In an experiment starting with `1` mol `C_(2)H_(5)OH, 1` mol `CH_(3)COOH`, and `1` mol of water, the equilibrium mixture mixture of analysis showa that `54.3%` of the acid is eaterified. Calculate `K_(c)`.

Explanation

`CH_(3)COOH(1)+C_(2)H_(5)OH(1)iffCH_(3)COOC_(2)H_(3)(1)+H_(2)O(1)`
`{:("Initial",1,1,0,1),("At equilbrium",1-x,1-x,x,1+x),(,1-0.543,1-0.543,0.543,1+0.543):} (54.3% "of" 1"mole"=(1xx54.3)/(100)=0.543 "mole")`
`"Hence given"x=0.543 "mole"`
Applying law of mass action:
`K_(c)=(["ester"]["water"])/(["acid"]["alcohol"])=(0.543xx1.543)/(0.457xx0.457)=4.0`
`PCl_(5)(g)iffPCl_(3)(g)+ Cl_(2)(g)`
`{:(therefore At, t=t_(eq), a-aalpha, aalpha, aalpha):}`
"Total no. of moles at equilibrium"`=a+aalpha=a(1+alpha)PCl_(5)=(a(1-alpha)P)/(2(1+alpha)), PCl_(3)=(a alpha .P)/(a(1+alpha))P_(Cl_(2)=(a alpha)/(a(1+alpha)).P`
`K_(P)=({((alphaP)/(1+alpha))}^(2))/(((1+alpha)/(1+alpha))P) , K_(P)=(alpha^(2)P)/(1-alpha^(2) ("Remember")`