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`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)` |
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Answer» `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` |
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