Question

$K_p=0.04$ atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of $C_2{H}_6$ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
$C_2{H}_6\left(g\right)\leftrightarrow C_2{H}_4\left(g\right)+H_2\left(g\right)$

Answer 1

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,
$\begin{array}{cccccc}& C_2{H}_{6\left(g\right)}& \leftrightarrow & C_2{H}_{4\left(g\right)}& +& H_{2\left(G\right)}\\ Initialcone.& 4.0atm& & 0& & 0\\ Atequilibrium& 4.0-p& & p& & p\end{array}$

We can write
$\frac{P_{c_2{H}_4}\times P_{H_2}}{P_{C_2{H}_6}}=K_p\Rightarrow \frac{p\times p}{4.0-p}=0.04\Rightarrow p^2=0.16-0.04p\Rightarrow p^2+0.04p-0.16=0$
Now, $p=\frac{-0.04\pm \sqrt{(0.04{)}^2-4\times 1\times (-0.16)}}{2\times 1}=\frac{-0.04\pm 0.80}{2}$
$=\frac{0.76}{2}$ (Taking positive value)
$=0.38$
Hence, at equilibrium,
$\left[C_2{H}_6\right]=4-p=4-0.38=3.62atm$