To write a balanced chemical equation for the standard formation reaction of liquid octanol ([tex]\(\text{C}_8\text{H}_{18}\text{O}\)[/tex]), we need to follow these steps:
1. Identify the elements in their standard states:
- Carbon in its standard state is graphite ([tex]\(\text{C}\)[/tex]).
- Hydrogen in its standard state is diatomic hydrogen gas ([tex]\(\text{H}_2\)[/tex]).
- Oxygen in its standard state is diatomic oxygen gas ([tex]\(\text{O}_2\)[/tex]).
2. Determine the formation reaction:
- The formation reaction entails forming one mole of the compound (in this case, octanol) from its constituent elements in their standard states.
3. Write the unbalanced equation:
- The unbalanced equation, representing the combination of elements to form one mole of liquid octanol ([tex]\(\text{C}_8\text{H}_{18}\text{O}\)[/tex]), is:
[tex]\[
\text{C} + \text{H}_2 + \text{O}_2 \rightarrow \text{C}_8\text{H}_{18}\text{O}
\][/tex]
4. Balance the equation:
- Balance carbon (C): Octanol has 8 carbon atoms, so we need [tex]\(8 \text{C}\)[/tex] (graphite).
[tex]\[
\text{8C} + \text{H}_2 + \text{O}_2 \rightarrow \text{C}_8\text{H}_{18}\text{O}
\][/tex]
- Balance hydrogen (H): Octanol has 18 hydrogen atoms, and since diatomic hydrogen ([tex]\(\text{H}_2\)[/tex]) contains 2 atoms per molecule, we need [tex]\( \frac{18}{2} = 9 \)[/tex] molecules of [tex]\(\text{H}_2\)[/tex].
[tex]\[
8\text{C} + 9\text{H}_2 + \text{O}_2 \rightarrow \text{C}_8\text{H}_{18}\text{O}
\][/tex]
- Balance oxygen (O): Octanol has 1 oxygen atom, so we need [tex]\(\frac{1}{2}\)[/tex] of a molecule of [tex]\(\text{O}_2\)[/tex] to provide 1 oxygen atom.
[tex]\[
8\text{C} + 9\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_8\text{H}_{18}\text{O}
\][/tex]
5. Final balanced equation:
- After ensuring that each type of atom is balanced on both sides of the equation, we have:
[tex]\[
8\text{C} (\text{graphite}) + 9\text{H}_2 (\text{g}) + \frac{1}{2}\text{O}_2 (\text{g}) \rightarrow \text{C}_8\text{H}_{18}\text{O} (\text{l})
\][/tex]
Hence, the balanced chemical equation for the standard formation reaction of liquid octanol ([tex]\(\text{C}_8\text{H}_{18}\text{O}\)[/tex]) is:
[tex]\[
8\text{C} (\text{graphite}) + 9\text{H}_2 (\text{g}) + \frac{1}{2}\text{O}_2 (\text{g}) \rightarrow \text{C}_8\text{H}_{18}\text{O} (\text{l})
\][/tex]
