The equation alpha = frac{D - d}{(n - 1)d} re | vedclass

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

Question

(B) The general formula for the degree of dissociation $\alpha$ is given by $\alpha = \frac{D - d}{(n - 1)d}$,where $n$ is the total number of moles o

Vedclass · Accepted Answer

$A \rightleftharpoons \frac{nB}{2} + \left( \frac{2n}{3} \right)C$

Vedclass · Answer

(B) The general formula for the degree of dissociation $\alpha$ is given by $\alpha = \frac{D - d}{(n - 1)d}$,where $n$ is the total number of moles of products formed from $1$ mole of reactant.For the reaction $A \rightleftharpoons xB + yC$,the total number of moles of products is $n = x + y$.Substituting $n = x + y$ into the formula,we get $\alpha = \frac{D - d}{(x + y - 1)d}$.Checking option $B$: $A \rightleftharpoons \frac{n}{3}B + \frac{2n}{3}C$.Here,the sum of stoichiometric coefficients is $x + y = \frac{n}{3} + \frac{2n}{3} = \frac{3n}{3} = n$.Substituting this into the denominator $(x + y - 1)$,we get $(n - 1)$,which matches the given equation.

The equation $\alpha = \frac{D - d}{(n - 1)d}$ represents the degree of dissociation. For which of the following reactions is this equation valid,where $D$ is the theoretical vapour density and $d$ is the observed vapour density?

Similar Questions

If in the reaction $N_2O_4 \rightleftharpoons 2NO_2$,$\alpha$ is the fraction of $N_2O_4$ that dissociates,then the total number of moles at equilibrium is:

For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

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....$\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module