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Solved: a. If the equilibrium constant for the reaction of
Chapter 16, Problem 85P(choose chapter or problem)
a. If the equilibrium constant for the reaction of acetic acid and ethanol to form ethyl acetate is \(4.02\), what will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with equal amounts of acetic acid and ethanol?
b. What will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with 10 times more ethanol than acetic acid? Hint: Recall the quadratic equation: for \(a x^{2}+b x+c=0\),
\(x=\frac{-b \pm\left(b^{2}-4 a c\right)^{1 / 2}}{2 a}\)
c. What will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with \(100\) times more ethanol than acetic acid?
Equation Transcription:
Text Transcription:
4.02
ax^2 +bx+c=0
x=-b plus minus (b^2 -4ac)^1/2 /2a
100
Questions & Answers
QUESTION:
a. If the equilibrium constant for the reaction of acetic acid and ethanol to form ethyl acetate is \(4.02\), what will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with equal amounts of acetic acid and ethanol?
b. What will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with 10 times more ethanol than acetic acid? Hint: Recall the quadratic equation: for \(a x^{2}+b x+c=0\),
\(x=\frac{-b \pm\left(b^{2}-4 a c\right)^{1 / 2}}{2 a}\)
c. What will be the concentration of ethyl acetate at equilibrium if the reaction is carried out with \(100\) times more ethanol than acetic acid?
Equation Transcription:
Text Transcription:
4.02
ax^2 +bx+c=0
x=-b plus minus (b^2 -4ac)^1/2 /2a
100
ANSWER:
Solution 85P
(a)
X2 - 2.248x + 1 = 0
By Solving the above quadratic equation
X = 0.610798, 1.6372 ignoring the value 1.6372
Therefore the equilibrium concentration of ethyl acetate is 0.610798M