Question

`K_(p)=0.04 atm` at `899 K` for the equilibrium shown below. What is the equilibrium concentration of `C_(2)H_(6)` when it is placed in a flask at `4.0 atm` pressure and allowed to come to equilibrium?
`C_(2)H_(6)(g) hArr C_(2)H_(4)(g)+H_(2)(g)`

Answer 1

The equilibrium in the reaction is :
`{:(,C_(2)H_(6)(g),hArr,C_(2)H_(4)(g),+,H_(2)(g)),("Initial pressure :", "4atm" ,,0,,0),("Eqm. molar conc.","(4-p)atm",,"p atm",,"p atm"):}`
`K_(p)=(Pc_(2)H_(4)xxP_(H_(2)))/(Pc_(2)H_(6))" or " 0.04= (P^(2))/((4-P))`
`P^(2)=0.04(4-P) or P^(2) + 0.04p-0.16=0`
`P= (-0.04)overset(+)(-)sqrt(0.0016 -4(-0.16))`
`=((-0.04)underset(-)(+) 0.8)/(2)= (0.76)/(2) =0.38`
Equilibrium pressure or concentration of `C_(2)H_(6) = (4 -0.38) =3.62 atm.`