Calculation of Equilibrium Concentrations from Initial Concentrations Acetic acid, CH,CO,H, reacts with ethanol, CH.OH, to form water and ethyl acetate, CH,CO,C,H, The equilibrium constant for this reaction 40 What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH.COH. 9.15 Min CHOH, 0.40 M in CHICO/Cts, and 6.40 M in H₂O? GH₂OH CH₂COH CHICO CH₂ + H₂O Step 1: Determine the direction the reaction proceeds. "Because you have a higher concentration of products, the reaction will proceed to the left Step 2: Develop RICE table. *Present all information in terms of a RICE table Reaction Initial conc Change in conc Equilibrium conc. CH₂OH 0.15 CH₂COH 0.15 -X [CH,CO,C.B_[8,O [C,H,OH][CH,CO₂H] CH₂CO/C₂H₂ + H₂O 0.40 0.40 Step 3: Solve for the change and the equilibrium concentrations. solve for 9x²+bx+c=0 **Now you do the moth!**
Calculation of Equilibrium Concentrations from Initial Concentrations Acetic acid, CH,CO,H, reacts with ethanol, CH.OH, to form water and ethyl acetate, CH,CO,C,H, The equilibrium constant for this reaction 40 What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH.COH. 9.15 Min CHOH, 0.40 M in CHICO/Cts, and 6.40 M in H₂O? GH₂OH CH₂COH CHICO CH₂ + H₂O Step 1: Determine the direction the reaction proceeds. "Because you have a higher concentration of products, the reaction will proceed to the left Step 2: Develop RICE table. *Present all information in terms of a RICE table Reaction Initial conc Change in conc Equilibrium conc. CH₂OH 0.15 CH₂COH 0.15 -X [CH,CO,C.B_[8,O [C,H,OH][CH,CO₂H] CH₂CO/C₂H₂ + H₂O 0.40 0.40 Step 3: Solve for the change and the equilibrium concentrations. solve for 9x²+bx+c=0 **Now you do the moth!**
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![Calculation of Equilibrium Concentrations from Initial Concentrations
Acetic acid, CH,CO₂H, reacts with ethanol, C₂H-OH. to form water and ethyl acetate, CH,CO,CHs. The equilibrium constant for this reaction
40 What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH,CO₂H, 0,15 M in C,H,OH, 0.40 M in CH,CO,Cs, and 0.40 M
in H₂O?
CHOH + CH,CO,H5CH
CO,CH, + HO
Step 1:
Determine the direction the reaction proceeds.
*Because you have a higher concentration of products, the
reaction will proceed to the left
Step 2:
Develop RICE table.
*Present all information in terms of a RICE table
Reaction
Initial conc.
Change in conc.
Equilibrium conc.
k-
C₂H₂OH + CH₂COH
0.15
0.15
Step 3:
Solve for the change and the equilibrium concentrations.
[CH,CO_C,H_][H,O
<-4.0
[C₂H₂OH][CH₂CO₂H]
You will need the quadratic equation:
ax² +bx+c=0
Solving for x you get:
CH₂CO/C₂H₂ + H₂O
0.40
0.40
-b± √b²-4ac
2a
Solve for
ax²+bx+c=0
**Now you do the math!**
Insert another page if needed!
var
Did you get:
[CH,COH] = 0.18 M
[CH₂OH]=0.18 M
| [CH,CO,CH] = 0.37 M
[H₂O) = 0.37 M
R. Moller Chemistry 18 Lecture 2-12](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9affadff-1dbc-4796-87cc-cd8c2c6f7c45%2F2ddfb896-13db-4396-a93d-2103b092b3fc%2Fqvcidu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Calculation of Equilibrium Concentrations from Initial Concentrations
Acetic acid, CH,CO₂H, reacts with ethanol, C₂H-OH. to form water and ethyl acetate, CH,CO,CHs. The equilibrium constant for this reaction
40 What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH,CO₂H, 0,15 M in C,H,OH, 0.40 M in CH,CO,Cs, and 0.40 M
in H₂O?
CHOH + CH,CO,H5CH
CO,CH, + HO
Step 1:
Determine the direction the reaction proceeds.
*Because you have a higher concentration of products, the
reaction will proceed to the left
Step 2:
Develop RICE table.
*Present all information in terms of a RICE table
Reaction
Initial conc.
Change in conc.
Equilibrium conc.
k-
C₂H₂OH + CH₂COH
0.15
0.15
Step 3:
Solve for the change and the equilibrium concentrations.
[CH,CO_C,H_][H,O
<-4.0
[C₂H₂OH][CH₂CO₂H]
You will need the quadratic equation:
ax² +bx+c=0
Solving for x you get:
CH₂CO/C₂H₂ + H₂O
0.40
0.40
-b± √b²-4ac
2a
Solve for
ax²+bx+c=0
**Now you do the math!**
Insert another page if needed!
var
Did you get:
[CH,COH] = 0.18 M
[CH₂OH]=0.18 M
| [CH,CO,CH] = 0.37 M
[H₂O) = 0.37 M
R. Moller Chemistry 18 Lecture 2-12
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