For the reaction N_2O_{4(g)} rightleftharpoon | vedclass

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

Question

(A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:Initial moles: $1$ $0$At equilibrium: $(1-\alpha)$ $2\alpha$Total moles at equilibriu

Vedclass · Accepted Answer

$\alpha=\sqrt{\frac{K_p}{K_p+4P}}$

Vedclass · Answer

(A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:Initial moles: $1$ $0$At equilibrium: $(1-\alpha)$ $2\alpha$Total moles at equilibrium $= 1-\alpha+2\alpha = 1+\alpha$.Partial pressure of $N_2O_4 = \frac{1-\alpha}{1+\alpha} P$.Partial pressure of $NO_2 = \frac{2\alpha}{1+\alpha} P$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\frac{2\alpha}{1+\alpha} P)^2}{\frac{1-\alpha}{1+\alpha} P} = \frac{4\alpha^2 P}{1-\alpha^2}$.Rearranging for $\alpha$:$K_p(1-\alpha^2) = 4\alpha^2 P$$K_p = \alpha^2(4P + K_p)$$\alpha^2 = \frac{K_p}{4P+K_p}$$\alpha = \sqrt{\frac{K_p}{4P+K_p}}$.

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

Similar Questions

For the equilibrium $PCl_{5_{(g)}} \rightleftharpoons PCl_{3_{(g)}} + Cl_{2_{(g)}}$,the observed vapour density of the mixture is $80$. Given atomic masses $P = 31$ and $Cl = 35.5$,the degree of dissociation of $PCl_{5_{(g)}}$ is approximately....$\%$

If $PCl_5$ undergoes $80\%$ dissociation at $250 \, ^\circ C$,what is its vapour density at this temperature?

What is the vapour density of pure ozone $(O_3)$?

At temperature,$T$,a compound $AB_{2(g)}$ dissociates according to the reaction; $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is small compared with unity. The expression for $K_p$,in terms of $x$ and the total pressure,$P$ is

$2SO_3 \rightleftharpoons 2SO_2 + O_2$. Initially,$3$ moles of $SO_3$ were heated. If the degree of dissociation of $SO_3$ is $40\%$,find the total number of moles at equilibrium.

Vedclass Test Series

Exam Paper Generator

Online Exam Module