# How would you balance the following equation: S8 + O2 --> SO3?

Nov 3, 2015

With ease! Or at least it's easy if you have a chemical industry.

#### Explanation:

${S}_{8} \left(s\right) + 12 {O}_{2} \left(g\right) \rightarrow 8 S {O}_{3} \left(g\right)$; (i)

Alternatively,
$\frac{1}{8} {S}_{8} \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$ (ii)

Note that this gives the stoichiometric conversion of $S \left(0\right)$ to $S \left(V I\right)$. Industrially sulfur trioxide (and what do you think this is used for?) derives from $S {O}_{2} \left(g\right)$.

i.e. $\frac{1}{8} {S}_{8} \left(s\right) + {O}_{2} \left(g\right) \rightarrow S {O}_{2} \left(g\right)$; (iii)

then $S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$ (iv)

In the third equation I used the $\frac{1}{8}$ coefficient to cope with the molecularity of the sulfur ring (then I used a $\frac{1}{2}$ coefficient in the fourth equation; I am certainly free to do this). I am not free to leave the equation unbalanced. Is it balanced? Don't trust my arithmetic. The fourth equation to give $S {O}_{3}$ uses (I think) a supported ${V}_{2} {O}_{5}$ catalyst of some sort. It must be an incredibly dirty and smelly process, yet it is undoubtedly vital to our civilization.

Note that $S {O}_{3} \left(g\right)$ is often added to sulfuric acid to give ${H}_{2} {S}_{2} {O}_{7}$, pyrosulfuric acid.