Question

$K_P=0.04$ atm at $899K$ for the iequilibrium 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?
$C_2{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+H_2\left(g\right)$

• A. $7.24$ atm
• B. $3.62$ atm
• C. $1$ atm
• D. $1.5$ atm

Answer 1

The correct option is B $3.62$ atm

$C_2{H}_6\left(g\right)\leftrightharpoons C_2{H}_4\left(g\right)+H_2\left(g\right)$

At t=0 4 0 0

At equilibrium 4-p p p

We know

$K_p=\frac{\left[P_{C_2{H}_4}\right]\left[P_{H_2}\right]}{\left[P_{C_2{H}_6}\right]}$

$K_p=\frac{p.p}{4-p}$

$0.04=\frac{p^2}{4-p}$

$\frac{1}{25}=\frac{p^2}{4-p}$

$4-p=25p^2$

$25p^2+p-4=0$

$p=\frac{-1\pm \sqrt{1+400}}{50}$

$p=-0.02\pm 0.4$

$p=0.38$

Hence, pressure of $C_2{H}_6=4-p=4-0.38=3.62atm$