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Experimental data are listed for the hypothetical reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\\{[\mathrm{~A}]} & 0.32 & 0.24 & 0.20 & 0.16 & 0.14 & 0.12 \\ \hline\end{array}$$ (a) Plot these data as in Figure \(11.2\). (b) Draw a tangent to the curve to find the instantaneous rate at \(30 \mathrm{~s}\). (c) Find the average rate over the 10 to \(40 \mathrm{~s}\) interval. (d) Compare the instantaneous rate at \(30 \mathrm{~s}\) with the average rate over the thirty-second interval.

Short Answer

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Question: Compare the instantaneous rate at 30 s and the average rate over the 30-second interval (from 10 s to 40 s) in a given reaction for the concentration of A.

Step by step solution

01

Plot the Data Points

Using the given data, plot concentration of A ([A]) versus time (s) on a graph paper or with a graphing software. Make sure to label the axes and add the data points properly.
02

Draw a Tangent at 30 s

On the plotted curve, draw a tangent line at time 30 s. To do this, you can use a ruler or a straight-edged object like an index card and make sure that the tangent line touches the curve only at the point corresponding to 30 s.
03

Determine the Instantaneous Rate at 30 s

The slope of the tangent line drawn at 30 s on the curve gives the instantaneous rate. To find the slope, choose any two points on the tangent line and use the formula: $$ slope = \frac{[A_2] - [A_1]}{t_2 - t_1} $$ where \([A_1]\) and \([A_2]\) are the concentrations at times \(t_1\) and \(t_2\). Record the value of the slope, which is the instantaneous rate at 30 s.
04

Calculate the Average Rate between 10 s to 40 s

To find the average rate over the interval 10 s to 40 s, use the following formula: $$ average \thinspace rate = \frac{[A_{40}] - [A_{10}]}{40 - 10} $$ where \([A_{10}]\) and \([A_{40}]\) are the concentrations at 10 s and 40 s, respectively. Using the data points given, calculate the average rate.
05

Compare the Instantaneous Rate and the Average Rate

Compare the values of the instantaneous rate at 30 s, which you found in step 3, and the average rate over the 30-second interval, which you found in step 4. Note any similarities or differences based on these values.

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Key Concepts

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

Instantaneous Rate
In reaction kinetics, understanding the concept of instantaneous rate is key for analyzing how quickly a reaction is proceeding at any given moment in time. The instantaneous rate specifically refers to the rate of reaction at a particular time. To determine this, we need to look at the slope of the tangent drawn to a concentration versus time plot at the specific point of interest.
For the given exercise, this is at 30 seconds. Drawing this tangent carefully ensures it only touches the curve at the desired time, which makes it different from other points.
To calculate the instantaneous rate, identify two points on the tangent line and apply the formula: \( ext{slope} = \frac{[A_2] - [A_1]}{t_2 - t_1} \). Here,
  • \([A_1]\) and \([A_2]\) represent the concentrations at times \(t_1\) and \(t_2\).
The resulting slope of this tangent gives you the rate at that precise moment, reflecting how fast the reactants are being used up or products formed at that specific time.
Average Rate
The average rate of a chemical reaction over a certain time interval gives us a general idea of how fast the reaction is proceeding. It does not reflect the changes that might occur at individual moments within the interval. Instead, it averages the overall change in concentration over the entire period.
From our exercise, we calculate the average rate from 10 seconds to 40 seconds using the formula: \[ ext{average rate} = \frac{[A_{40}] - [A_{10}]}{40 - 10} \] Here,
  • \([A_{40}]\) is the concentration of A at 40 seconds.
  • \([A_{10}]\) is its concentration at 10 seconds.
By determining this value, you obtain the average rate of reaction over the entire interval, showing how much the concentration has generally changed from the beginning to the end of the specified period.
Concentration versus Time Plot
A concentration versus time plot is a graphical representation that shows how the concentration of reactants or products changes over time in a chemical reaction. This kind of plot is an essential tool for visualizing reaction kinetics.
It forms the basis for analyzing both the instantaneous rate (through tangents) and the average rate (over intervals). To create this plot, one must place time on the x-axis and concentration on the y-axis, then accurately plot the data points provided.
For example, in the scenario from our exercise, you plot points based on given times and concentrations, such as 10 s and [A] = 0.24. Connecting these points helps illustrate the process and gives insight into how the concentration of the reactants or products decreases or increases over time.
This plot provides a visual tool to physically interpret where the reaction speeds up, slows down, or remains constant, which can be crucial for both beginners and advanced studies in chemical reactions.

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Most popular questions from this chapter

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

For the decomposition of HI, the activation energy is \(182 \mathrm{~kJ} / \mathrm{mol}\). The rate constant at \(850^{\circ} \mathrm{C}\) is \(0.0174 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{h}\). (a) What is the rate constant at \(700^{\circ} \mathrm{C} ?\) (b) At what temperature will the rate constant be a fourth of what it is at \(850^{\circ} \mathrm{C} ?\)

Consider the following hypothetical reaction: $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad \Delta H=-125 \mathrm{~kJ}$$ Draw a reaction-energy diagram for the reaction if its activation energy is \(37 \mathrm{~kJ} .\)

The greatest increase in the reaction rate for the reaction between \(\mathrm{A}\) and \(\mathrm{C}\), where rate \(=k[\mathrm{~A}]^{1 / 2}[\mathrm{C}]\), is caused by (a) doubling [A] (b) halving [C] (c) halving [A] (d) doubling [A] and [C]

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 M\) to \(0.00350 \mathrm{M} ?\) (c) What is the half-life of the hormone?

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