Question

The equilibrium constant $K_p$ for the reaction $A\rightleftharpoons 2B$ is related to degree of dissociation $\alpha$ of $A$ and total pressure $P$ is:

• A. $\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$
• B. $\frac{4{\alpha }^2{P}^2}{1-{\alpha }^2}$
• C. $\frac{4{\alpha }^{2}P}{1-\alpha }$
• D. none of these

Answer 1

The correct option is A $\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$

$A⟶2B$

moles

at $t=0$ $1$ -

at $eq{b}^m$ $1-\alpha$ $2\alpha$

$K_P=K_n{\left[\frac{P}{n}\right]}^{\Delta ng}$ $P⟶$ Total pressure, $n$= Total moles, $n=1-\alpha +2\alpha =1+\alpha$

$\Rightarrow K_P=\frac{[nB{]}^2}{n_A}{\left[\frac{P}{n}\right]}^{2-1}$

$\Rightarrow K_P=\frac{[2\alpha {]}^2}{[1-\alpha ]}\left[\frac{P}{1+\alpha }\right]$

$\Rightarrow K_P=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$