The industrial solvent ethyl acetate is produced by the reac-tion of acetic acid with ethanol: CH3CO2H1soln2 + CH3CH2OH1soln2 ∆ CH3CO2CH2CH31soln2 + H2O1soln2 Ethyl acetate (b) A solution prepared by mixing 1.00 mol of acetic acid and 1.00 mol of ethanol contains 0.65 mol of ethyl ace- tate at equilibrium. Calculate the value of Kc. Explain why you can calculate K without knowing the volume of the solution.

Video transcript

Hello in this problem we are told the reaction of formic acid and methanol can produce method format. If a mixture of one mole of formic acid and one mole of methanol has 10. moles of method format at equilibrium it is our equilibrium constant. And how can we calculate our equilibrium constant without information about the volume of the solution? Given that we have an equilibrium problem. Let's begin by writing our equilibrium prostate expression. So that would be step one. So equilibrium constant expression then is the concentration of our products. So I'll raise to the one power based on the coefficient in the balanced reaction equation divided by the concentration of our reactant. Also raised to the one power based on the coefficient in the vast reaction. So note that water is included. That's because it is a product of the reaction and not solvent. Step two. Then we're going to show that we can find the equilibrium constant without volume. So we're gonna show that our volume units cancel. So step two, then we're gonna take our equilibrium constant and we're gonna look at the units. So we have moles of that's all for mate leader and molds of water for later. Bye bye, moles of formic acid later and then moles of methanol per meter. So we have been leaders here that we'll move into our numerator and then leaders in our denominator. So leaders squared in our numerator and leaders squared in our denominator units of volume cancel. So we can calculate Casey without our unit for volume. So step three, then we're going to make an ice table and so we'll write out our reaction and then we have initial and change in equilibrium so initially then we have One mole of formic acid, one mole of methanol. And we have none of our products present are changed in, We are consuming reacting, so that's -1. Both of those. And we are producing our products. So those are both plus X. We are now combined the initial and the change to come up with the equilibrium Get 1 -6 and then X and X. So up above we are told that the equilibrium amount of our method format is 0.78 at equilibrium. So that means then the x equals 0.78. So step four. Then we will now plug things back into our equilibrium constant expression based on what's provided an ice table. So we have our products, Those are both .78. And we have our reactant which are -1, which is .78. And that works out to 13. So our equilibrium constant For this reaction between formic acid and methanol is 13. And because our units cancel, our volume units cancel, we can calculate our equilibrium constant with just moles. Thanks for watching. Hope this help