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The following data were collected for a reaction known to be pseudofirst order in analyte, \(A\), during the time in which the reaction is monitored. $$ \begin{array}{cc} \text { time }(s) & {[A]_{t}(\mathrm{mM})} \\ \hline 2 & 1.36 \\ 4 & 1.24 \\ 6 & 1.12 \\ 8 & 1.02 \\ 10 & 0.924 \\ 12 & 0.838 \\ 14 & 0.760 \\ 16 & 0.690 \\ 18 & 0.626 \\ 20 & 0.568 \end{array} $$ What is the rate constant and the initial concentration of analyte in the sample?

Short Answer

Expert verified
The rate constant \(k\) is determined by the slope of the \(\ln [A]_t\) vs. time plot, and the initial concentration \([A]_0\) is found from the y-intercept.

Step by step solution

01

Understand Pseudo-first Order Kinetics

A pseudo-first order reaction occurs when one reactant is in excess, causing the reaction rate to depend only on the concentration of the other reactant. For such reactions, the rate law is given by the equation, \(\ln [A]_t = \ln [A]_0 - kt\), where \([A]_t\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
02

Plot \( \ln [A]_t \) versus Time (s)

Convert the concentration values, \([A]_t\), into their natural logarithms. Next, plot \( \ln [A]_t \) on the y-axis against time on the x-axis. This should yield a straight line if the reaction is pseudo-first order.
03

Calculate the Slope of the Line

The slope of the line obtained in Step 2 is equal to \(-k\) for the linear equation \(\ln [A]_t = \ln [A]_0 - kt\). Use two points from the line to find the slope, say at times \(t_1 = 2 \text{ s}\) and \(t_2 = 4 \text{ s}\). Compute \(slope = \frac{\ln [A]_{t_2} - \ln [A]_{t_1}}{t_2 - t_1}\).
04

Extrapolate to Find Initial Concentration

The y-intercept of the line in the plot corresponds to \(\ln [A]_0\), from which you can find \([A]_0\). Use the line equation obtained in Step 3 to solve for \([A]_0\).
05

Compute the Rate Constant

From the slope calculated in Step 3, derive the rate constant \(k = -\text{(slope)}\). This gives the reaction rate's rate constant.

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Key Concepts

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

Rate Constant
The rate constant, often represented by the symbol \(k\), is a crucial factor in determining the speed of a chemical reaction. In pseudo-first order kinetics, where one reactant is in large excess, the reaction depends primarily on the concentration of the other reactant.
The rate constant gives us insight into how swiftly the reaction proceeds. It is calculated using the slope of the line from a plot of \(\ln [A]_t\) against time, as described in the step by step solution.
If the plot yields a straight line, the slope of that line directly correlates to the negative rate constant \(-k\).
  • Calculate \(-k\) as the slope from the plot.
  • The steeper the slope, the larger the rate constant, indicating a faster reaction.
To find \(k\), simply take the negative of this slope, turning what you gathered from plotting and applying linear regression into a concrete measure of reaction speed.
Initial Concentration
The initial concentration, \([A]_0\), refers to the concentration of the reactant \(A\) at the very start of the reaction. It's fundamental in establishing the baseline for understanding how concentrations change as the reaction proceeds.
In pseudo-first order reactions, determining the initial concentration is key because it allows you to see how much of the reactant is there to contribute to the reaction rate.
The y-intercept of the graph from the plot of \(\ln [A]_t\) versus time represents \(\ln [A]_0\). This helps us back-calculate \([A]_0\) by taking the exponential of the y-intercept:
  • Use the equation \( [A]_0 = e^{\ln [A]_0} \).
  • This calculation allows for determining \([A]_0\) accurately from the linear regression line.
Understanding \([A]_0\) will give you a clear picture of where the reaction began, providing context for interpreting how the kinetics unfold over time.
Reaction Rate
The reaction rate describes how quickly or slowly a reaction occurs. In the case of pseudo-first order kinetics, the reaction rate depends on the concentration of the analyte \(A\), since the other reactant is in excess and does not significantly affect the rate.
This thin slice of chemistry reflects how often reactant molecules collide with enough energy to react, and how rapidly their concentration reduces over time.
The pseudo-first order rate law is expressed as:\[ ext{rate} = k[A] \]where:
  • \(k\) is the rate constant.
  • \([A]\) is the concentration of the reactant.
The dependence of the reaction rate on \([A]\) means that as you measure the reaction, you can predict and track the concentration of the reactant \(A\) over time.
This makes pseudo-first order kinetics a powerful concept for understanding a wide variety of chemical processes, allowing scientists and students alike to predict how fast a reaction will proceed with given reactant concentrations.

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Most popular questions from this chapter

Malmstadt and Pardue developed a variable time method for the determination of glucose based on its oxidation by the enzyme glucose oxidase. \({ }^{22}\) To monitor the reaction's progress, iodide is added to the samples and standards. The \(\mathrm{H}_{2} \mathrm{O}_{2}\) produced by the oxidation of glucose reacts with \(\mathrm{I}^{-}\), forming \(\mathrm{I}_{2}\) as a product. The time required to produce a fixed amount of \(I_{2}\) is determined spectrophotometrically. The following data was reported for a set of calibration standards $$ \begin{array}{rrrr} \text { [glucose] (ppm) } & & \text { time }(s) & \\ \hline 5.0 & 146.5 & 150.0 & 149.6 \\ 10.0 & 69.2 & 67.1 & 66.0 \\ 20.0 & 34.8 & 35.0 & 34.0 \\ 30.0 & 22.3 & 22.7 & 22.6 \\ 40.0 & 16.7 & 16.5 & 17.0 \\ 50.0 & 13.3 & 13.3 & 13.8 \end{array} $$ To verify the method a standard solution of 20.0 ppm glucose was analyzed in the same way as the standards, requiring \(34.6 \mathrm{~s}\) to produce the same extent of reaction. Determine the concentration of glucose in the standard and the percent error for the analysis.

The steady state activity for \({ }^{14} \mathrm{C}\) in a sample is 13 cpm per gram of carbon. If counting is limited to \(1 \mathrm{hr}\), what mass of carbon is needed to give a percent relative standard deviation of \(1 \%\) for the sample's activity? How long must we monitor the radioactive decay from a 0.50 -g sample of carbon to give a percent relative standard deviation of \(1.0 \%\) for the activity?

The enzyme acetylcholinesterase catalyzes the decomposition of acetylcholine to choline and acetic acid. Under a given set of conditions the enzyme has a \(K_{m}\) of \(9 \times 10^{-5} \mathrm{M}\) and a \(k_{2}\) of \(1.4 \times 10^{4} \mathrm{~s}^{-1}\). What is the concentration of acetylcholine in a sample if the reaction's rate is \(12.33 \mu \mathrm{M} \mathrm{s}^{-1}\) in the presence of \(6.61 \times 10^{-7} \mathrm{M}\) enzyme? You may assume the concentration of acetylcholine is significantly smaller than \(K_{m}\).

The enzyme urease catalyzes the hydrolysis of urea. The rate of this reaction is determined for a series of solutions in which the concentration of urea is changed while maintaining a fixed urease concentration of \(5.0 \mu \mathrm{M}\). The following data are obtained. $$ \begin{array}{cc} \text { [urea }](\mu \mathrm{M}) & \text { rate }\left(\mu \mathrm{M} \mathrm{s}^{-1}\right) \\ \hline 0.100 & 6.25 \\ 0.200 & 12.5 \\ 0.300 & 18.8 \\ 0.400 & 25.0 \\ 0.500 & 31.2 \\ 0.600 & 37.5 \\ 0.700 & 43.7 \\ 0.800 & 50.0 \\ 0.900 & 56.2 \\ 1.00 & 62.5 \end{array} $$ Determine the values of \(V_{\max }, k_{2}\), and \(K_{m}\) for urease.

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