
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
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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 (Nreactant and number of moles of product (Nproduct) 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 $2 \mathrm{NO}_{2}$.
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 (Nreactant and number of moles of product (Nproduct) 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 $2 \mathrm{NO}_{2}$.
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