# Why is octasulfur present in equation for standard enthalpy of formation of sulfuric acid?

Write a balanced chemical equation corresponding to the standard enthalpy of formation of $$\ce{H2SO4}.$$

\begin{align} \ce{1/8 S8 (s) + H2(g) + 2 O2(g) &-> H2SO4(l)} \label{rxn:r1}\tag{R1}\\ \ce{S(s) + H2(g) + 2 O2(g) &-> H2SO4(l)} \label{rxn:r2}\tag{R2} \end{align}

How can I know I have to use $$\ce{S8}$$ rather than simply $$\ce{S}$$?

• Your answer would have been correct if $\ce{S}$ was available as a pure substance. The trouble is that the free sulfur atom tends to join to other sulfur atoms at room temperature to make $\ce{S8}$ octogones. Nobody knows why. So $\ce{S(s)}$ does not exist. Apr 27 at 5:24
• It is important to know what is and what isn't important. Writing S as S8, for one thing, isn't. Apr 27 at 6:25
• You know it the same way you know to write $\ce{O2}$ instead of $\ce{O}$: from knowledge of the chemistry. And it is important to know this, since the free energy for that reaction includes breaking the covalent bonds within $\ce{S8}$, which wouldn't be the case if it were instead $\ce{S}$. Apr 27 at 6:47
• Well, IMHO, if we cannot use S(s) with implied whatever structure the solid sulfur has, we cannot use NaCl(s) nor NaHCO3(s) either, as there is no NaCl nor NaHCO3 in these solid salts. Apr 27 at 10:27
• Apr 27 at 10:29

So according to it, the constituent elements should be in their standard states. Now coming to question, the standard state of sulphur is rhombic sulphur and it exists as $$\ce{S8}$$ molecule in nature but not simply as individual atoms.