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?

C_{2}H_{6} (g) ⇌ C_{2}H_{4} (g) + H_{2} (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