At temperature T,compound AB_{2(g)} dissociat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The dissociation reaction is: $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-x),

Vedclass · Accepted Answer

$\sqrt[3]{\frac{2 K_p^2}{p}}$

Vedclass · Answer

(C) The dissociation reaction is: $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-x), x, \frac{x}{2}$Total moles at equilibrium: $1-x+x+\frac{x}{2} = 1+\frac{x}{2}$Partial pressures: $P_{AB_2} = \frac{1-x}{1+x/2}P$,$P_{AB} = \frac{x}{1+x/2}P$,$P_{B_2} = \frac{x/2}{1+x/2}P$Since $x \ll 1$,we approximate $1+x/2 \approx 1$ and $1-x \approx 1$.Thus,$P_{AB_2} \approx P$,$P_{AB} \approx xP$,$P_{B_2} \approx \frac{x}{2}P$.$K_p = \frac{P_{AB} \cdot (P_{B_2})^{1/2}}{P_{AB_2}} = \frac{(xP) \cdot (xP/2)^{1/2}}{P} = x \cdot (xP/2)^{1/2} = x^{3/2} \cdot (P/2)^{1/2}$.Squaring both sides: $K_p^2 = x^3 \cdot \frac{P}{2}$.Solving for $x$: $x^3 = \frac{2 K_p^2}{P} \Rightarrow x = \left( \frac{2 K_p^2}{P} \right)^{1/3}$.

At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is

Similar Questions

$AB_{3(g)}$ dissociates as:
$AB_{3(g)} \rightleftharpoons AB_{2(g)} + \frac{1}{2} B_{2(g)}$. When the initial pressure of $AB_3$ is $800 \ torr$ and the total pressure developed at equilibrium is $900 \ torr$,what percentage of $AB_{3(g)}$ is dissociated?

The vapor density of a gas is $16$. If the ratio of specific heat at constant pressure to specific heat at constant volume is $1.4$,find its atomic weight.

The vapour density of undecomposed $N_2O_4$ is $46$. When heated,the vapour density decreases to $24.5$ due to its dissociation into $NO_{2(g)}$. The percentage dissociation of $N_2O_4$ is:

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module