Equilibrium Lab 2 PDF

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Equilibrium Lab Report 2...

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Chemical Equilibrium: Iron (III) with SCN Lab Report Introduction: The purpose of this lab is to observe the chemical equilibria between the reaction of Fe3 + ions and SCN- ions by looking at absorbance values through a spectrophotometer. A spectrophotometer emits a beam of light through a sample and then measures the intensity of light through the solution. This information can be used to calculate the absorbance and can be then used with Lambert-Beer’s law to calculate the concentration of a solution tested. Those concentration values can help predict what product was formed from a certain reaction, and can help determine equilibrium concentrations of products and reactants + Keq  values. Experimental: No deviations were made in experimental procedures. Data Analysis:

[Fe3+  ]i

Solution

[SCN]i

A

A/Ai

1

0.1M

0.001M

.295

1.000

2

0.01M

0.001M

.283

0.959

3

0.005M

0.001M

.232

0.786

4

0.0025M

0.001M

.157

0.532

5

0.00125M

0.001M

.077

0.261

The above table lists the amounts of each ion present in each beaker along with the absorbances and relative absorbances. Potential Product of ( FeSCN)2 2+ 

If (FeSCN)22+ was the product formed from the reaction, then the equation of the reaction would be as follows:

Fe3+ + SCN- ⇌ (FeSCN)22+ ICE charts can be used to determine the equilibrium concentrations of the ions involved in the reaction. An example calculation is below for Solution 1:

I

Fe3+ (M)

SCN- (M)

0.1

0.001

(FeSCN)22+ (M)

0

C

-X

-X

+X

E

0.1 - X

0

X

This calculation allowed for X to be easily determined as being equal to 1.00x10-3  , equivalent to the initial concentration of thiocyanate (we assume it is all consumed, as iron is in excess). Due to this, the equilibrium concentration of Fe3+  was determined to be 0.099 M, while that of [FeSCN]2+ is 0.001M.

The following equation can be used with the relative absorption values to determine the concentrations of (FeSCN)22+ in the other beakers: A 1/ A  2 =  C 1/ C  2 Example calculation for Solution 2: (1.00)/(.959) = (.001/C2) → C2 = .000959; this was repeated for beakers 3-5...

The equilibrium concentrations of Fe3+  and SCN- can be determined by subtracting the values of [FeSCN]2+ concentrations calculated from Lambert-Beer’s law from their respective initial concentrations.

The equilibrium constant, Keq, can be determined using the following equation:

Keq = [(FeSCN)2 2+]eq/([Fe3+]eq*[SCN-]eq)

Solution

A/A1

[(FeSCN)2 2+]eq

[Fe3+  ]eq

[SCN- ]eq

Keq

1

1.00

.001M

.099M

0.00

…

2

0.959

0.000959M

.00904M

.0000410

2590

3

0.786

0.000786M

.00421M

.000214

872

4

0.532

0.000532M

.00197M

.000468

577

5

0.261

0.000261M

.000989M

.000739

357

Stdev of Keq = 1020, 260 (without Solution 2) Mean of Keq = 1099, 602 (without Solution 2)

We assumed that all of the SCN- was consumed in Solution 1, and we must test this assumption now:

SCN- = [(FeSCN)22+]/Keq(avg)*[Fe3+] = (.001)/(.099)(602) = 1.68*10-5 % Unreacted: (1.68*10-5  )/(.001) x 100 = 1.68%

Potential Product of Fe(SCN) 3 Below is the chemical equation for equilibrium production of Fe(SCN)3: Fe3+ + 3SCN- ⇋ Fe(SCN)3 ICE charts were again useful for assessing this reaction. An example calculation for the reaction in Solution #1 is seen below:

Fe3+ (M)

SCN- (M)

Fe(SCN)3 (M)

I

.10

.001

0

C

-X

-3X

+X

E

.10-X

0

X

Given that .001-3X is 0 (an assumption made because Fe3+  was in excess), X = .00033. Thus, the equilibrium concentration of Fe(SCN)3 +  is .00033M. The following equation can be used

with the relative absorption values to determine the concentrations of Fe(SCN)3  in the  other beakers: A 1/ A  2 =  C 1/ C  2 Example for Solution 2: (1.000)/(.959) = (.00033)/(C2) → C2 = .000316; this was repeated for beakers 3-5... The equilibrium concentrations of Fe3+  and SCN- can be determined by subtracting the values of Fe(SCN)3 concentrations calculated from Lambert-Beer’s law from their respective initial concentrations.

The equilibrium constant, Keq, can be determined using the following equation:

Keq = [Fe(SCN)3]/([Fe3+  ]eq*[SCN- ]3 eq)

For possible product Fe(SCN)3: Solution

A/A1

[Fe(SCN)3]eq

[Fe3+  ]eq

[SCN- ]eq

Keq

1

1.00

.000333M

.00967

0.00

…

2

0.959

0.000316M

.00968

0.000052 2.32E10

3

0.786

0.000262M

.00474

.000921

7.08E7

4

0.532

0.000177M

.00232

.000469

7.40E8

5

0.261

0.00008616M

.00116

.000742

1.82E10

Stdev of Keq = 1.19E10 Mean of Keq = 1.06E10 Discussion: The mean of equilibrium constants of the possible product FeSCN)2 2+ is 1020 and the standard deviation is 260 (if we exclude the outlier in solution 2). The mean of the constants for Fe(SCN)3 was 1.06E10 and the standard deviation was 1.19E10. Thus, F  eSCN)2 2+ has to be the equilibrium product because it has more consistent equilibrium constants. Our setting the value of SCN- as zero in the first solution was accurate because-- after excluding beaker 2-- we obtained a percent unused of 1.68%, a value very close to 0.0%. Therefore, there was very little residual, however, there was some  residual thiocyanate left. The additional Fe3+  would not affect the absorbance values, because SCN- was a limiting reagent and the absorbance is dependant on the amount of red product formed by the reaction. Thus, the addition of further surplus iron would not affect the absorbances.

Conclusion: From this lab, we applied the principles of Lambert-Beer’s Law and photospectrometry to determine that the actual product of the reaction between iron(III) and thiocyanate to be Fe(SCN)22+ and not Fe(SCN)3. This was because the reactions between iron(III) and thiocyanate yielded a smaller standard deviation in equilibrium constants for potential product validates our conclusion because Fe(SCN)22+ than that of the potential product Fe(SCN)3. This  the same reaction should have similar equilibrium constants. There was some error in this lab, as we assumed all of the thiocyanate reacts in solution #1, however, there was 1.68% residual....