Question

K_p = 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?

$$\mathrm{C}{\phantom{A}}_2\mathrm{H}{\phantom{A}}_6(\mathrm{g})\stackrel{-\rightharpoonup }{\leftharpoondown -}\mathrm{C}{\phantom{A}}_2\mathrm{H}{\phantom{A}}_4(\mathrm{g})+\mathrm{H}{\phantom{A}}_2(\mathrm{g})$$

Answer 1

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

	C2H6(g)	↔	C2H4(g)	+	H2(g)
Initial conc.	4.0 atm		0		0
At equilibrium	4.0 - p		p		p

We can write,

${\displaystyle \frac{{\text{p}}_{{\text{C}}_2{\text{H}}_4}\times {\text{p}}_{{\text{H}}_2}}{{\text{p}}_{{\text{C}}_2{\text{H}}_6}}={\text{K}}_{\text{p}}}$

${\displaystyle \Rightarrow \frac{\text{p}\times \text{p}}{4.0-\text{p}}=0.04}$

${\displaystyle \Rightarrow {\text{p}}^2+0.16-0.04\text{p}}$

${\displaystyle \Rightarrow {\text{p}}^2+0.04\text{p}-0.16=0}$

Now ${\displaystyle \text{p}=\frac{-0.04\pm \sqrt{{\left(0.04\right)}^2-4\times 1\times (-0.16)}}{2\times 1}}$

${\displaystyle =\frac{-0.04\pm 0.80}{2}}$

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

= 0.38

Hence, at equilibrium,

${\displaystyle \left[{\text{C}}_2{\text{H}}_6\right]-4-\text{p}}$

= 4 - 0.38

= 3.62 atm