`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

3.62 atm

Explanation

`{:(,C_2H_6(g)hArr, C_2H_4(g) + , H_2(g)),("Initial conc.","4.0 atm",0,0),("At equi.",4-p,p,p):}`
`K_p=(pxxp)/(4-p)`=0.04
`p^2` =0.16-0.04 p
Solving for p , we get p =0.38
`therefore` Conc of `C_2H_6` at eq. =4-0.38 =3.62 atm