`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)`

Explanation

Suppose the partial pressure of `C_2H_6` decreases by `rho` atm will equilibrium is attained.
`{:(,C_2H_6(g),hArr, C_2H_4(g) ,+,H_2(g)) , ( " Initial pressure ", 4.0 " atm " ,, 0,,0),(" Press at Eq.", 4.0 -rho ,, rho ,,rho ):}`
` therefore " " K_rho =(rho _(C_2H_4) .rho _(H_2))/(rho_(C_2H_5)) = (rho xx rho )/(4.0 - rho ) `
or ` " " 0.04 = (rho ^(2))/(4.0 - rho ) `
` or " " rho ^(2) = 0.16 - 0.04 rho `
` or rho ^(2) +0.04 rho - 0.16 =O`
` therefore " " rho = ( -0.04 +-sqrt( (0.04)^(2) ) -4xx 1xx (-0.16))/(2xx 1)`
` " " = (-0.04+-0.801)/(2)`
Taking positive value , `rho =0.38 `
` therefore " " [C_2H_6(g)_(eq) = 4.0 - rho =4.0 -0.38 = 3.62` atm .