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Chapter 19: Q17P (page 488)

Simulating a Job’s plot. Consider the reaction A+2B⇌AB2,forwhichK=[AB2]/[A][B]2. Suppose that the following mixtures of A and B at a fixed total concentration of

10-4 M are prepared:

(b)Prepare a graph by the method of continuous variation in which you plot[AB2]versus mole fraction offor each equilibrium constant. Explain the shapes of the curves.

Short Answer

Expert verified

It is the ratio of number of mole of one component to the total number of moles in given mixture.

Step by step solution

01

Define mole fraction.

It is the ratio of number of mole of one component to the total number of moles in given mixture.

02

 Calculate mole fraction.

We will calculate mole fraction by=[A][A]+[B]

03

Draw spreadsheet of mole fraction.

04

Draw graph.

With that, we conclude that stoichmetry is1:2(AB2), so the maximum mole fraction will be 0.33. The higher the equilibrium constant, the steeper the curve and this will results in hig extent of reaction.

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Most popular questions from this chapter

What is the advantage of a time-resolved emission measurement with Eu3+versus measurement of fluorescence from organic chromophores?

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for CO2in the ocean was designed to operate without the need for calibration.33


The sensor compartment is separated from seawater by a silicone membrane through which CO2, but not dissolved ions, can diffuse. Inside the sensor, CO2equilibrates with HCO3and CO32. For each

measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can

write two mass balances:

[HIn−]+[ln2−]=FIn=50.0μMand[Na+]=FNα=50.0μM+42.0μM=92.0μM

has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio RA=A620/A434reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:

(a).From Beer’s law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that

[ln2−][Hln−]=RAε434HHn−−ε6,20Hln−ε620ln2−−RAε434ln2−=Rln (A)

(b) From the mass balance (1) and the acid dissociation constant

, show that

[Hln−]=F1nRln+1 (B)

[ln2−]=KlnFln[H+](Rln+1) (C)

(c) Show that H+=Kln/Rln (D)

(d) From the carbonic acid dissociation equilibria, show that

[HCO3−]=K1[CO(aq)]E[H+][CO32−]=K1K2[CO(aq)]F[H+]2

(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln,In2-,[HCO3], and[CO32−]

(f) Suppose that the various constants have the following values:

ε4344HHn−=8.00×103M−1cm−1    K1=3.0×10−7ε6620Hn−=0    K2=3.3×10−11ε434ln2=1.90×103M−1cm−1    Kln=2.0×10−7ε620ln2−=1.70×104M−1cm−1    Kw=6.7×10−15

From the measured absorbance ratio=2.84, findin the seawater.

(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?

When are isosbestic points observed and why?

The end of Box 19-2 states that “the advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.” Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

The spreadsheet gives the productεbfor four pure compounds and a mixture at infrared wavelengths. Modify Figure 19-4 to solve four equations and find the concentration of each compound. You can treat the coefficient matrix as if it were molar absorptivity because the path length was constant (but unknown) for all measurements.

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