Question

For the reaction $AB\left(g\right)\rightleftharpoons A\left(g\right)+B\left(g\right)$, AB is 33% dissociatated at a total pressure of P. Then

• A. $P=K_p$
• B. $P=4K_p$
• C. $P=3K_p$
• D. $P=8K_p$

Answer 1

The correct option is D $P=8K_p$

$$\begin{array}{ccccc}AB\left(g\right)& \rightleftharpoons & A\left(g\right)& +& B\left(g\right)\\ 1& & 0& & 0\\ 1-\frac{1}{3}=\frac{2}{3}& & \frac{1}{3}& & \frac{1}{3}atequilibrium\end{array}$$
$\text{total number of moles}=$
$(\Sigma n{)}_{eqlm}=\frac{2}{3}+\frac{1}{3}+\frac{1}{3}=\frac{4}{3}\text{Partial pressure of A=B=}\frac{1/3}{4/3}\times P=\frac{P}{4}$
Partial pressure of AB= $\frac{2/3}{4/3}\times P=\frac{P}{2}$
From the equation,
$AB\left(g\right)\rightleftharpoons A\left(g\right)+B\left(g\right)$
$K_p$ can be given as:
$K_p=\frac{P_A{P}_B}{P_{AB}}$
Substituting the values, we get
$K_p=\frac{P/4\times P/4}{P/2}$
$K_p=\frac{P}{8}$
$P=8K_P$