Chapter 9, Problem 9.52PAE

9.52 Write the formation reaction for each of the following substances. (a) ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ , (b) ${\text{C}}_{\text{3}}\text{H}\left(l\right)$ , (c) HCI(g), (d) ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}\text{O6}\left(\text{s}\right)$ , (e) NaF(s)

Interpretation Introduction

Interpretation: The formation reaction for each of the given substances has to be written.

Concept introduction:

• Formation reaction is the reaction for the formation of exactly one mole of a substance from constituent elements in their standard states.
• These are the benchmark reactions if the enthalpy of these benchmark reactions are calculated, enthalpies of the other reactions which can be constructed from these benchmark reactions can also be calculated.
• Certain rules are followed for writing the formation reaction.
• Only 1 mole of the product should be formed. For example, ${\text{H}}_{\text{2}}{\text{+Cl}}_{\text{2}}\to \text{2HCl}$ is not a formation reaction because 2 moles of the product is formed. The formation reaction is: $\frac{1}{2}{\text{H}}_{\text{2}}\text{+}\frac{1}{2}{\text{Cl}}_{\text{2}}\to \text{HCl}$.
• The reactants should be in their elementary states. For example, ${\text{CaO + CO}}_{\text{2}} \to {\text{CaCO}}_{\text{3}}$

  ${\text{H}}_{\text{2}}{\text{+Cl}}_{\text{2}}\to \text{2HCl}$ is not a formation reaction because it is produced from two other compounds and not its elements.

Answer to Problem 9.52PAE

Solution:

(a) $\text{C}\left(\text{s}\right){\text{ + 2 H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)$

(b) $\text{3C}\left(\text{s}\right){\text{ + 4 H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{3}}{\text{H}}_8(l)$

(c) $\frac{1}{2}{\text{H}}_{\text{2}}(g)\text{+}\frac{1}{2}{\text{Cl}}_{\text{2}}(g)\to \text{HCl}(g)$

(d) $\text{6C}{(s)}_{ }+ {\text{ 6H}}_{\text{2}}(g) +{\text{ 3O}}_{\text{2}}(g) \to {\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)$

(e) $\text{Na}{(s)}_{ }+ \frac{1}{2}{\text{F}}_{\text{2}}(g)\to \text{ NaF}(s)$

(a) The formation reaction for ${\text{CH}}_{\text{4}}\text{(g)}$

Explanation of Solution

The formation reaction for ${\text{CH}}_{\text{4}}\text{(g)}$ is written as: $\text{C}\left(\text{s}\right){\text{ + 2 H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{4}}\left(\text{g}\right)$

This is a proper formation reaction because only one mole of the product is produced and the reactants are in their elemental standard states.

(b) The formation reaction for ${\text{C}}_{\text{3}}{\text{H}}_8(l)$

The formation reaction for ${\text{C}}_{\text{3}}{\text{H}}_8(l)$ is written as: $\text{3C}\left(\text{s}\right){\text{ + 4 H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{3}}{\text{H}}_8(l)$

This is a proper formation reaction because only one mole of the product is produced and the reactants are in their elemental standard states.

(c) The formation reaction for $\text{HCl}(g)$

The formation reaction for $\text{HCl}(g)$ is written as: $\frac{1}{2}{\text{H}}_{\text{2}}(g)\text{+}\frac{1}{2}{\text{Cl}}_{\text{2}}(g)\to \text{HCl}(g)$

This is a proper formation reaction because only one mole of the product is produced and the reactants are in their elemental standard states.

(d) The formation reaction for ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6(s)$

The formation reaction for ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_6(s)$ is written as: $\text{6C}{(s)}_{ }+ {\text{ 6H}}_{\text{2}}(g) +{\text{ 3O}}_{\text{2}}(g) \to {\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)$

This is a proper formation reaction because only one mole of the product is produced and the reactants are in their elemental standard states.

(e) The formation reaction for $\text{NaF}(s)$

The formation reaction for $\text{NaF}(s)$ is written as: $\text{Na}{(s)}_{ }+ \frac{1}{2}{\text{F}}_{\text{2}}(g)\to \text{ NaF}(s)$

This is a proper formation reaction because only one mole of the product is produced and the reactants are in their elemental standard states

Conclusion

The proper formation reactions for the given compounds have been determined.