Determining an Equilibrium Constant

 

Introduction

            It is frequently assumed that reactions go to completion; that all of the reactants are converted into products.  Most chemical reactions do not go to completion because they are equilibrium systems where the reaction proceeds in both directions.  As the reactants are used up, the rate of the forward reaction  decreases.  Conversely, as the concentrations of the products increase, the rate of the reverse reaction increases.  Eventually, the rate of the forward reaction equals the rate of the reverse reaction and the concentrations  of the reactants and the products stay constant. The system has reached a state of dynamic equilibrium. At equilibrium, both the forward and reverse reactions are occuring, but no net change is observed. 

 

For the general reaction:

 

aA  +  bB  ↔  cC  +   dD

 

where a,b,c and d are the stoiochiometric coefficicents. 

( 1 )

 

Experimental evidence shows that the ratio of products to reactants (raised to their stiochiometric coefficients) is a constant at equilibrium. 

 

At equilibrium, the equilibrium constant, K, is equal to:

                                                                                               

K =      where the brackets [ ] imply molarity.

( 2 )

             

The equilibrium constant  measures the extent to which a chemical  reaction occurs.  The larger the value for K, the greater the tendency  for the reaction to go to completion is and more product will be formed.

 

In this experiment you will determine the equilibrium constant for the  following reaction:

 

Fe 3+ (aq)  +  HSCN (aq)  ↔  FeSCN 2+ (aq)   +   H + (aq)

( 3 )

 

                                                                                   

K = 

( 4 )

 

            Solutions of  Fe3+  and HSCN will be mixed and will react to form some FeSCN 2+ and H+.  The initial amounts of  Fe3+  and HSCN can be calculated.  The  equilibrium concentration of FeSCN 2+  will be found using its spectroscopic  properties.  FeSCN 2+  , a blood red complex, absorbs the blue-green wavelengths of visible light.  Its absorbance is directly proportional its concentration.  The abosrbance (an measure of the amount of light absorbed) will be measured by a Bausch & Lomb Spectrophotometer 1001.  For a 1 cm cuvette (a square test tube), the absorbance, A, is equal to the “extinction coefficient”, ε (epsilon) times the molar concentration, C.

 

A = εC.                                                                                                                            ( 5 )

 

For the FeSCN 2+ , ε equals 4,400.  So the equilibrium concentration of  FeSCN 2+ will equal your measured absorbance divided by 4,400.  Using the stoichiometric coefficients (which are all one) in reaction ( 3 ) and an equilibrium (ICE) chart, the equilibrium  concentrations of Fe 3+  and  HSCN are then calculated.  Finally, the equilibrium concentrations are put into equation ( 4 ) to find the equilibrium  constant.

 

Note:  All of the solutions are made in 0.20M HNO3(aq).  This will hold the concentration of [H+] constant at 0.20M throughout the experiment.

 

Equipment

(2) 50 mL volumetric flasks

3 test tubes

50 mL beaker

4 cuvettes

stirring rod

Thermometer

(2) 10 mL graduated cylinders

Spectronic  1001

 

 

           

Chemicals

0.20 M HNO3(aq),  nitric acid

0.10 M KSCN(aq), potassium thiocyanate, (source of HSCN)

Acidified 0.10 M Fe(NO3)3(aq), iron (III) nitrate dissolved in 0.20M HNO3(aq).  (Source of Fe3+)

 

Spill/Disposal                    

Spill/Disposal Reaction Mixture : A 

0.20 M HNO3  (Nitric Acid) Spill/Disposal: B1

0.010 M KSCN (Potassium thiocyanate) Spill/Disposal : A

acidified 0.10 M Fe(NO3)3   (Ferric Nitrate) Spill/Disposal : A          

 

 

Procedure

 

1.  Preparing the HSCN(aq) solution: 

Clean a 50 mL volumetric flask by rinsing with deionized water.  Rinse it several times with 5 mL  portions of 0.20 M HNO3 . Use the provided KSCN dispenser

(make sure that the tip  is full) to add 5.0 mL of 1.0 x 10-2 M KSCN into the flask.  Add 0.20 M HNO3  until the solution reaches the etched ring on the neck of  the flask. The H+ reacts with the SCN― to make 1.0 x 10-3  M HSCN in 0.20 M H+. Thiocyanic acid (HSCN) decomposes so it must be made fresh. Swirl the solution to be sure that it is well mixed.

 

2.  Preparing the Fe(NO3)3(aq) solution: 

Clean a second 50 mL volumetric flask with deionized water.  Rinse it several times with 5 mL  portions of 0.20 M HNO3(aq) . Use the provided 0.10 M Fe(NO3)3(aq) dispenser (make sure that the tip  is full) to add 5.0 mL of 0.10 M Fe(NO3)3(aq) into the flask.  Add 0.20 M HNO3(aq) until the solution reaces the etched ring on the neck of  the flask.  This new solution will be 0.010 M Fe(NO3)3(aq) in 0.2 M HNO3(aq). 

 

3.  Preparing the three solutions for the equilibrium concentration determination:

Label three clean test tubes 1, 2, & 3.  Use a clean 10 mL graduated  cylinder to pour 5.0 mL of the 0.010 M Fe(NO3)3 (aq) that  you have just made into test tube #1.

 

4.   Pour 5.0 mL of the 0.010 M Fe(NO3)3(aq)  into a clean 50 mL beaker.  Add 5.0 mL of 0.20 M HNO3(aq) and stir.  You have just diluted the Fe+3 concentration in half. 

( milliliters ) ( molarity )   =    ( milliliters )  ( molarity )                               (6)

Pour 5.0 mL of  this new solution in test tube #2.

 

5.  To the remaining 5 mLs of 0.010 M Fe(NO3)3(aq) still in  the beaker, add 5.0 mL of  0.20 M HNO3(aq)  and stir. Again, you have diluted the Fe(NO3)3(aq) concentration by half.  Pour 5.0 mL of this third solution  into test tube #3.   

 

6.  To each of the three test tubes from steps #3 – 5, add 5.0 mL of the 1.0 x 10-3 M HSCN(aq)  from step #1.  Stir well. The concentration of HSCN(aq) will be the same for all three test tubes.  By adding the 5.0mL of HSCN(aq) to each of the Fe(NO3)3(aq) solutions, the initial concentrations of the HSCN(aq) and Fe(NO3)3(aq) solutions have been halved.  Measure the temperature of each solution.  (K is temperature dependent.)

 


7.  Specroscopy:

Obtain 4 cuvettes.  Fill one (to the mark) with 0.20 M HNO3(aq).  This will be used as the standard (all solutions are in 0.20 M HNO3(aq), and the absorbance of of  0.20 M HNO3(aq) will be set to zero.  One by one, transfer each of the three solutions (in test tubes 1 – 3) to cuvettes and record the absorbance of [FeSCN2+] of each using the Spectronic 1001.  The cuvettes should be filled to the same level each time.  The Spectronic 1001 has been set at 450 nm.  Each lab group will need to rezero the aborbance needs using their sample of 0.20 M HNO3(aq).  Your lab instructor will show you how to use this spectrometer.

 

 

8.  Calculations:

Taking into account the dilutions in steps 3 – 6, calculate the initial amounts of [Fe 3+] and [HSCN ] (the NO3― is a spectator.)  Remember that the [H+] is constant at 0.20M.  Record the initial concentrations in the PreLab. Using the absorbance values obtained and the value of ε, 4400, calculate the equilibrium concentrations of [FeSCN2+] in each cuvette.  Use the stoichiometric relationships in equation (3) and ICE charts to find the equilibrium concentrations of [Fe 3+]  and[HSCN ]. ([H+] will always  be 0.20 M.)  Plug these values into the equilibrium expression (4) and calculate k.  Do this for all three solutions.  

 

Disposal

The contents of all test tubes and volumetric flasks may be disposed of in the sink.

Determining an Equilibrium Constant Pre Lab

 

 

Name:______________________________________________________________

 

1.     List health and safety hazards associated with HNO3 and KSCN . 

 

 

 

 

 

 

 

2.     Calculate the initial concentrations for Fe(NO3)3(aq), [Fe3+] since nitrate is a spectator, and [HSCN] in this experiment, and enter them in the chart below.  Don't forget the dilutions! 

 

Test Tube

[Fe3+]

[HSCN]

1

 

 

 

 

 

2

 

 

 

 

 

3

 

 

 

 

 

 

 

 

3.  You start with 4.00 mole of pure SO3 in an 8.00 L flask.  At equilibrium 0.50 mole of O2 has been formed.  What is the value of K for the reaction?

 

2 SO3 (g)  Û  2 SO2 (g)  +  O2 (g)

Equilibrium Constant Worksheet                                        

 

Name:____________________________________

 

Test Tube                   [Fe3+] initial              [HSCN] initial

 

 

1                                  ______________        _____________

 

 

 

2                                  ______________        _____________

 

 

 

3                                  ______________        ______________

 

 

 

Test Tube                   Absorbance                            [FeSCN2+] eq   (C = A/4400) 

 

 

1                                  _______________                  _________________________ 

 

 

 

2                                  _______________                  _________________________

 

 

 

3                                  _______________                  _________________________

 

 

 

 

Temperature of equilibrium mixture:

 

Test Tube 1  ____________________

 

Test Tube 2    ____________________

 

Test Tube 3    ____________________

 

 

 

 

 

ICE Charts

 

1.  From Test Tube 1

 

 

Fe3+      +          HCSN             ↔        FeSCN2+         +          H+

 

 

I           _________      _________                  _________                  _________                 

initial

C         _________      _________                  _________                  _________

change

 

E         _________      _________                  _________                  _________

equilibrium

 

 

 

K=

 

 

 

 

 

 

 

 

2.    From Test Tube 2

 

Fe3+      +          HCSN             ↔        FeSCN2+         +          H+

 

 

I           _________      _________                  _________                  _________                 

 

C         _________      _________                  _________                  _________

 

 

E         _________      _________                  _________                  _________

 

 

 

 

 

K=

 

3.  .  From Test Tube 3

 

 

Fe3+      +          HCSN             ↔        FeSCN2+         +          H+

 

 

I           _________      _________                  _________                  _________                 

 

C         _________      _________                  _________                  _________

 

 

E         _________      _________                  _________                  _________

 

 

 

 

 

K=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Average K= ___________________

 

 

 

Partner’s names:_________________________________________________________