Question

For the reaction: \[ N_2O_3(g) \rightarrow NO(g) + NO_2(g) \] If the degree of dissociation is $\alpha = 0.2$, find $\Delta n_g$ and total number of moles at equilibrium (starting with 1 mole).

Hint:

For dissociation reactions, total moles increase if $\Delta n_g>0$. Always use: initial moles → change using $\alpha$ → equilibrium moles.

Answer 1

Correct Answer: 1.2

Step 1: Find $\Delta n_g$.
Number of gaseous moles on product side = 2
Number of gaseous moles on reactant side = 1
\[ \Delta n_g = 2 - 1 = 1. \]
Step 2: Initial moles.
Assume 1 mole of $N_2O_3$ initially.
Initial moles:
\[ N_2O_3 = 1, NO = 0, NO_2 = 0. \]
Step 3: Change in moles.
If degree of dissociation = $\alpha = 0.2$, then moles dissociated:
\[ = 1 \times 0.2 = 0.2. \]
Thus:
\[ N_2O_3 \text{ left} = 1 - 0.2 = 0.8. \]
\[ NO = 0.2. \]
\[ NO_2 = 0.2. \]
Step 4: Total moles at equilibrium.
\[ n_{total} = 0.8 + 0.2 + 0.2 = 1.2. \]
Final Answers:
\[ \boxed{\Delta n_g = 1} \] \[ \boxed{n_{total} = 1.2} \]