K_p= 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_2H_6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ? C_2H_(6(g)) hArr C_2H_(4(g)) + H_(2(g))

K_p= 0.04 atm at 899 K for the equilibrium shown below. What is the eq

1.	K_p= 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_2H_6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ? C_2H_(6(g)) hArr C_2H_(4(g)) + H_(2(g))
Answer» Solution :`{:("Reaction equilibrium:",C_2H_(6(g)) hArr, C_2H_(4(g)) +, H_(2(g))),("Initial partial pressure :","4.0 ATM","0 atm","0 atm"),("CHANGE in pressure","-p atm","+p atm","+p atm"),("Partial pressure at equilibrium :", "(4-p)atm","p atm","p atm"):}` `K_p=((p_(C_2H_4))(p_(H_2)))/((p_(C_2H_6)))` `therefore 0.04 = ((p)(p))/((4-p))` `therefore p^2+ 0.04 p - 0.16 =0` In this quadratic equation put B= +0.04 , a=1 and c=-0.16 So, `p=(-bpmsqrt(b^2-4ac))/(2a)` `=(-(0.04)pmsqrt((+0.04)^2-4(1)(-0.16)))/2` `=(-(0.04)PM sqrt(0.0016 +0.64))/2` `=(-(0.04) pm sqrt0.6416)/2` `=(-0.04)pm0.8009)/2` OR `(-0.4-0.80009)/2` p = 0.3805 atm = 0.38 atm OR - 0.6005 atm it is possible OR impossible `therefore p_(C_2H_6)`=(4-p)=(4-0.3805)=3.62 atm