Faraday Constant and Avogadro Number Voltaic Cell & Electrodes Lab Report

Lab: Chemical Equilibrium

 

OBJECTIVE

 

  1. To determine the shift in equilibrium position of a chemical reaction with applied stress.
  2. To calculate the equilibrium constant for the reaction.

 

INTRODUCTION

 

CHEMICAL EQUILIBRIUM

No chemical reaction goes to completion. When a reaction stops, some reactants remain. For example, although we write

 

2 CO2(g) word image 2592 2 CO(g) + O2(g) (reaction 1)

 

as though it goes entirely to products, at 2000 K only 2% of the CO2 decomposes. A chemical reaction reaches equilibrium when the concentrations of the reactants and products no longer change over time. The position of the equilibrium describes the relative amounts of reactants and products that remain at the end of a chemical reaction. The position of the equilibrium for reaction 1 is said to lie with the reactants, or to the left, because at equilibrium very little of the carbon dioxide has reacted. On the other hand, in the reaction

 

2 H2(g) + O2(g) word image 2593 2 H2O(g) (reaction 2)

 

the equilibrium position lies very far to the right since only very small amounts of H2 and O2 remain after the reaction reaches equilibrium. Since chemists often wish to maximize the yield from a reaction, it is vital to determine how to control the position of the equilibrium.

 

The equilibrium position of a reaction may shift if an external stress is applied. The stress may be in the form of a change in temperature, pressure, or the concentration of one of the reactants or products. For example, consider a flask with an equilibrium mixture of CO2, CO, and O2, as in reaction 1. If a small amount of CO is then injected into the flask, the concentration of CO2 increases. Here the external stress is the increase in concentration of CO. The system responds by reacting some of the added CO with O2 to yield an increased amount of CO2. That is, the position of equilibrium shifts to the left, yielding more reactant.

 

Because reaction 1 involves gaseous reactants and products, it also shifts with changes in pressure. Starting with reaction 1 at equilibrium, an increase in pressure causes the position of equilibrium to shift to the side of the reaction with the smaller number of moles of gas. That is, by shifting the equilibrium position to the left, the reaction decreases the number of moles of gas, thereby decreasing the pressure in the flask. In so doing, some of the applied stress is

1

 

relieved. On the other hand, an increase in pressure for reaction 2 shifts the equilibrium position to the right to decrease the number of moles of gas. The response of a reaction at equilibrium to changes in conditions is summarized by Le Châ e e c e:

A system perturbed from equilibrium shifts its equilibrium position to relieve the applied stress.

 

For reactions that undergoes a net change in energy, the net energy (symbolized by q) absorbed (in endothermic reactions) or released (in exothermic reactions) can be considered as part of reactants and products, respectively. The decomposition of CO2 into CO and O2 is endothermic, while the formation of ammonia (NH3) is exothermic.

 

Endothermic reaction: q + 2 CO2(g) word image 2594 2 CO(g) + O2(g) (reaction 1)

 

Exothermic reaction: N2(g) + 3 H2(g) word image 2595 2 NH3(g) + q (reaction 3)

 

For an increase in temperature, the reaction shifts in the endothermic direction to relieve the stress. The decomposition of CO2 (reaction 1) is endothermic in the forward direction. Upon an increase in temperature (i.e. adding heat q to the reaction), the equilibrium position shifts in the forward direction (making more CO and O2) to minimize the temperature increase by using up some of the heat added.

 

By contrast, a decrease in temperature causes heat to be removed from a reaction at equilibrium. The reaction will shift in the exothermic direction to produce more heat to minimize the heat loss. For reaction 1, a drop in temperature causes the reaction to shift left, forming more CO2 while decreasing the amounts of CO and O2.

 

THE IRON-THIOCYANATE EQUILIBRIUM

When potassium thiocyanate (KSCN) is mixed with iron(III) nitrate (Fe(NO3)3) in solution, an equilibrium mixture of iron(III) ion (Fe3+), thiocyanate ion (SCN ), and the iron thiocyanate complex ion ([FeSCN]2+) is formed:

 

Fe3+(aq) + SCN (aq) word image 2596 Fe(SCN)2+(aq)

 

yellow colorless blood red

 

The solution also contains the ions K+ and NO3 , but these are merely spectator ions and do not participate in this reaction. The relative amounts of the various ions participating in the reaction can be judged from the color of the solution since in neutral or slightly acidic solutions, Fe3+ is light yellow, SCN is colorless, and [FeSCN]2+ complex ion is blood red. If the solution is initially reddish, and the equilibrium shifts to the right (more [FeSCN]2+), the solution becomes darker red, while if the equilibrium shifts to the left (less [FeSCN]2+), the solution becomes lighter red or perhaps straw-yellow.

 

 

PART I DETERMINE THE SHIFT IN EQUILIBRIUM BASED ON LE CH TELIER S PRINCIPLE

Assume you start with an iron thiocyanate reaction system at equilibrium:

 

Fe3+(aq) + SCN (aq) word image 2597 Fe(SCN)2+(aq)

 

yellow colorless blood red

 

An external stress is applied to the equilibrium, and you will be provided with any necessary information regarding how one or more of the chemical species is affected. For each reaction, you will predict what would happen to the iron-thiocyanate equilibrium: shift left, shift right, or no effect, and the resulting change in color of the initial equilibrium solution: solution turns more yellow, solution turns more red, or no change to solution color.

 

As an example, if you added a small amount of concentrated HCl to the iron thiocyanate solution at equilibrium, the solution color lightens (becomes more yellow). This change in color indicates that the [FeSCN]2+ concentration decreases because the equilibrium shifts left. To explain this result, it is necessary to know that in the presence of a large excess of Cl (as a result of adding HCl), Fe3+ forms complex ions with Cl :

 

word image 2598 Fe3+(aq) + 6 Cl (aq) FeCl 63─(aq)

 

The increase in Cl reduces the free Fe3+ concentration as it reacts with Cl to form FeCl63─, so in accord with Le Ch e e principle, some [FeSCN]2+ dissociates to replace some of the Fe3+ removed by reaction with Cl . This result is summarized in the table below:

 

Stress

Observation

Explanation

 

Add conc. HCl

 

Solution turns yellow

 

Excess Cl due to addition of conc. HCl react with Fe3+ :

 

Fe3+ + 6 Cl œ FeCl63─

This results in a decrease in [Fe3+], causing some [FeSCN]2+ to dissociate and the solution turns more yellow (less red):

 

Fe3+ + SCN œ Fe(SCN)2+(aq)

 

yellow blood-red

 

 

Your assignment: For each of the stresses (ao f) given in Table A on page 4:

 

  1. Predict what the observation on the solution color would be: (a) solution turns more yellow, (b) solution turns more red, or (c) no effect on solution color.

 

  1. Provide a detailed explanation for the predicted observation you made in (1).

 

Use the example in the table above as a guide to how you should structure your answers.

 

 

 

 

TABLE A: OPERATIONS TO INTRODUCE AN EXTERNAL STRESS

 

Stress

Observation

Explanation

 

a) Add some Fe(NO3)3

 

 

 

 

 

 

 

 

 

 

 

b) Add some

KSCN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c) Add some SnCl2

(see note below)

 

 

 

 

 

 

 

 

 

 

 

 

 

Tin(II) chloride (SnCl2) produces Sn2+ ion, which reduces iron(III) ion to iron(II) ion:

 

Sn2+(aq) + 2 Fe3+(aq) œ Sn4+(aq) + 2 Fe2+(aq)

 

 

d) Add some AgNO3

(see note below)

 

 

 

 

 

 

 

 

 

 

 

Silver nitrate (AgNO3) produces Ag+ ion, which reacts with thiocyanate ion to give a white precipitate of silver thiocyanate: Ag+(aq) + SCN(aq) œ AgSCN(s)

 

 

e) Add some Na2HPO4

(see note below)

 

 

 

 

 

 

 

 

 

 

 

 

Sodium hydrogen phosphate (Na2HPO4) produces hydrogen phosphate ion (HPO42─), which reacts with Fe3+ to form a complex ion: Fe3+(aq) + HPO42─(aq) œ [FeHPO4]+(aq)

 

TABLE A (CONTINUED)

 

Stress

 

Observation

Explanation

 

f) Add some NH3

(see note below)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ammonia (NH3) is a base that reacts with water to form hydroxide ion (OH) :

 

NH3(aq) + H2O(l) œ NH4+(aq) + OH(aq)

 

The hydroxide ion in turns reacts with Fe3+ to form a precipitate:

 

Fe3+(aq) + 2 OH(aq) œ Fe(OH)3(s)

 

 

TABLE B: EFFECTS OF TEMPERATURE AND PRESSURE ON EQUILIBRIUM

The iron-thiocyanate reaction has a standard enthalpy of reaction shown below:

 

Fe3+(aq) + SCN (aq) word image 2599 [FeSCN]2+(aq) ‘Hq = ─26 kJ/mol

 

yellow colorless blood red

 

Answer each of the following questions by circling the correct choice:

 

  1. Increasing the temperature of the reaction will result in a shift in the reaction equilibrium in

 

the endothermic / exothermic direction, resulting in the formation of more Fe3+ / Fe(SCN)2+

(circle the answer) (circle the answer)

 

  1. Decreasing the temperature of the reaction will cause the color of the solution to turn

 

more yellow / more red / no effect on solution color

(circle the answer)

 

  1. If the reaction solution were enclosed in a piston-cylinder apparatus, and the piston is now pressed such that the solution is placed under higher pressure, the color of the solution will

 

turn more yellow / more red / no effect on solution color

(circle the answer)

 

 

 

 

 

PART II DETERMINE THE FORMATION CONSTANT (Kf) OF THE IRON THIOCYANATE COMPLEX ION

Quantitatively, the relative amounts of the two reactants and the product are related by the equilibrium constant of the reaction; in this case, the formation constant (Kf) :

 

Fe3+(aq) + SCN (aq) word image 2600 Fe(SCN)2+(aq)

2+]eq

[Fe(SCN)

Kf =

[Fe3+]eq [SCN ] eq

To completely characterize this reaction, it is necessary to know the value for Kf. Kf can be calculated through an experimental determination of the equilibrium concentration of the complex ion, [Fe(SCN)2+]eq, in equilibrium with [Fe3+]eq and [SCN ]eq. You will use a standard curve to determine the [Fe(SCN)2+]eq. The equilibrium concentrations of the other two ions is determined using the 1:1 stoichiometry of the reaction:

 

Fe3+(aq) + SCN (aq) word image 2601 Fe(SCN)2+(aq)

 

Initial [Fe3+]i [SCN ]i

 

Change [Fe(SCN)2+]eq [Fe(SCN)2+]eq + [Fe(SCN)2+]eq

 

Equilibrium [Fe3+]i [Fe(SCN)2+]eq [SCN ]i [Fe(SCN)2+]eq [Fe(SCN)2+]eq

 

Where: [Fe3+]i and [SCN ]i are the initial concentrations

[Fe3+]eq, [SCN ]eq, and [Fe(SCN)2+]eq are the concentrations at equilibrium

 

Part II of this lab consists of two assignments:

 

  1. The first assignment is to generate a standard curve (see below) by plotting the absorbance versus the concentration of the red iron(III) thiocyanate complex using Excel.

 

  1. The second assignment is to use the linear regression equation generated by Excel to determine the equilibrium concentrations of Fe3+, SCN , and Fe(SCN)2+ ions from which the formation constant (Kf) for the reaction will be determined.

 

ASSIGNMENT 1: STANDARD CURVE AND BEER-LAMBERT S LAW

Use of a standard curve is a common experimental strategy in chemistry to determine the concentration of an unknown solution. In this technique, a series of solutions with known concentrations is prepared and then a parameter such as absorbance is measured. This parameter is then plotted against concentration to yield the standard curve, which is often a straight line, with some degree of scatter caused by experimental error (see Fig 1 on page 7). Regression analysis of the data using the method of least squares allows determination of the best fit line. Curve fitting is easily accomplished with Excel, which not only provides the equation of the best-fit line but also provides information as to the quality of the straight line with a regression coefficient, R2. The R2 value ranges from 0 to 1.0, with 1.0 indicating a perfect fit.

Subsequent measurement of the absorbance in an unknown sample allows determination of the unknown concentration through the equation of the line.

 

Beer s La plo of Absorbance ers s Concen ra ions

Absorbance

word image 2602

1.2

0

 

 

 

1.0

0

 

 

0.80

 

 

0.60

 

 

0.40

 

 

0.20

 

 

0

 

 

5.0 10.0 15.0 20.0 25.0

 

Concentration (M)

 

Figure 1: A ical anda d c e ba ed on Bee La . Bee La i mo eliable fo ab o bance values between 0.1 and 1. Unknown concentrations of a given chromophore can be determined using the equation of the linear trend line.

 

Example: Le assume the standard curve shown in Fig 1 is that of absorbance produced by a series of Co2+ standard solutions at 25 qC at the indicated concentrations. The plot is generated by Excel with the best fit equation of line given as y = 0.0416 x ─ 0.0066. If a solution of Co2+ ion with unknown concentration gives an absorbance of 0.75, its concentration can be determined using the equation of line as follows:

 

y = 0.0416 x ─ 0.0066 translates into: A = 0.0416 [Co2+] ─ 0.0066

 

[Co2+] = = A + 0.0066 0.75 + 0.0066

0.0416 0.0416

 

[Co2+] = 18 M

 

To learn more about Bee La and spectrophotometry, please view the following 8-minute YouTube video: https://www.youtube.com/watch?v=QPHo5lFWgT0

 

For a quick tutorial on how to make a best-fit line graph on Excel, please view the following 6minute YouTube video: https://www.youtube.com/watch?v=d65jx4BhslA

 

 

Beer’s law has many forms, the most common of which is: A = ˜l˜c

Where: A is the absorbance (no units) is the molar absorption coefficient (also called molar absorptivity) with units

M─1˜cm─1

l is the path length of the cuvette with units usually in cm. For many commonly used lab cuvettes, l = 1 cm c is the concentration in mol/L (M)

 

Note that A = ˜l˜c essentially has the form of an equation of straight line (y = m˜x + b) in which the y-intercept (b) is equal to 0 (i.e. passing through the origin) and the slope is . Hence, the slope of a plot of absorbance vs. concentration is the molar absorption coefficient. The y-

intercept for Bee La d be a he g 0 ce a 0 M c ce a , he ab orbance is

theoretically zero. [Considering A has no units and c has units of M, what is the units of the slope ? ]

 

The standard curve that you will generate is based on the red color of the Fe(SCN)2+eq ion. By making up a series of solutions containing known amounts of this chromophore (a substance that absorbs light) and using Beer-La be La ( Bee La , f h ), e ca relate the resulting experimentally determined values of absorbance to the known concentrations. The intensity of the red color Fe(SCN)2+ ion is directly proportional to its equilibrium concentration and is measured using a spectrophotometer.

 

To plot a standard curve, the concentration of Fe(SCN)2+ in a series of solutions must be accurately known. The concentration of Fe(SCN)2+ in a solution can be calculated using the formation constant, Kf. However, the formation constant is not known in this case. Rather, the purpose of this experiment is to calculate the formation constant, which leaves us in a bit of a quandary. How can we create a series of solutions of known Fe(SCN)2+ concentrations without knowing the equilibrium constant ? We can use our knowledge of Le Ch e e c e determine how to force the equilibrium position essentially to completion, so that the stoichiometry of the reaction alone can be used to find the concentrations of Fe(SCN)2+. Le look again at the equilibrium reaction for the formation of iron thiocyanate ion:

 

Fe3+(aq) + SCN (aq) word image 2603 Fe(SCN)2+(aq)

 

What stress can we apply to force the equilibrium very far to the right, such that one of the reactants with a known initial concentration is converted to completion to product ? This can be accomplished by making one reactant the limiting reagent and setting the other reactant at a large excess amount. For example, if the initial concentration of Fe3+ is set at 100 times or higher than that of SCN , the equilibrium will be forced almost completely to product, and [Fe(SCN)2+]eq [SCN ]i because of the 1:1 stoichiometry. See example at the top of page 9 that illustrates how to determine the [Fe(SCN)2+]eq from a solution mixture.

 

 

Example: You set up a reaction for the formation of Fe(SCN)2+ by mixing 5.00 mL of 0.100 M Fe(NO3)3 solution with 1.00 mL of 0.00100 M KSCN solution at 25 qC. You then add distilled water to the solution mixture to make a total final volume of 10.00 mL. Calculate the equilibrium concentration of the iron-thiocyanate complex ion that forms at 25 qC.

 

Solution: Recall that in aqueous solution, Fe(NO3)3 dissociates to form Fe3+ and NO3 ions, and KSCN forms K+ and SCN ions. K+ and NO3 ions are spectator ions and can be ignored. Fe3+ and SCN ions react to form the iron-thiocyanate complex Fe(SCN)2+. Because the solution mixture has a total volume of 10.00 mL, we first determine the initial concentrations of Fe3+ and SCN ions by simple dilution calculations:

 

[Fe3+]i = 5.00 mL x 0.100 M = 0.0500 M Fe3+ (3 sig figs)

10.00 mL

 

[SCN]i = 1.00 mL x 0.00100 M = 0.000100 M SCN (3 sig figs)

10.00 mL

 

We now can set up an ICE table that shows what is happening in an equilibrium reaction:

 

Fe3+ SCN Fe(SCN)2+

 

Initial 0.0500 M 0.000100 M 0

 

Change 0.000100 M 0.000100 M + 0.000100 M

 

Equilibrium 0.0499 M ~ 0 ~ 0.000100 M

 

Because [Fe3+]i is 500 times (0.0500 M / 0.000100 M = 500) larger than [SCN]i, the equilibrium lies far to the right and it can be assumed that the limiting reactant (SCN) is completely converted to product. Hence, at equilibrium: [Fe(SCN)2+]eq [SCN]i = 0.000100 M

[For the formation of iron-thiocyanate reaction, if the initial concentration of one reactant is at least 100 times larger than that of the other reactant, we can assume that the reaction is driven to completion to the right.]

 

Your assignment 1: Create a standard curve of Absorbance versus standard solutions of Fe(SCN)2+. Table C on page 10 shows a list of six solution mixtures and their corresponding data, with proper sig figs, that you will need to perform the calculations required to complete Table D. (Solution 1 does not contain Fe(SCN)2+ and is used as a blank to zero the spectrophotometer). You will then use the data entered in Table D to plot the standard curve using Microsoft Excel. All calculation data must have proper units and rounded off to correct sig figs for full credit. (0.2 pt is deducted for each mistake in significant figs and for any number, where applicable, without proper units.)

 

 

TABLE C: STANDARD CURVE DATA

 

Solution Mixture

 

 

Volume of

0.200 M Fe(NO3)3

 

Volume of

0.00200 M KSCN

 

Volume of distilled water

 

Total volume

1

2.50 mL

0.00 mL

7.50 mL

10.00 mL

2

2.50 mL

0.50 mL

7.00 mL

10.00 mL

3

2.50 mL

0.75 mL

6.75 mL

10.00 mL

4

2.50 mL

1.00 mL

6.50 mL

10.00 mL

5

2.50 mL

1.25 mL

6.25 mL

10.00 mL

6

2.50 mL

1.50 mL

6.00 mL

10.00 mL

 

Solution 1 has only Fe3+ and no SCN (no product Fe(SCN)2+ is formed). It is used to blank the spectrophotometer.

Under actual experimental conditions, the solutions are acidified with nitric acid to prevent some side reactions of Fe3+ from occurring. For simplicity purpose in this virtual lab, we can ignore this requirement.

 

TABLE D: CONCENTRATIONS FOR STANDARD CURVE

 

Solution Mixture

 

 

[Fe3+]i (= initial concentration of Fe3+)

 

[SCN ]i (= initial concentration of SCN )

 

[Fe(SCN)2+]eq (= equilibrium concentration of Fe(SCN)2+])

 

Absorbance

 

1

 

 

 

 

0.0 M

 

0.0 M

 

0.0000

 

2

 

 

 

 

 

0.4390

 

3

 

 

 

 

 

0.6543

 

4

 

 

 

 

 

0.8700

 

5

 

 

 

 

 

1.0860

 

6

 

 

 

 

 

1.3011

 

Your tasks:

 

  1. Using data in Table C, calculate the initial concentrations of Fe3+ and SCN ions, and the equilibrium concentration of Fe(SCN)2+ complex ion. Enter the results in Table D. For full credit, set the answers in correct sig figs and with proper units.

 

  1. On page 11, provide sample calculations in the blank space provided using data of Solution Mixture 2 in Table C to show how you obtained the results you entered in Table D.

 

  1. Use the data in Table D and Excel to plot a Standard Curve of Absorbance versus [Fe(SCN)2+]eq (similar to the graph shown on page 7). Note that Absorbance data are given in the table.

 

Your plot must have the following:

 

    1. A proper title

 

    1. All axes must be properly labelled with correct units. (Label y-axis as Absorbance and x-axis as [Fe(SCN)2+] with units M.) Note that [0.0] should be included as a data point.

 

    1. Display of the best-fit line.

 

    1. The equation of the best-fit line for the plot.

 

If you are unfamiliar with Excel plotting software, see YouTube video link: https://www.youtube.com/watch?v=d65jx4BhslA for a quick tutorial on how to plot a best-fit curve using Excel.

 

Print out a copy of the Standard Curve and attach it to your lab report, or save it and attach an electronic copy with your lab report when you send it in.

 

Sample Calculations Using Data of Solution Mixture 2 for [Fe3+]i, [SCN ]i, and

[Fe(SCN)2+]eq.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ASSIGNMENT 2: DETERMINATION OF THE FORMATION CONSTANT, Kf, FOR THE IRONTHIOCYANATE COMPLEX ION, Fe(SCN)2+.

After a standard curve is produced, the conditions will be altered so that the concentrations of each of the two reacting species (Fe3+ and SCN ) will be the same order of magnitude (~0.00200 M each). Because the concentrations will be so similar, the system will no longer be forced all the way to the right (towards the product) and you will be able to determine an equilibrium constant from the data. The concentration of Fe(SCN)2+ at equilibrium will be determined based on its spectrophotometrically measured absorbance and the use of the bestfit equation of the standard curve. Since for every mole of the red complex Fe(SCN)2+ produced, one mole of Fe3+ and one mole of SCN will have reacted, the equilibrium concentrations (unreacted species) of Fe3+ and SCN can be determined by subtracting the concentration of Fe(SCN)2+ formed from the initial concentrations before the reaction took place. We can set up an ICE table, find the equilibrium concentrations for each of the three species, and solve for Kf. See sample calculation in the example shown below.

 

Example: Assume a solution mixture containing an initial concentration of Fe3+ = 0.00100 M and an initial concentration of SCN = 0.000600 M. The spectrophotometrically measured absorbance of the formed Fe(SCN)2+ in this solution mixture is 0.3094 at 25 qC. Le a assume the best-fit equation of the standard curve is y = 1250 x + 0.0288. Determine the equilibrium concentrations of Fe3+, SCN , and Fe(SCN)2+, and the formation constant, Kf, of the iron-thiocyanate reaction.

 

Solution: y = 1250 x + 0.0288 translates into: A = 1250 [Fe(SCN)2+]eq + 0.0288

 

[Fe(SCN)2+]eq = A 0.0288 = 0.3094 0.0288

1250 1250

 

[Fe(SCN)2+]eq = 2.24 x 10 4 M

 

Knowing [Fe(SCN)2+]eq, we now set up an ICE table to determine the equilibrium concentrations of Fe3+ and SCN :

 

Fe3+ SCN Fe(SCN)2+

 

Initial 0.00100 M 0.000600 M 0

 

Change 2.24 x 10 4 M 2.24 x 10 4 M + 2.24 x 10 4 M

 

Equilibrium 7.76 x 10 4 M 3.76 x 10 4 M 2.24 x 10 4 M

 

We can now determine the formation constant, Kf, for the reaction:

SCN)

=

Kf = [Fe3+](eq [SCN2+]eq] eq (7.76 x 102.244 M) (3.7 x 1064 x M10 4 M) = 768

[Fe

Note: This is not the actual Kf value for the formation of Fe(SCN)2+. The data given here are only for illustration.

 

DATA FOR ASSIGNMENT 2

Five solution mixtures are set up containing a constant amount of Fe3+ and varying amounts of SCN . The total volume of the solutions is 10.00 mL, as shown in Table E below.

 

TABLE E: EQUILIBRIUM DATA

 

Solution Mixture

 

 

Volume of

0.00200 M Fe(NO3)3

 

Volume of

0.00200 M KSCN

 

Volume of distilled water

 

Total volume

1

5.00 mL

1.00 mL

4.00 mL

10.00 mL

2

5.00 mL

2.00 mL

3.00 mL

10.00 mL

3

5.00 mL

3.00 mL

2.00 mL

10.00 mL

4

5.00 mL

4.00 mL

1.00 mL

10.00 mL

5

5.00 mL

5.00 mL

0.00 mL

10.00 mL

 

Your tasks:

 

  1. Using data of Table E, calculate the initial concentrations of Fe3+ and SCN in the 10-mL solution mixtures. (Note: you will need to perform dilution calculations.) Enter the results in Table F below. For full credit, set the answers in correct sig figs and with proper units.

 

  1. Using the absorbance data provided in Table F and the best-fit equation of line that you obtained for your standard curve (from assignment 1), calculate the equilibrium concentrations of Fe(SCN)2+ for the solution mixtures and enter the results in Table F.

 

On the top half of page 14, provide sample calculations using data of solution mixture 1 to show how you obtained the [Fe3+]i, [SCN ]i, and [Fe(SCN)2+]eq.

 

TABLE F: CONCENTRATIONS FOR EQUILIBRIUM CALCULATIONS

 

Solution Mixture

 

 

[Fe3+]i (= initial concentration of Fe3+)

 

[SCN ]i (= initial concentration of SCN )

 

[Fe(SCN)2+]eq (= equilibrium concentration of Fe(SCN)2+])

 

Absorbance

 

1

 

 

 

 

 

0.1080

 

2

 

 

 

 

 

0.2071

 

3

 

 

 

 

 

0.3031

 

4

 

 

 

 

 

0.3959

 

5

 

 

 

 

 

0.4828

 

 

 

  1. Show sample calculations for [Fe3+]i and [SCN ]i using data of Solution Mixture 1 from Table E.
  2. Use provided absorbance value for Solution Mixture 1 in Table F to show how you obtained [Fe(SCN)2+]eq.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EQUILIBRIUM RESULTS AND CALCULATION OF THE FORMATION CONSTANT, Kf.

Your tasks: 1) For each of the Solution Mixtures 1 o 5, an ICE table is provided as a first step to help you determine the equilibrium concentrations of Fe3+ and SCN . Note that the equilibrium concentration of Fe(SCN)2+ has been previously determined and tabulated in Table F. 2) In the space below the table, calculate the formation constant, Kf, using the equilibrium data. Round off all numerical values to 3 sig figs.

 

ICE Table Solution Mixture 1

 

 

[Fe3+]

 

 

[SCN ]

 

[Fe(SCN)2+]

 

Initial

 

 

 

 

0.00 M

 

Change

 

 

 

 

 

Equilibrium

 

 

 

 

 

Calculation of Kf #1:

 

 

 

 

ICE Table Solution Mixture 2

 

 

[Fe3+]

 

 

[SCN ]

 

[Fe(SCN)2+]

 

Initial

 

 

 

 

0.00 M

 

Change

 

 

 

 

 

Equilibrium

 

 

 

 

 

Calculation of Kf #2:

 

 

 

 

 

ICE Table Solution Mixture 3

 

 

[Fe3+]

 

 

[SCN ]

 

[Fe(SCN)2+]

 

Initial

 

 

 

 

0.00 M

 

Change

 

 

 

 

 

Equilibrium

 

 

 

 

 

Calculation of Kf #3:

 

 

 

 

 

ICE Table Solution Mixture 4

 

 

[Fe3+]

 

 

[SCN ]

 

[Fe(SCN)2+]

 

Initial

 

 

 

 

0.00 M

 

Change

 

 

 

 

 

Equilibrium

 

 

 

 

 

Calculation of Kf #4:

 

 

 

 

ICE Table Solution Mixture 5

 

 

[Fe3+]

 

 

[SCN ]

 

[Fe(SCN)2+]

 

Initial

 

 

 

 

0.00 M

 

Change

 

 

 

 

 

Equilibrium

 

 

 

 

 

Calculation of Kf #5:

 

 

 

 

 

 

Average formation constant, Kf.

Enter the calculated values of Kf in the table below. Then calculate the average value of Kf.

 

Kf #1

 

 

Kf #2

 

Kf #3

 

Kf #4

 

Kf #5

 

Average Kf

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

POST-LAB QUESTION

 

Consider the following reaction: 3 N2H4(g) + 4 ClF3(g) œ 12 HF(g) + 3 N2(g) + 2 Cl2(g)

 

A mixture, initially consisting of 0.880 M N2H4(g) and 0.880 M ClF3(g), reacts at a certain temperature. At equilibrium, the concentration of N2(g) is 0.525 M. Calculate the concentrations of all the other reactant and product species (N2H4, ClF3, HF, and Cl2) at equilibrium. For credit, show all calculation set ups leading to the answers. Your answers should have correct units and sig figs.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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