. In performing the lab reaction, the student made a mixture of 3 reaction components in a beaker. This was done by mixing 13.00 mL of 0.00250 M Fea) with 11.00 mL of 0.00250 M SCN (a) and 6.00 ml. of deionized H₂O. Upon mixing these solutions (total volume of 30.00 mL), the final solution turned a deep- red color. The absorbance of this equilibrium mixture was measured and the concentration of Fe(SCN)² was determined to be 3.00 x 105 M with the use of the Beer's Law constant (k). Determine the equilibrium constant (K) for the reaction from these values by completing the ICE table below. Be sure to report your final value of K at the bottom. (5 points) Fe³+ (aq) +SCN- (aq) → Fe(SCN)² (aq) 3/15 Initial (mmol) Change (mmol) Equilibrium (mmol) Equilibrium (M) Fe (aq) 0.008 -3X10² 9.7X10 5 SCN (aq) = Fe(SCN)²(aq) 0.02. 0 3x10-5 +3x10-5 7.7x15% 3x105 3.00 × 10³ M
. In performing the lab reaction, the student made a mixture of 3 reaction components in a beaker. This was done by mixing 13.00 mL of 0.00250 M Fea) with 11.00 mL of 0.00250 M SCN (a) and 6.00 ml. of deionized H₂O. Upon mixing these solutions (total volume of 30.00 mL), the final solution turned a deep- red color. The absorbance of this equilibrium mixture was measured and the concentration of Fe(SCN)² was determined to be 3.00 x 105 M with the use of the Beer's Law constant (k). Determine the equilibrium constant (K) for the reaction from these values by completing the ICE table below. Be sure to report your final value of K at the bottom. (5 points) Fe³+ (aq) +SCN- (aq) → Fe(SCN)² (aq) 3/15 Initial (mmol) Change (mmol) Equilibrium (mmol) Equilibrium (M) Fe (aq) 0.008 -3X10² 9.7X10 5 SCN (aq) = Fe(SCN)²(aq) 0.02. 0 3x10-5 +3x10-5 7.7x15% 3x105 3.00 × 10³ M
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![. In performing the lab reaction, the student made a mixture of 3 reaction components in a beaker. This was
done by mixing 13.00 mL of 0.00250 M Fe³(aq) with 11.00 mL of 0.00250 M SCN (aq)
and 6.00 ml. of
deionized H₂O. Upon mixing these solutions (total volume of 30.00 mL), the final solution turned a deep.
red color. The absorbance of this equilibrium mixture was measured and the concentration of Fe(SCN)2
was determined to be 3.00 x 10 M with the use of the Beer's Law constant (k). Determine the equilibrium
constant (K) for the reaction from these values by completing the ICE table below. Be sure to report your
final value of K at the bottom. (5 points)
3/15
Initial (mmol)
Fe³+
Equilibrium (M)
(aq)
Fe (aq)
0.008
Change (mmol)
-3X105
Equilibrium (mmol)X106
+ SCN- (aq)
Fet][SCN)
3X105
SCN (aq) =
Fe(SCN)2- (aq)
Fe(SCN)² (aq)
0
+3×105 X
0.02
3x05
7.7x1556 3 x 105
3.00 x 105 M
2+
Fe ² (aq) + SON (ag) → Fe (SCN) ²+ (aq)
(fe+³) = 13X 0.002 = 0.008
30
SCN³) = 30X0.002 0.02
30
Fet ³ (aq) + SeNi (ag) → FeSCN³²2 (aq)
K= FeSCN1²]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4549f807-3428-422e-9b66-cc76881554de%2Fe322d9d0-de2e-4e41-9f3f-9e4b9cb56330%2F6se0vgq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:. In performing the lab reaction, the student made a mixture of 3 reaction components in a beaker. This was
done by mixing 13.00 mL of 0.00250 M Fe³(aq) with 11.00 mL of 0.00250 M SCN (aq)
and 6.00 ml. of
deionized H₂O. Upon mixing these solutions (total volume of 30.00 mL), the final solution turned a deep.
red color. The absorbance of this equilibrium mixture was measured and the concentration of Fe(SCN)2
was determined to be 3.00 x 10 M with the use of the Beer's Law constant (k). Determine the equilibrium
constant (K) for the reaction from these values by completing the ICE table below. Be sure to report your
final value of K at the bottom. (5 points)
3/15
Initial (mmol)
Fe³+
Equilibrium (M)
(aq)
Fe (aq)
0.008
Change (mmol)
-3X105
Equilibrium (mmol)X106
+ SCN- (aq)
Fet][SCN)
3X105
SCN (aq) =
Fe(SCN)2- (aq)
Fe(SCN)² (aq)
0
+3×105 X
0.02
3x05
7.7x1556 3 x 105
3.00 x 105 M
2+
Fe ² (aq) + SON (ag) → Fe (SCN) ²+ (aq)
(fe+³) = 13X 0.002 = 0.008
30
SCN³) = 30X0.002 0.02
30
Fet ³ (aq) + SeNi (ag) → FeSCN³²2 (aq)
K= FeSCN1²]
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