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 one starts the decomposition reaction with $4$ moles of $P_{2}Q$ and value of equilibrium constant $K_p$ is numerically 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 $n_{P_2}=\left(8/3\right)$
• C. degree of dissociation is $\alpha =\left(2/3\right)$
• D. total number of moles of product $(P_2$ & $Q_2)$ 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 $n_{P_2}=\left(8/3\right)$
C degree of dissociation is $\alpha =\left(2/3\right)$
D total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $4$

$2P_{2}Q\rightleftharpoons 2P_2+Q_2$

$t=0400$

$t=eq4(1-\alpha )4\alpha 2\alpha$

$K_p=\frac{{(\frac{4\alpha }{4+2\alpha }\times P)}^2(\frac{2\alpha }{4+2\alpha }\times P)}{{(\frac{4(1-\alpha )}{4+2\alpha }\times P)}^2}=\frac{2{\alpha }^3}{(1-\alpha {)}^2(4+2\alpha )}\times P$

But $K_p=P$ (given) $\therefore 2{\alpha }^3=(1-\alpha {)}^2(4+2\alpha )$

$\therefore -6\alpha +4=0\therefore \alpha =\left(2/3\right)$

Total moles at eq$=4+2\alpha =\left(16/3\right)$

$n_{P_{2}Q}=n_{Q_2}=\left(4/3\right),n_{P_2}=\left(8/3\right)$

$\therefore$ Total number of moles of products $=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4$