Determination of an Equilibrium Constant
Determination of an Equilibrium Constant
Purpose A spectrophotometer will be used to attain an absorbance values for part A and B of this lab. In part A students are to calculate the concentration of FeSCN2+ions that formed. A graph will also be made from absorbance values and FeSCN2+concentrations. For the second part of the lab students are given the concentrations of KSCN and Fe(NO3)3 and is expected find the unknown equilibrium concentrations of Fe3+,SCN-, and FeSCN2+ and the K value for equilibrium of Rxn: Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq) . Procedure Please refer to: Chemistry 1E03 Laboratory Manual; McMaster University: Hamilton, ON, 2013; p 43-47. Given Info & useful equations for calculations: FeSCN 2+¿ ¿ eq ¿ [Concentration Products]eq Fe3+¿ Keq= = [Concentration Reactants ]eq ¿ SCN −¿ eq ¿ ¿
C[KSCN]= 0.002 mol/L
C[Fe(NO3)3]=0.200mol/L(part A)&0.002mol/L(partB)
C1V1=C2V2
A=y[conc] (where ‘y’ is K & ‘A’ is absorbance value) VTotal=0.050L
Rxn: Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq)
Observations and Sample Calculation Part A: Table 1:(absorp.Values) Flask#
Vol. KSCN
Vol.Fe(NO3)3(mL)
Absorp.Value(at wavelength 447 nm)
(mL) 1
1.00
49.0
0.247
2
2.00
48.0
0.445
3
3.00
47.0
0.659
Sample Calc. for Flask #1 Using: C1V1=C2V2
V [KSCN]=0.001L (Flask#1)
C2=
(0.002 mol/ L)(0.001 L) (0.05 L)
C2=4.00∙10-5mol/L is the concentration of FeSCN2+ions formed(also the initial concentration of SCN- ions) for flask # 1 There is less moles of KSCN than Fe(NO3)3, so students can assume that all the SCN- ions in the KSCN solution will be completely used up to make FeSCN2+ ions, thus students may conclude that the KSCN initial concentration is equal to the concentration of FeSCN2+ of ions formed. Table 2:(values of calculations in Part A) Flask #
Concent. Initial SCN- (in mol/L)
1
4.00e-5
2
8.00e-5
3
1.20e-4
Graph: (using values from Table 1 & 2)
Observation and Sample Calculation Part B: Table 3:(absorp.values) Test Tube #
1
2
3
4
5
KSCN(mL)
1.00
2.00
3.00
4.00
5.00
Fe(NO3)3(mL) 5.00
5.00
5.00
5.00
5.00
H2O (mL)
4.00
3.00
2.00
1.00
----
Absorbance
0.116 0.229
0.35
0.45
0.561
7
1
Value (nm)
For Test Tube#1 (sample calculation) a) Using: C1V1=C2V2
C2=
[ ( C 1 )( V 1 ) ] V2
(0.002mol/L)(1mL)=C2(10mL) C2=2.00e-4(mol/L) is the initial concentration of SCNConvert to mol b) Using: C1V1=C2V2
(0.002mol/L)(5mL)= C2(10mL) C2=1.00e-3(mol/L) is the initial concentration of Fe3+ c) Using: A=y[conc.]
A=0.116(absorption value for test tube# 1 (at 447 nm)) y=5560.7 (from slope of calibration curve part A ‘y=5560.7x’ from the equation y=mx+b) 0.116=5560.7[conc. EQ. of FeSCN2+] [Conc.EQ. FeSCN2+]=2.086e-5 ≈ 2.07e-5 (mol/L) Table 5:Ice table general format for Determination of an EQ Constant Lab Fe3+(aq) I(nitial)
SCN-(aq)
⇌ FeSCN2+(aq)
Fe3+initial
SCN-initial
----
-x
-x
+x
Fe3+initial-x
SCN-initial-x
x=calculated from absorption equation
C(oncentration) E(quilibrium)
+
Table 6: Sample Ice table Test Tube #1 Fe3+(aq) I(nitial) C(oncentration) E(quilibrium)
SCN-(aq)
⇌ FeSCN2+(aq)
1e-3
2e-4
----
-x
-x
+x
(1e-3)-x
(2e-4)-x
2.086e-5
d) K calculation from ice table information.
Keq=[FeSCN2+]eq/[[ Fe3+initial-x]eq[ SCN-initial-x]eq] For Test Tube #1 Keq=[2.086e-5]/[[(1e-3)- 2.086e-5][(2e-4)- 2.086e-5]] Keq=118.9260355 ≈ 119
+
Table 7: Keq. values attained from Part B calculations (**note: kept extra digits for ice for more accurate Keq calc.) Test Tube #
Fe3+initial
SCN-initial
FeSCN2+ conc. Eq.
Keq
1
1.00e-3
2.00e-4
2.086e-5
119
2
1.00e-3
4.00e-4
4.118e-5
120
3
1.00e-3
6.00e-4
6.420e-5
128
4
1.00e-3
8.00e-4
8.110e-5
123
5
1.00e-3
1.00e-3
1.009e-4
125
Equilibrium Average Keq avg.=(119+120+128+123+125)/5 Keq avg.=123 Discussion for Part A&B: Part A: Students are to calculate the initial concentration of SCN – ions. This is done by using the formula C1V1=C2V2 to solve for C2; where V2 is constant at 50mL, V1 is 0.002mol/L and C1 is the varying concentration of KSCN . For every mol of FeSCN2+ produced, one mol of SCN- and one mol of Fe3+are required. By reacting a huge excess of ferric nitrate with small amounts of potassium thyocinate at 3 different volumes it allows the students to conclude that the initial concentration of SCN- must be equal to the concentration of FeSCN2+ions formed since it is assumed that the reaction will go to completion. Absorbance values measured at 447nm are attained for the three test tubes in part A from the spectrophotometer. Then absorbance values are graphed against the calculated concentrations of FeSCN2+.The slope of the line of best fit for this graph is the constant ‘y’ in the A=y[Conc] formula and represents the proportional relationship between absorbance and FeSCN2+ concentrations. Part B: The goal of part B is to attain equilibrium values for Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq) reaction but students need to know the initial concentrations of Fe3+ ,SCN- and FeSCN2+ equilibrium concentration according to the
FeSCN 2+¿ ¿ eq ¿ Fe3+¿ equation Keq= . These concentrations can be attained from calculations from experimental the ¿ SCN −¿ eq ¿ ¿ values. Students react a constant volume of 5mL of Fe(NO3)3 at 0.002mol/L with five varying amounts of KSCN (at the same concentration as Fe(NO3)3) and varying amounts of distilled water adding up to a volume of 10mL(there will be five test tubes of varying amounts of reactant). Absorbance values are then attained for the five test tubes where each absorbance value represents the ‘A’ in the A=y[Conc] formula and allows the students to calculate the [Conc] of FeSCN2+ at equilibrium by using the constant proportionality value ‘y’ from part A. To figure out the initial concentration of Fe3+and SCN- students use the formula C1V1=C2V2 as in part A to solve for C2. The concentration of Fe3+ and SCN- at equilibrium should be their initial concentration ‘C2’ minus the [Conc] of FeSCN2+ at equilibrium because one mol of Fe3+ and SCN-(where both reactants have same M) are used to make one mol of FeSCN2+. Placing all these values in an ICE table students realize that they now know all the required to solve for Keq. Sources of Error: One source of error could be that there may have been water residue left inside of the test tubes diluting the solutions placed into the test tube hence resulting in inaccurate absorbance value readings from the spectrophotometer. The equilibrium constant is dependent upon temperature, it may also be an error because students made an assumption that all the reactants reacts at constant temperature (room temp.).This assumption may not have true since there may have been external sources of heat like solutions were near a cold air vent or the thermal energy from a student. Reacting reactants at a higher temperatures or lower temperatures will hence affect the equilibrium constant. Ideally the Keq calculations for part B should be the same but because of these errors they were not. However the values were fairly close to each other in the low 100 region and therefore gave a good approximation of the equilibrium constant. Conclusion: Overall the average Keq constant for this experiment is approximately 123 for the reaction of Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq)
Purpose A spectrophotometer will be used to attain an absorbance values for part A and B of this lab. In part A students are to calculate the concentration of FeSCN2+ions that formed. A graph will also be made from absorbance values and FeSCN2+concentrations. For the second part of the lab students are given the concentrations of KSCN and Fe(NO3)3 and is expected find the unknown equilibrium concentrations of Fe3+,SCN-, and FeSCN2+ and the K value for equilibrium of Rxn: Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq) . Procedure Please refer to: Chemistry 1E03 Laboratory Manual; McMaster University: Hamilton, ON, 2013; p 43-47. Given Info & useful equations for calculations: FeSCN 2+¿ ¿ eq ¿ [Concentration Products]eq Fe3+¿ Keq= = [Concentration Reactants ]eq ¿ SCN −¿ eq ¿ ¿
C[KSCN]= 0.002 mol/L
C[Fe(NO3)3]=0.200mol/L(part A)&0.002mol/L(partB)
C1V1=C2V2
A=y[conc] (where ‘y’ is K & ‘A’ is absorbance value) VTotal=0.050L
Rxn: Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq)
Observations and Sample Calculation Part A: Table 1:(absorp.Values) Flask#
Vol. KSCN
Vol.Fe(NO3)3(mL)
Absorp.Value(at wavelength 447 nm)
(mL) 1
1.00
49.0
0.247
2
2.00
48.0
0.445
3
3.00
47.0
0.659
Sample Calc. for Flask #1 Using: C1V1=C2V2
V [KSCN]=0.001L (Flask#1)
C2=
(0.002 mol/ L)(0.001 L) (0.05 L)
C2=4.00∙10-5mol/L is the concentration of FeSCN2+ions formed(also the initial concentration of SCN- ions) for flask # 1 There is less moles of KSCN than Fe(NO3)3, so students can assume that all the SCN- ions in the KSCN solution will be completely used up to make FeSCN2+ ions, thus students may conclude that the KSCN initial concentration is equal to the concentration of FeSCN2+ of ions formed. Table 2:(values of calculations in Part A) Flask #
Concent. Initial SCN- (in mol/L)
1
4.00e-5
2
8.00e-5
3
1.20e-4
Graph: (using values from Table 1 & 2)
Observation and Sample Calculation Part B: Table 3:(absorp.values) Test Tube #
1
2
3
4
5
KSCN(mL)
1.00
2.00
3.00
4.00
5.00
Fe(NO3)3(mL) 5.00
5.00
5.00
5.00
5.00
H2O (mL)
4.00
3.00
2.00
1.00
----
Absorbance
0.116 0.229
0.35
0.45
0.561
7
1
Value (nm)
For Test Tube#1 (sample calculation) a) Using: C1V1=C2V2
C2=
[ ( C 1 )( V 1 ) ] V2
(0.002mol/L)(1mL)=C2(10mL) C2=2.00e-4(mol/L) is the initial concentration of SCNConvert to mol b) Using: C1V1=C2V2
(0.002mol/L)(5mL)= C2(10mL) C2=1.00e-3(mol/L) is the initial concentration of Fe3+ c) Using: A=y[conc.]
A=0.116(absorption value for test tube# 1 (at 447 nm)) y=5560.7 (from slope of calibration curve part A ‘y=5560.7x’ from the equation y=mx+b) 0.116=5560.7[conc. EQ. of FeSCN2+] [Conc.EQ. FeSCN2+]=2.086e-5 ≈ 2.07e-5 (mol/L) Table 5:Ice table general format for Determination of an EQ Constant Lab Fe3+(aq) I(nitial)
SCN-(aq)
⇌ FeSCN2+(aq)
Fe3+initial
SCN-initial
----
-x
-x
+x
Fe3+initial-x
SCN-initial-x
x=calculated from absorption equation
C(oncentration) E(quilibrium)
+
Table 6: Sample Ice table Test Tube #1 Fe3+(aq) I(nitial) C(oncentration) E(quilibrium)
SCN-(aq)
⇌ FeSCN2+(aq)
1e-3
2e-4
----
-x
-x
+x
(1e-3)-x
(2e-4)-x
2.086e-5
d) K calculation from ice table information.
Keq=[FeSCN2+]eq/[[ Fe3+initial-x]eq[ SCN-initial-x]eq] For Test Tube #1 Keq=[2.086e-5]/[[(1e-3)- 2.086e-5][(2e-4)- 2.086e-5]] Keq=118.9260355 ≈ 119
+
Table 7: Keq. values attained from Part B calculations (**note: kept extra digits for ice for more accurate Keq calc.) Test Tube #
Fe3+initial
SCN-initial
FeSCN2+ conc. Eq.
Keq
1
1.00e-3
2.00e-4
2.086e-5
119
2
1.00e-3
4.00e-4
4.118e-5
120
3
1.00e-3
6.00e-4
6.420e-5
128
4
1.00e-3
8.00e-4
8.110e-5
123
5
1.00e-3
1.00e-3
1.009e-4
125
Equilibrium Average Keq avg.=(119+120+128+123+125)/5 Keq avg.=123 Discussion for Part A&B: Part A: Students are to calculate the initial concentration of SCN – ions. This is done by using the formula C1V1=C2V2 to solve for C2; where V2 is constant at 50mL, V1 is 0.002mol/L and C1 is the varying concentration of KSCN . For every mol of FeSCN2+ produced, one mol of SCN- and one mol of Fe3+are required. By reacting a huge excess of ferric nitrate with small amounts of potassium thyocinate at 3 different volumes it allows the students to conclude that the initial concentration of SCN- must be equal to the concentration of FeSCN2+ions formed since it is assumed that the reaction will go to completion. Absorbance values measured at 447nm are attained for the three test tubes in part A from the spectrophotometer. Then absorbance values are graphed against the calculated concentrations of FeSCN2+.The slope of the line of best fit for this graph is the constant ‘y’ in the A=y[Conc] formula and represents the proportional relationship between absorbance and FeSCN2+ concentrations. Part B: The goal of part B is to attain equilibrium values for Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq) reaction but students need to know the initial concentrations of Fe3+ ,SCN- and FeSCN2+ equilibrium concentration according to the
FeSCN 2+¿ ¿ eq ¿ Fe3+¿ equation Keq= . These concentrations can be attained from calculations from experimental the ¿ SCN −¿ eq ¿ ¿ values. Students react a constant volume of 5mL of Fe(NO3)3 at 0.002mol/L with five varying amounts of KSCN (at the same concentration as Fe(NO3)3) and varying amounts of distilled water adding up to a volume of 10mL(there will be five test tubes of varying amounts of reactant). Absorbance values are then attained for the five test tubes where each absorbance value represents the ‘A’ in the A=y[Conc] formula and allows the students to calculate the [Conc] of FeSCN2+ at equilibrium by using the constant proportionality value ‘y’ from part A. To figure out the initial concentration of Fe3+and SCN- students use the formula C1V1=C2V2 as in part A to solve for C2. The concentration of Fe3+ and SCN- at equilibrium should be their initial concentration ‘C2’ minus the [Conc] of FeSCN2+ at equilibrium because one mol of Fe3+ and SCN-(where both reactants have same M) are used to make one mol of FeSCN2+. Placing all these values in an ICE table students realize that they now know all the required to solve for Keq. Sources of Error: One source of error could be that there may have been water residue left inside of the test tubes diluting the solutions placed into the test tube hence resulting in inaccurate absorbance value readings from the spectrophotometer. The equilibrium constant is dependent upon temperature, it may also be an error because students made an assumption that all the reactants reacts at constant temperature (room temp.).This assumption may not have true since there may have been external sources of heat like solutions were near a cold air vent or the thermal energy from a student. Reacting reactants at a higher temperatures or lower temperatures will hence affect the equilibrium constant. Ideally the Keq calculations for part B should be the same but because of these errors they were not. However the values were fairly close to each other in the low 100 region and therefore gave a good approximation of the equilibrium constant. Conclusion: Overall the average Keq constant for this experiment is approximately 123 for the reaction of Fe3+(aq)+SCN-(aq) ⇌FeSCN2+(aq)
