Question

The equation $\alpha = \dfrac{{D - d}}{{(n - 1)d}}$ is not correctly matched for?
A. $A(g) \rightleftharpoons (n/2)B(g) + (n/3)C(g)$
B. $A(g) \rightleftharpoons (n/3)B(g) + (2n/3)C(g)$
C. $A(g) \rightleftharpoons (n/2)B(g) + (n/4)C(g)$
D. $A(g) \rightleftharpoons (n/2)B(g) + C(g)$

Answer 1

A chemical equation is the symbolic representation of chemical reaction in which we write reactant entities on the left-hand side and product entities on right-hand side. The coefficients written before the symbol are the absolute value of stoichiometric numbers, which are supposed to be 1 for reactants and n for products in total.

Complete step by step answer:
Degree of dissociation is defined as the fraction of a mole of the reactant undergoing dissociation and can be represented in terms of vapour density as a formula. Vapour density is the density of a vapour with respect to hydrogen.
$\alpha = \dfrac{{D - d}}{{(n - 1)d}}$ is the formula for the representation of degree of freedom in the terms of vapour density where $\alpha$ is the degree of dissociation, D and d are the vapour densities before and after dissociation respectively and n is the number of moles of gaseous product for one mole of reactant. We can further rearrange it as
$\alpha = \dfrac{{D - d}}{{(n - 1)d}}$

$$\Rightarrow (n - 1)\alpha = \dfrac{D}{d} - 1 \\\ \Rightarrow \dfrac{D}{d} = 1 + (n - 1)\alpha \\\$$

$\dfrac{d}{D} = \dfrac{1}{{1 + (n - 1)\alpha }} = \dfrac{1}{{(1 - \alpha + n\alpha )}}$
The above equation states that in any equation, the total number of moles on the reactant side is one and the total number of moles on the product side is n.
Now, let us check for all the four options.
For A, $A(g) \rightleftharpoons (n/2)B(g) + (n/3)C(g)$ , we can see that the total number of moles of reactants are 1. But on product side, if we add the number of moles of both B and C, we get n = $\dfrac{n}{2} + \dfrac{n}{3} = \dfrac{{5n}}{6} \ne n$ .
For B, $A(g) \rightleftharpoons (n/3)B(g) + (2n/3)C(g)$ , we can see that the total number of moles of reactants are 1. But on product side, if we add the number of moles of both B and C, we get n = $\dfrac{n}{3} + \dfrac{{2n}}{3} = \dfrac{{6n}}{6} = n$
For C, $A(g) \rightleftharpoons (n/2)B(g) + (n/4)C(g)$ , we can see that the total number of moles of reactants are 1. But on product side, if we add the number of moles of both B and C, we get n = $\dfrac{n}{2} + \dfrac{n}{4} = \dfrac{{2n}}{4} \ne n$
For D, $A(g) \rightleftharpoons (n/2)B(g) + C(g)$ , we can see that the total number of moles of reactants are 1. But on product side, if we add the number of moles of both B and C, we get n = $\dfrac{n}{2} + 1 = \dfrac{{n + 1}}{2} \ne n$
From these calculations, we can say that the equation $\alpha = \dfrac{{D - d}}{{(n - 1)d}}$ correctly matches with option B, but not correctly matches with the rest of the three options.

Hence, the correct answer is (A), (C) and (D).

Note:
This method is only valid for reactions whose ${K_P}$ exists i.e., reactions having at least one gas and having no solution. This means that a homogeneous equilibrium exists in which everything in the equilibrium mixture is present in the same phase. That’s why we use ${K_P}$ here, in which everything (reactants and products) must be a gas.