Question

The reaction is acetic acid plus isopentyl alcohol with an acid catalyists gives isopentyl acetate (banana oil).

a. In your reaction you started with a solution that was approximately 8 M acetic acid and 4 M isopentyl alcohol. Use the equilibrium expression for this reaction to calculate the final concentrations of water, isopentyl acetate, isopentyl alcohol, and acetic acid (you' ll need to use the quadratic equation to solve this expression) assuming that the equilibrium constant is 4.2. Calculate the percent yield of isopentyl acetate to which this value would correspond.

b. In order to give a better yield, it is common to carry the reaction out under conditions where the water can be removed. Let's say you carried out the reaction using the same initial concentrations of starting materials but removed the water as the reaction proceeded. What would be the final concentration of isopentyl acetate if the final water concentration was 0.25 M? What would the percent yield of this reaction be?

Answer 1

Ka = 4.2 = ([isopentylacetate][H2O])/([Aceticacid][isopentyl alcohol])

at equilibrium, 4.2 = x²/(8-x)(4-x)

forming a quadratic equation, 3.2x²-50.4x+134.4 = 0 and solving it for x, x = 1/2a(-b +/- (b²-4ac)^1/2) for an equation of type ax² + bx + c = 0

x = 12.35M and 3.4 M, and excluding the unrealistic 12.35M solution, we have x = 3.4 M

Concentration of isopentyl acetate = 3.4 M, water = 3.4 M, acetic acid = 4.6 M and isopentyl alcohol is 0.6 M.

yield = (3.4/4)*100 = 85%

2) 4.2 = (3.4+x)0.25/(4.6-x)(0.6-x), simplifying this to get a quadratic equation, 4.2x2 - 22.09x + 10.74 = 0

Solving for x and using the realistic value, x = 0.54. Therefore, after removing water to have a final concentration of 0.25 M, [isopentyl acetate] = 3.94 M

and the yield is = 98.54%