Question

1.0 mol of ethanol and 1.0 mol of acetic acid are dissolved in water and kept at 100? degrees C. The volume of the solution is 250 mL. At equilibrium, 0.25 mol of acetic acid has been consumed in producing ethyl acetate. Calculate K_c? at 100 degrees C for the reaction: C₂H₅OH_?(aq) + CH₃CO₂H_(aq) $\rightleftharpoons$ CH₃CO₂C₂H_5(aq) + H₂O_?(l)

Answer 1

The reaction is given as follows: C,H,OH(aq) + CH,CO,H(aq) = CH,CO,C,Hz (aq) + H2O(1) In ICE table the concentrations are given, therefore, firstly calculate the concentrations of C,H,OH and CH,COH as follows: molarity (C Hz moles volume lmol 250 x 10-L = 4M moles molarity (CH,CO,H)=- volume Imol 250x10-'L = 4M At equilibrium, the moles of acetic acid consumed are 0.25 mol. Converting the moles consumed at equilibrium into molarity as follows:

moles Molarity (CH,CO,H) = volume 0.25 mol 250x10-'L =1M = The ICE table is given as follows: Initial (M) Change Equilibrium C,H,OH(aq) + 4 - x 4 -x CH,CO,H(aq) = L 4 -X 4 -x CH,CO,C,Hz(aq) +H2O(1) L +x +x According to the given information, the value of x=1M, therefore, at equilibrium, the concentration of C,H,OH is 3 M and the concentration of CH,CO,H is also 3 M, and the concentration of CH,CO,C,H, is 1 M. Here, H2O(1) is present as liquid, therefore, it will not contribute in the equilibrium constant equation.

The K, for the reaction is given as follows: Bi-Tan [CH CO,C,H,(aq)| [C,H,OH(aq)][CH,CO,H(aq)] - (1) (3) = 0.11 Therefore, the K for the reaction is 0.11.