Saved Bookmarks
| 1. |
`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 came in equilibrium ? ` C_(2) H_(6) (g) hArr C_(2) H_(4) (g) + H_(2) (g)` |
|
Answer» ` {:(,C_(2)H_(6)(g),hArr,C_(2)H_(4)(g),+,H_(2)(g)),("Intial pressure",4*0"atm",,0,,0),("At eqm".,4-p,,p,,p):}` `K_(p) = (p_(C_(2)H_(4))xx p_(H_(2)))/(p_(C_(2)H_(6))):. 0*04 = p^(2)/(4-p) or p^(2) = 0*16 - 0*04 p ` or ` p^(2) + 0*04 p - 0*16 = 0 :. p = (-0*04 pm sqrt(0*0016 - 4(-0*16)))/2 = (0*04 pm 0*89)/2` aking positive value, ` p= (0*80)/2 = 0*40 :. [C_(2) H_(6) ]_(eq) = 4 - 0*40 "atm" = 3* 60 "atm " ` |
|