Solid is vaporized and decomposes according to at . At equilibrium, the total pressure is and the de — Chemical Equilibrium Chemistry Question
Question
Solid $SO_3$ is vaporized and decomposes according to $SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$ at $1000\text{ K}$. At equilibrium, the total pressure is $1.642\text{ atm}$ and the density of the gaseous mixture is $1.28\text{ g/L}$. What is the degree of dissociation of $SO_3$? (Use $R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1}$)
💡 Solution & Explanation
First, find the observed molar mass ($M_{obs}$) using $M_{obs} = \frac{dRT}{P} = \frac{1.28 \times 0.0821 \times 1000}{1.642} = \frac{105.088}{1.642} = 64\text{ g/mol}$. The theoretical molar mass of $SO_3$ is $32 + 3 \times 16 = 80\text{ g/mol}$. For the reaction, 1 mole of $SO_3$ gives $1 + 0.5 = 1.5$ moles of products, so $n = 1.5$. Using $\alpha = \frac{M_{th} - M_{obs}}{(n-1)M_{obs}} = \frac{80 - 64}{(1.5 - 1) \times 64} = \frac{16}{0.5 \times 64} = \frac{16}{32} = 0.5$.