Question

For the reaction N$_2$ + O$_2$ $\rightleftharpoons$ 2NO, equilibrium constant K$_c$= 2. Degrees of dissociation of N$_2$ and O$_2$ are:
A) $\frac{1}{1+\sqrt{2}}$, $\frac{1}{1-\sqrt{2}}$
B) $\frac{1}{1-\sqrt{2}}$, $\frac{1}{1+\sqrt{2}}$
C) Both are $\frac{1}{1+\sqrt{2}}$
D) $\frac{2}{1+\sqrt{2}}$, $\frac{2}{1-\sqrt{2}}$

Answer 1

Hint: The dissociation degree is the fraction of original solute molecules that have dissociated. The degree of dissociation of a weak electrolyte is inversely proportional to the inverse square root of the concentration, or square root of the dilution.

Complete answer: Firstly, the equilibrium constant (K$_c$) is given i.e. 2
We have to find the degree of dissociation; it is represented by the $\alpha$.
Let us write the chemical reaction N$_2$ + O$_2$ $\rightleftharpoons$ 2NO, consider the initial concentration of N$_2$, and O$_2$ be c.
Thus,
N$_2$ + O$_2$ $\rightleftharpoons$ 2NO
 c c 0 (initial)
            c(1-$\alpha$) c(1-$\alpha$) 2$\alpha$c (equilibrium)
In this equations we have represented the two terms, one is for initial reaction, and other is for the final chemical equilibrium.
Thus, from the above data, we can write the equilibrium constant in terms of concentration, i.e.
            K$_c$ = (2$\alpha$c)$^{2}$/ c$^{2}$(1-$\alpha$)$^{2}$
              2 = (2$\alpha$ / 1-$\alpha$) $^{2}$
By solving this equation for the term $\alpha$ = $\frac{1}{1+\sqrt{2}}$, or we can say the degree of dissociation is 0.414.
The degree of dissociation for both N$_2$, and O$_2$ is $\frac{1}{1+\sqrt{2}}$.

The correct option is (C).

Note: Don’t get confused about why we considered the same value for the initial, and the final concentration of , and . This equation is a balanced chemical equation and is having reversible dissociation in chemical equilibrium. The greater the degree of dissociation; the equilibrium constant doesn’t change upon dilution, whereas the concentration of each species decreases.