Question

The equilibrium constant $K_p$ for the reaction $A\leftrightharpoons 2B,$ is related to degree of dissociation of A and total pressure $P_T$ as

• A. $\frac{4{\alpha }^2{P}_T}{1-{\alpha }^2}$
• B. $\frac{4{\alpha }^2{P}_T^2}{1-{\alpha }^2}$
• C. $\frac{4{\alpha }^2{P}_T^2}{1-\alpha }$
• D. $\frac{4{\alpha }^2{P}_T}{1-\alpha }$

Answer 1

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

$A\leftrightharpoons 2BP-P\alpha 2P\alpha K_p=\frac{(2\alpha P{)}^2}{(1-\alpha )P}=\frac{4{\alpha }^2{P}^2}{(1-\alpha )P}{K}_P=\frac{P4{\alpha }^2}{1-\alpha }$

$P_T$ =$P-P\alpha$ + $2P\alpha$

$P_T$ =$P+P\alpha$

$P=\frac{P_T}{1+\alpha }$

$K_P=\frac{P_{T}4{\alpha }^2}{(1-\alpha )\times (1+\alpha )}$

$K_P=\frac{P_{T}4{\alpha }^2}{1-{\alpha }^2}$