Question

An arbitrary compound $P_{2}Q$ decomposes according to reaction :
$2P_{2}Q\left(g\right)\rightleftharpoons 2P_2\left(g\right)+Q_2\left(g\right)$
If initially, decomposition reaction starts with 4 moles of $P_{2}Q$ and value of equilibrium constant $K_p$ is numercially equal to total pressure at equilibrium. Then which option is/are correct at equilibrium :

• A. moles of $n_{P_{2}Q}=n_{Q_2}$
• B. moles of $n_{P_2}=\left(\frac{8}{3}\right)$
• C. degree of dissociation is $\alpha =\left(\frac{2}{3}\right)$
• D. total number of moles of products $\left(P_2\text{and}{Q}_2\right)$ at equilibrium is 4

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A moles of $n_{P_{2}Q}=n_{Q_2}$
B moles of $n_{P_2}=\left(\frac{8}{3}\right)$
C degree of dissociation is $\alpha =\left(\frac{2}{3}\right)$
D total number of moles of products $\left(P_2\text{and}{Q}_2\right)$ at equilibrium is 4

$\begin{array}{ccccccc}& 2P_{2}Q& \rightleftharpoons & 2P_2& +& Q_2& \\ t=0& 4& & 0& & 0& \\ t=\text{eq}& 4(1-\alpha )& & 4\alpha & & 2\alpha & \end{array}$
$K_p=\frac{{(\frac{4\alpha }{4+2\alpha }\times P)}^2(\frac{2\alpha }{4+2\alpha }\times P)}{{(\frac{4(1-\alpha )}{4+2\alpha }\times P)}^2}=\frac{2{\alpha }^3}{(1-\alpha {)}^2(4+2\alpha )}\times P$
But $K_p=P$(given) $\therefore 2{\alpha }^3=(1-\alpha {)}^2(4+2\alpha )$
$\therefore -6\alpha +4=0\therefore \alpha =\left(\frac{2}{3}\right)$
$\text{Total moles at equilibrium}=4+2\alpha =\left(\frac{16}{3}\right)$
$n_{P_{2}Q}=n_{Q_2}=\left(\frac{4}{3}\right),n_{P_2}=\left(\frac{8}{3}\right)$
$\therefore$ Total number of moles of products $=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4$