Question

The equation, $\alpha =\frac{D-d}{(n-1)d}$ is correctly matched for:

• A. $A\rightleftharpoons nB/2+nC/3$
• B. $A\rightleftharpoons nB/3+\left(2n/3\right)C$
• C. $A\to \left(n/2\right)B+\left(n/4\right)C$
• D. $A\rightleftharpoons \left(n/2\right)B+C$

Answer 1

Answer: (B)

$A\rightleftharpoons nB/3+\left(2n/3\right)C$

$\alpha =\frac{(D-d)}{(n-1)d}\to \left(1\right)$

$\alpha =$Degree of dissociation

$d=$Vapor density

$n=$ no of moles of gaseous product for $1$ mole of reactant

Any equation in which the total number of moles on product side $=n$ and on reactant side $=1$.

$A)$ $A\rightleftharpoons nB/2+nC/3$

here at reactant side $n=1$

But product side $n=\frac{n}{2}+\frac{n}{3}=\frac{5n}{6}\ne n$

$B)A\rightleftharpoons nB/3+(2n/3)C$

here at reactant side $n=1$

product side $n=\frac{n}{2}+\frac{2n}{3}=\frac{6n}{6}=n$

$C)A\to (n/2)B+(n/4)C$

here at reactant side $n=1$

But product side $n=\frac{n}{2}+\frac{n}{}=\frac{2n}{4}\ne n$

$D)A\rightleftharpoons (n/2)B+C$

here at reactant side $n=1$

But product side $n=\frac{n}{2}+1=\frac{n+1}{2}\ne n$

Hence the correct answer is option $B$.