Question

If in the reaction $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}$, x is that part of $\mathrm{N}_{2} \mathrm{O}_{4}$ which dissociates, then the number of moles at equilibrium will be:
A. 1
B. 3
$C. \ (1+x)$
$\mathrm{D. }(1+\mathrm{x})^{2}$

Answer 1

Hint: As per the given equation, one $\mathrm{N}_{2} \mathrm{O}_{4}$ is in equilibrium with two moles $\mathrm{NO}_{2}$. At the initial state, number of moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ is $1_{\text {and }}$ and number of moles of $2 \mathrm{NO}_{2}$ is zero. At the equilibrium state, number of moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ is decreased by x after dissociation and on the other hand, moles of $2 \mathrm{NO}_{2}$ are increased by x. To find the total number of moles at equilibrium, add the number of moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ and the number of moles of $2 \mathrm{NO}$, at the equilibrium point.

Complete Step by Step Answer:
Given chemical equation,
$\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}$
Initially, no part of $\mathrm{N}_{2} \mathrm{O}_{4}$ has been dissociated, thus the number of moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ is 1 , and 0 moles of $\mathrm{NO}_{2}$ 5uch as
$\mathrm{N}_{2} \mathrm{O}_{a} \rightleftharpoons 2 \mathrm{NO}_{2}$
1 mole 0 moles (initially)

After dissociation of $x$ part of $\mathrm{N}_{2} \mathrm{O}_{4}$, moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ has decreased by $x$ amount whereas moles of $2 \mathrm{NO}_{7}$ increase by the same amount, $x$. As at equilibrium, there will be the formation of $\mathrm{Z}$ moles of $\mathrm{NO}_{2}$ from $\mathrm{N}_{2} \mathrm{O}_{4}$ thus,
$$
\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}
$$
$1-x$ moles $2 x$ moles (at equilibrium)

Now, to find the total number of moles at equilibrium, the simple addition of number of moles of reactant (N_reactant and number of moles of product (N_product) individually take place such as
$\mathrm{N}_{\text {reactant }}+\mathrm{N}_{\text {product }}$

The number of moles of $\mathrm{N}_{2} \mathrm{O}_{4}$ (reactant) is 1 - $x$ and the number of moles of $2 \mathrm{NO}_{2}(p r o d u c t)$ is $2 x$, thus, the total number of moles at equilibrium is
$\mathrm{N}_{\mathrm{N} 2 \mathrm{O} 4}+\mathrm{N}_{2 \mathrm{NOO}}$
$(1-x)+2 x=1+x$
Thus, the correct option is C.

Note: It is important to note that, as two moles of $2 \mathrm{NO}_{2}$ has been formed from one mole of $\mathrm{N}_{2} \mathrm{O}_{4}$ thus $\mathrm{x}$ amount which has been dissociated from it, is added to two of the moles of $\mathrm{NO}_{2}$ such as:
$\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{NO}_{2}+\mathrm{NO}_{2}$
1 mole 0 mole 0 mole (initially)
$1-x \quad 0+x \quad 0+x \quad$ (at equilibrium)
$1-x \quad x \quad x$
Now the number of moles at the reactant is equal to $1-x$ and the number of moles on the product side is $x+x=2 x$ moles for $2 \mathrm{NO}_{2}$.