How do you write the formation equation for $\left( i \right)C{H_3}COOH$ and $\left( {{\text{ii}}} \right){\text{NaHS}}$?

Hint: The formation reactions are the reactions where the products are formed from the reactant elements in their standard states. One can use the elements required in their standard state to form the $C{H_3}COOH$ and ${\text{NaHS}}$

Complete step by step answer:
1) First of all, as we need to form the products as $C{H_3}COOH$ and ${\text{NaHS}}$ the reaction is called the formation reaction. In the case of formation reaction, the reactant elements which are required to form the products are taken in their standard states that means at temperature ${25^o}C$ and pressure of ${\text{1 atm}}$.
2) Now let us see the balanced formation reaction of the $C{H_3}COOH$ as below,
${C_{(s)}} + 2{H_{2(g)}} + {O_{2(g)}} \to C{H_3}COO{H_{(l)}}$
In the above reaction, the carbon atom is in solid-state in its standard state. Hydrogen and oxygen elements are in their gaseous state in their standard state. The product formed is in the liquid state in its standard state. The acetic acid is present in the liquid state in its pure form which has the melting point around ${\text{290 K}}$.
3) Now let us see the balanced formation reaction of the ${\text{NaHS}}$ as below,
$8N{a_{(s)}} + 4{H_{2(g)}} + {S_{8(ortho)}} \to 8NaH{S_{(s)}}$
In the above reaction, sodium metal is taken as solid-state and the hydrogen element is taken as gas as per their respective standard states. The sulfur element is taken in its ortho state which is also called orthorhombic sulfur. The sodium hydrosulfide has the melting point as ${350^o}C$ and is present in solid-state in its standard form.

Note:
The sodium hydrosulfide is an ionic compound hence it has a very high melting point. Hydrogen and oxygen are the diatomic gases in their standard form. One must remember to balance the whole formation reaction. Whenever the values $\Delta {H^0}_f = 0$ and $\Delta {G^0}_f = 0$ for an element then that is the standard state of that element.