# How do you write the formation equation for (i) CH_3COOH and (ii) NaHS?

Apr 11, 2017

I assume you mean the formation reactions, i.e. the reactions that form the products from their elements in their standard states; that is, at ${25}^{\circ}$ $C$ and $1$ $a t m$.

If you are unsure which phases to use, look in the Appendix of your textbook and find your element... if you see $\Delta {H}_{f}^{\circ} = 0$ or $\Delta {G}_{f}^{\circ} = 0$, you've found the standard state for that element.

Well, acetic acid consists of $\text{C}$, $\text{H}$, and $\text{O}$, so determine their standard states. Carbon is graphite at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, whereas hydrogen and oxygen are diatomic gases. Hence, we begin with the UNBALANCED reaction:

$\text{C"(gr) + "H"_2(g) + "O"_2(g) -> "CH"_3"COOH}$

(Your book might denote carbon as $C \left(s\right)$, though evidently carbon diamond is also a solid, so $g r$ is clearer in my opinion.)

Balance to get:

$\textcolor{b l u e}{2 \text{C"(gr) + 2"H"_2(g) + "O"_2(g) -> "CH"_3"COOH} \left(l\right)}$

(I know that acetic acid in its standard state is a liquid, because that is how it is when it is in its pure form; its melting point is around $\text{290 K}$, so by $\text{298 K}$, it is a liquid.)

It should be easy to point to what elements are in sodium hydrosulfide: $\text{Na}$, $\text{H}$, and $\text{S}$. Now identify their standard states, which are again, how you see them in real life at room temperature and pressure, ${25}^{\circ} \text{C}$ and $\text{1 atm}$.

The only tricky element here is sulfur, whose elemental state is "orthorhombic sulfur", which we denote as ${\text{S}}_{8} \left(o r t h o\right)$ (it is the sulfur analog of cyclohexane).

The UNBALANCED reaction is therefore:

$\text{Na"(s) + "H"_2(g) + "S"_8(o rtho) -> "NaHS} \left(s\right)$

($\text{NaHS}$ has a melting point of ${350}^{\circ} \text{C}$, so clearly it is a solid at room temperature. Even without knowing its melting point, it is a fairly ionic compound, so its melting point should be reasonably high.)

I'll leave it up to you to inspect this answer properly and balance this second reaction yourself. Note that it is entirely reasonable to have fractional stoichiometric coefficients for formation reactions.