reactants are converted to products does not change over time, as long as there is some reactant present.
02

Determine the Integrated Rate Law

For a zero-order reaction, the integrated rate law is:\[[A] = [A]_0 - kt\]where \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
03

Analyze the Graph Characteristics

The formula \([A] = [A]_0 - kt\) indicates a linear relationship between concentration \([A]\) and time \(t\) with a slope of \(-k\) and a y-intercept of \([A]_0\). Thus, the graph of concentration vs. time for a zero-order reaction is a straight line with a negative slope.
04

Choose the Correct Option

Since the graph has a negative slope (because of \(-k\)) and a non-zero intercept (\([A]_0\)), the correct choice is:(c) negative slope with non-zero intercept.

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Key Concepts

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

Integrated Rate Law
The integrated rate law provides a mathematical relationship that helps to describe the concentration of a reactant over time in a chemical reaction. For zero-order reactions, this relationship is particularly straightforward. The reaction rate is constant regardless of the concentration of reactants. The formula used to express the integrated rate law for a zero-order reaction is:\[[A] = [A]_0 - kt\]- \[A\] is the concentration of the reactant at time \(t\).- \[A_0\] represents the initial concentration of the reactant.- \(-k\) is the reaction rate constant with a negative sign showing the decrease in concentration over time.This equation reflects how, in a zero-order reaction, the concentration decreases linearly over time. As such, it provides a simple and effective way to predict the concentration of reactants at any given time during the reaction.
Reaction Rate Constant
The reaction rate constant, denoted as \(k\), is a crucial factor in the study of chemical kinetics. It essentially determines the speed of a reaction. In the context of zero-order reactions, \(k\) is constant. This means that the reaction rate remains unaltered as the reactants are consumed. The constant nature of \(k\) in zero-order reactions indicates that external factors like temperature or catalyst presence may affect the rate, but the concentration of reactants does not. In the integrated rate law, \(-k\) acts as the slope when concentration is plotted against time. This slope's constant value is what guarantees the linear decrease in concentration in a zero-order reaction, making it predictable and easy to calculate.Understanding how \(k\) functions can help you predict how quickly a reaction will proceed and how long it will take for a reactant to be used up under certain conditions.
Graph of Concentration vs. Time
In zero-order reactions, the graph of concentration vs. time is unique and simple to interpret due to its linear pattern. The integrated rate law equation \[A] = [A]_0 - kt\] clearly indicates a linear decrease in reactant concentration over time.
  • The graph is a straight line, which makes it easier to identify and differentiate from graphs of other reaction orders.
  • The slope of the line is \(-k\), highlighting the consistent reaction rate irrespective of concentration changes.
  • The y-intercept is \[A]_0\], showing the initial concentration from which the reaction begins.
The graph's linearity simplifies the task of tracking how reactant concentration decreases and allows for easy estimation and prediction of future points in the reaction timeline. This clarity is what makes zero-order reaction kinetics particularly approachable for anyone trying to master chemical kinetics.

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Most popular questions from this chapter

The experimental data for the reaction \(2 \mathrm{~A}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) is \(10.2\) Table \begin{tabular}{llll} \hline Exp. & [A] & [B_ ] & Rate \(\left(\mathrm{Ms}^{-1}\right)\) \\ \hline 1. & \(0.50 \mathrm{M}\) & \(0.50 \mathrm{M}\) & \(1.6 \times 10^{-4}\) \\ \(2 .\) & \(0.50 \mathrm{M}\) & \(1.00 \mathrm{M}\) & \(3.2 \times 10^{-4}\) \\ \(3 .\) & \(1.00 \mathrm{M}\) & \(1.00 \mathrm{M}\) & \(3.2 \times 10^{-4}\) \\ \hline \end{tabular} the rate equation for the above data is (a) rate \(=\mathrm{k}\left[\mathrm{B}_{2}\right]\) (b) rate \(=k\left[\mathrm{~B}_{2}\right]^{2}\) (c) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}\) (d) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)

The energy of activation and specific rate constant for a first order reaction at \(25^{\circ} \mathrm{C}\) are \(100 \mathrm{~kJ} / \mathrm{mol}\) and \(3.46\) \(\times 10^{-5} \mathrm{sec}^{-1}\) respectively. Determine the temperature at which half life of reaction is 2 hour. \(2 \mathrm{~N}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) \(\begin{array}{ll} \left.\text { (in } \mathrm{CCl}_{4}\right) & \left.\text { (in } \mathrm{CCl}_{4}\right)\end{array}\) (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)

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) \(313 \mathrm{~mm} \mathrm{Hg}\)

The following data pertains to the reaction between \(\mathrm{A}\) and \(\underline{B}\) \begin{tabular}{llll} \multicolumn{4}{c} { Table \(10.5\)} \\ \hline S. No. & {\([\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}\)} & Rate mol \(\mathrm{L}^{-1} \mathrm{~S}^{-1}\) \\ \hline 1. & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ 2\. & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ 3\. & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3

For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of \(10 \mathrm{~min}\) from the start the total optical, rotation of the system was \(50^{\circ} \mathrm{C}\) and when the reaction is complete it was \(100^{\circ}\). Assuming that only \(\mathrm{B}\) and \(\mathrm{C}\) are optically active and dextro rotator, the rate constant of this first order reaction would be (a) \(6.9 \mathrm{~min}^{-1}\) (b) \(0.069 \mathrm{~min}^{-1}\) (c) \(6.9 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(0.69 \mathrm{~min}^{-1}\)
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