Step 1: Understanding the Topic
This question deals with chemical equilibrium in the gas phase. It requires calculating the change in the number of moles of gas ($\Delta n_g$) for the reaction and determining the total number of moles at equilibrium, given the initial moles and the degree of dissociation ($\alpha$).
Step 2: Key Approach
First, we will calculate $\Delta n_g$ by comparing the stoichiometry of the gaseous products and reactants. Second, we will use an ICE (Initial, Change, Equilibrium) table approach to track the number of moles of each species as the reaction proceeds to equilibrium based on the given degree of dissociation.
Step 3: Detailed Calculation
A. Calculating $\Delta n_g$:
$\Delta n_g$ is the total moles of gaseous products minus the total moles of gaseous reactants.
\[
\Delta n_g = (\text{moles of } NO(g) + \text{moles of } NO_2(g)) - (\text{moles of } N_2O_3(g))
\]
From the balanced equation, the stoichiometric coefficients are:
\[
\Delta n_g = (1 + 1) - 1 = 1
\]
B. Calculating Total Moles at Equilibrium:
Let's set up an ICE table, starting with 1 mole of $N_2O_3$. The degree of dissociation, $\alpha = 0.2$, means that 20% of the initial reactant dissociates.
Initial moles (I): $n(N_2O_3) = 1$, $n(NO) = 0$, $n(NO_2) = 0$.
Change in moles (C): The amount of $N_2O_3$ that dissociates is $1 \times \alpha = 1 \times 0.2 = 0.2$ moles. So, the change is $-0.2$ for the reactant. Based on stoichiometry, the change for the products is $+0.2$ for NO and $+0.2$ for $NO_2$.
Equilibrium moles (E):
$n(N_2O_3)_{eq} = 1 - 0.2 = 0.8$
$n(NO)_{eq} = 0 + 0.2 = 0.2$
$n(NO_2)_{eq} = 0 + 0.2 = 0.2$
The total number of moles at equilibrium is the sum of the moles of all species present:
\[
n_{total} = n(N_2O_3)_{eq} + n(NO)_{eq} + n(NO_2)_{eq}
\]
\[
n_{total} = 0.8 + 0.2 + 0.2 = 1.2 \text{ moles}
\]
Step 4: Final Answer
The calculated values are:
\[
\boxed{\Delta n_g = 1}
\]
\[
\boxed{n_{total} = 1.2}
\]