Question

If a sample of pure $SO_3$ gas is heated to 600°C, it dissociates into $SO_2$ and $O_2$ gases upto 50%. If the average molar mass of the final sample is $M_{avg}$ find value of $(\frac{M_{avg}}{100})$.

Answer 1

0.64

Answer 2

The dissociation reaction of $SO_3$ is:

$SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$

Let's assume we start with 1 mole of pure $SO_3$ gas. The initial molar mass of $SO_3$ ($M_{initial}$) is:

$M_{SO_3} = \text{Atomic mass of S} + 3 \times \text{Atomic mass of O}$ $M_{SO_3} = 32 + 3 \times 16 = 32 + 48 = 80 \text{ g/mol}$

The degree of dissociation ($\alpha$) is given as 50%, which means $\alpha = 0.5$.

Let's set up an ICE table for the reaction:

Initial moles:
$SO_3$: 1 mole
$SO_2$: 0 moles
$O_2$: 0 moles

Change in moles due to dissociation:
$SO_3$: $- \alpha$
$SO_2$: $+ \alpha$
$O_2$: $+ \frac{1}{2}\alpha$

Equilibrium moles:
$SO_3$: $1 - \alpha$
$SO_2$: $\alpha$
$O_2$: $\frac{1}{2}\alpha$

Substitute the value of $\alpha = 0.5$:
Moles of $SO_3 = 1 - 0.5 = 0.5$ moles
Moles of $SO_2 = 0.5$ moles
Moles of $O_2 = \frac{1}{2}(0.5) = 0.25$ moles

Total moles at equilibrium ($n_{total}$):
$n_{total} = (1 - \alpha) + \alpha + \frac{1}{2}\alpha = 1 + \frac{1}{2}\alpha$
$n_{total} = 1 + \frac{1}{2}(0.5) = 1 + 0.25 = 1.25$ moles

The total mass of the system remains constant because mass is conserved. The initial mass is the mass of 1 mole of $SO_3$, which is 80 g.
So, Total mass at equilibrium = 80 g.

The average molar mass of the final sample ($M_{avg}$) is calculated as:
$M_{avg} = \frac{\text{Total mass}}{\text{Total moles at equilibrium}}$
$M_{avg} = \frac{80 \text{ g}}{1.25 \text{ moles}}$
$M_{avg} = \frac{80}{\frac{5}{4}} = \frac{80 \times 4}{5} = \frac{320}{5} = 64 \text{ g/mol}$

Alternatively, for a reaction where 1 mole of reactant dissociates to form 'n' moles of products, the average molar mass ($M_{avg}$) is related to the initial molar mass ($M_{initial}$) and degree of dissociation ($\alpha$) by the formula:

$M_{avg} = \frac{M_{initial}}{1 + (n-1)\alpha}$

For the reaction $SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$, 1 mole of $SO_3$ gives $1 + 0.5 = 1.5$ moles of gaseous products. So, $n = 1.5$.
$M_{initial} = 80 \text{ g/mol}$
$\alpha = 0.5$
$M_{avg} = \frac{80}{1 + (1.5 - 1) \times 0.5}$
$M_{avg} = \frac{80}{1 + (0.5) \times 0.5}$
$M_{avg} = \frac{80}{1 + 0.25}$
$M_{avg} = \frac{80}{1.25} = 64 \text{ g/mol}$

The question asks for the value of $(\frac{M_{avg}}{100})$.
$(\frac{M_{avg}}{100}) = \frac{64}{100} = 0.64$