Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C2H6 (g) ⇌ C2H4 (g) + H2 (g)
Solution
Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
Now, according to the reaction,
`C_2H_(6(g)) ↔ C_2H_(4(g)) + H_(2(g))`
Initial conc 4.0 atm 0 0
At equilibrium 4.0 - p p p
We can write,
`(p_(C_2H_4) xx p_(H_2))/p_(C_2H_6) = K_p`
`=> (p xx p)/(4.0 - p) = 0.04`
`=> p^2 + 0.16 - 0.04 p`
`=> p^2 + 0.04 p - 0.16 = 0`
Now `p = (-0.04 +- sqrt((0.04)^2 - 4xx1xx (-0.16)))/(2xx1)`
`= (-0.04 +- 0.80)/2`
`= 0.76/2` (Taking positive value)
= 0.38
Hence, at equilibrium,
`[C_2H_6] - 4 - p`
= 4 - 0.38
= 3.62 atm
