Question

For the reaction : $C_2{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+H_2\left(g\right)$
$K_p$ is ${5\times 10}^{-2}$ atm. Calculate the mole percent of $C_2{H}_6\left(g\right)$ at equilibrium if pure $C_2{H}_6$ at $1$ atm is passed over a suitable catalyst at $900K$:

• A. $20$
• B. $33.33$
• C. $66.66$
• D. none of these

Answer 1

The correct option is C $66.66$

Let 1 mole of ethane is passed over a suitable catalyst.
Let x moles of ethane dissociate to reach the equilibrium.
The equilibrium number of moles of ethane, ethylene and hydrogen are 1-x, x and x moles respectively

$K_p=\frac{P_{ethylene}{P}_{hydrogen}}{P_{ethane}}$

$\frac{x^2}{1-x}=5\times {10}^{-2}$

$x^2+0.05x-0.05=0$

$x=\frac{-0.05+\sqrt{(0.05{)}^2+4\times 0.05}}{2}=0.20atm$

The partial pressure of ethane is the product of mole fraction and total pressure.

$0.80=\text{mole fraction}\times 1.2$

Mole percentage of ethane $=\frac{0.8}{1.2}\times 100=66.66$