Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$A\rightleftharpoons 2B$
Initial conc. 1 0
Equilibrium conc. $1-\alpha$ $2\alpha$
Total no. of moles at equilibrium= $1-\alpha +2\alpha =1+\alpha$
$P_A=\frac{(1-\alpha )P}{(1+\alpha )},P_B=\frac{\left(2\alpha \right).P}{(1+\alpha {)}^2}{K}_p=\frac{{\left[\frac{\left(2\alpha \right).P}{(1+\alpha {)}^2}\right]}^2}{\left[\frac{1-\alpha .P}{1+\alpha }\right]}{K}_p=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$