Question

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?
$C_2H_6 (g) ⇋ C_2H_4 (g) + H_2 (g)$

Answer 1

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,
$\frac {P_{C_2H_4} × P_{H_2}}{P_{C_2H_6}} = K_p$

⇒$\frac {p × p}{4.0 - p} = 0.04$
⇒ $P^2 = 0.16 - 0.04 p$
⇒ $p^2 + 0.04 p - 0.16 = 0$
Now, $p = \frac {-0.04 ± \sqrt { (0.04)^2 - 4×1×(-0.16)}}{2×1}$

= $\frac {-0.04 ± 0.80}{2}$

= $\frac {0.76}{2}$ (Taking positive value)

$= 0.38$
Hence, at equilibrium,
$[C_2H_6] -4 - p = 4 - 0.38$
$= 3.62\ atm.$