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

`C_(2)H_(6)(g) hArr C_(2)H_(4)(g)+H_(2)(g)`
`{:("Initial Pressure",4.0 atm,,0,0),(At eq,4-p,,p,p):}`
`K_(p)=(p_(C_(2)H_(2))xxp_(H_(2)))/p_(C_(2)H_(6))`
`:. 0.04=p^(2)/(4-p)`
or `p^(2)=0.16-0.04p`
or `p^(2)+0.04p-0.16=0`
`:. P=(-0.04+-sqrt(0.0016-4(-0.16)))/(2)=(-0.04+-0.89)/(2)`
Taking positive value,
`p=0.80/2=0.40`
`:. [C_(2)H_(6)]_(eq)=4-0.40 atm =3.60 atm`