For which of the following reactions,the degr | vedclass

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

Question

(A) The degree of dissociation $(\alpha)$ can be calculated from vapour density data using the formula $\alpha = \frac{D-d}{(n-1)d}$,where $D$ is the

Vedclass · Accepted Answer

$I$ and $III$

Vedclass · Answer

(A) The degree of dissociation $(\alpha)$ can be calculated from vapour density data using the formula $\alpha = \frac{D-d}{(n-1)d}$,where $D$ is the theoretical vapour density,$d$ is the observed vapour density,and $n$ is the number of moles of products formed from $1$ mole of reactant.This method is applicable only when there is a change in the number of moles during the reaction,i.e.,$\Delta n_g 
eq 0$.Let us calculate $\Delta n_g$ for each reaction:$I. \ 2HI_{(g)} ightleftharpoons H_{2(g)} + I_{2(g)} \Rightarrow \Delta n_g = (1+1) - 2 = 0$$II. \ 2NH_{3(g)} ightleftharpoons N_{2(g)} + 3H_{2(g)} \Rightarrow \Delta n_g = (1+3) - 2 = 2$$III. \ 2NO_{(g)} ightleftharpoons N_{2(g)} + O_{2(g)} \Rightarrow \Delta n_g = (1+1) - 2 = 0$$IV. \ PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)} \Rightarrow \Delta n_g = (1+1) - 1 = 1$Since $\Delta n_g = 0$ for reactions $I$ and $III$,the degree of dissociation cannot be calculated using the vapour density method for these reactions.

Similar Questions

At $444\,^oC$,the equilibrium constant $K$ for the reaction $2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $\frac{1}{64}$. The degree of dissociation of $AB$ will be .....$\%$

$AB_{3(g)}$ 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?

Ammonium carbonate,when heated to $200\, ^oC$,dissociates into a mixture of $NH_3$ and $CO_2$ vapors with a vapor density of $13.0$. What is the degree of dissociation of ammonium carbonate?

In a closed vessel,the dissociation of phosphorus pentachloride occurs as: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If the total pressure at equilibrium is $P$ and the degree of dissociation of $PCl_5$ is $x$,then the partial pressure of $PCl_3$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module