$K_{p} = 0.04 \ atm$ at $899 \ K$ for the equilibrium shown below. What is the equilibrium pressure of $C_{2}H_{6}$ when it is placed in a flask at $4.0 \ atm$ pressure and allowed to come to equilibrium?
$C_{2}H_{6(g)} \longleftrightarrow C_{2}H_{4(g)} + H_{2(g)}$

(D) Let $p$ be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
According to the reaction:
$C_{2}H_{6(g)} \longleftrightarrow C_{2}H_{4(g)} + H_{2(g)}$
Initial pressure: $4.0 \ atm$,$0$,$0$
At equilibrium: $(4.0 - p) \ atm$,$p \ atm$,$p \ atm$
We can write the expression for $K_{p}$ as:
$K_{p} = \frac{p_{C_{2}H_{4}} \times p_{H_{2}}}{p_{C_{2}H_{6}}}$
Substituting the values:
$\frac{p \times p}{4.0 - p} = 0.04$
$p^{2} = 0.16 - 0.04p$
$p^{2} + 0.04p - 0.16 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$p = \frac{-0.04 \pm \sqrt{(0.04)^{2} - 4 \times 1 \times (-0.16)}}{2 \times 1}$
$p = \frac{-0.04 \pm \sqrt{0.0016 + 0.64}}{2}$
$p = \frac{-0.04 \pm \sqrt{0.6416}}{2}$
$p \approx \frac{-0.04 + 0.801}{2} \approx 0.3805 \ atm$
Hence,at equilibrium,the pressure of $C_{2}H_{6}$ is:
$p_{C_{2}H_{6}} = 4.0 - 0.3805 = 3.6195 \ atm \approx 3.62 \ atm$