Question

ASSERTION: If the order of reaction is zero, then degree of dissociation will be independent of initial concentration.
REASON: The degree of dissociation of zero order reaction is given by $\alpha =\dfrac{kt}{{{c}_{{\mathrm O}}}}$.
(A) Both assertion and reason are correct and reason is the correct explanation for assertion.
(B) Both assertion and reason are correct but the reason is not the correct explanation for assertion.
(C) Assertion is correct but reason is incorrect.
(D) Assertion is incorrect but reason is correct.

Answer 1

Order of a reaction defines the relationship between the rate of a reaction and the concentration of the reactants involved. Order of any reaction can be determined using rate law equation which is given by the formula
$r=k{{[A]}^{x}}{{[B]}^{y}}$
Where r is the rate of the reaction, k is the rate constant, [A] is the concentration term of species A and x is its order of reaction, and [B] is the concentration term of species B and y is its order of reaction.

Complete answer:
We know that a zero-order reaction is a reaction in which the rate of the reaction is independent of the concentration of the species involved. This means that there is no change in the rate of the reaction even if there is an increase or decrease in the concentration of the reactants.
Now degree of dissociation is defined as the amount of solute dissociated into ions per mole. Its value is always between 0 and 1.
$degree\text{ }of\text{ }dissociation\text{ }(\alpha )=\dfrac{moles\text{ }dissociated}{initial\text{ }moles}$
For a zero-order reaction $\text{A}\to \text{Products}$, the rate law equation will be $r=k{{[A]}^{0}}$, or r=k.
Let us assume that initial concentration of reactant A is ${{c}_{{\mathrm O}}}$, and the initial concentration of product P is 0.
After some time, t, let the concentration of the product P be x, and hence the concentration of the remaining reactant A after time t, is $({{c}_{{\mathrm O}}}-x)$.
Now we know that rate of a reaction can also be defined by
$r=\dfrac{-\partial [A]}{\partial t}$
Since r=k, this can be re-written as
$k=\dfrac{-\partial [A]}{\partial t}$
Using putting the values of concentration of reactant A initially and after time t in the above equation, we get that

$$\Rightarrow k=\dfrac{-\partial [({{c}_{{\mathrm O}}}-x)-{{c}_{{\mathrm O}}})}{\partial t} \\\ \Rightarrow k=\dfrac{\partial x}{\partial t} \\\ \Rightarrow \partial x=k\times \partial t \\\ \text{integrating both sides, we get} \\\ \Rightarrow \int\limits_{0}^{x}{\partial x}=\int\limits_{0}^{t}{k\times \partial t} \\\ \Rightarrow \left[ x \right]_{0}^{x}=k[t]_{0}^{t} \\\ \Rightarrow x=kt\text{ (i)} \\\$$

Where k is the rate constant of the equation and x is the concentration of the product P after some time t.
Now using the degree of dissociation formula, we can say that

$$\alpha =\dfrac{x}{{{c}_{{\mathrm O}}}} \\\ \text{or }x=\alpha {{c}_{{\mathrm O}}} \\\$$

From this and equation (i), we can say that

$$\alpha {{c}_{{\mathrm O}}}=kt \\\ or\text{ }\alpha =\dfrac{kt}{{{c}_{{\mathrm O}}}} \\\$$

Hence it can be concluded that degree of dissociation of a zero-order reaction is
$\alpha =\dfrac{kt}{{{c}_{{\mathrm O}}}}$

Where $\alpha$= degree of dissociation,
k = rate constant of the reaction,
t = time of the reaction,
and ${{c}_{{\mathrm O}}}$ = initial concentration of reactants.
So, the assertion that if the order of reaction is zero, then degree of dissociation will be independent of initial concentration, is false, but the reason is true. Answer to the question is option (D) Assertion is incorrect but reason is correct.

Note: This question can be confusing as one could assume that in a zero-order reaction the degree of dissociation will also be independent of the concentration of the reactants, since the rate of the reaction is independent of the concentration of the reactants. But this assumption would be false as the degree of dissociation is indeed dependent on the initial concentration of the reactants involved.