Chapter 15, Problem 107A

Differentiate between the enthalpy of formation of $H_{2}O\left(l\right)and{H}_{2}O\left(g\right)$ . Why is it necessary to specify thephysical state of water in the following thermochemicalequation $C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(lorg\right)\Delta H=?$

Interpretation Introduction

Interpretation:

The difference between enthalpy of formation of ${\text{H}}_{\text{2}}\text{O(l)}\text{and}{\text{H}}_{\text{2}}\text{O(g)}$ needs to be explained and the reason due to which it is necessary to state the physical state of water in equation needs to be explained.

$\begin{array}{l}{\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+2O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{l}\text{or}\text{g}\right)\\ \end{array}$

Concept introduction:

Hess’s law: According to this law, total enthalpy change of a reaction is the sum of all the reactions regardless of the steps or stages of reaction.

Explanation of Solution

The water in liquid state will have higher change in enthalpy as compared to water in gaseous state due to change in intermolecular interactions.

In the process the combustion of methane:

${\text{CH}}_{\text{4}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{CO}}_2(g)+{\text{H}}_{\text{2}}\text{O(g)}$

Combustion in any reaction is exothermic reaction. So heat is produced that means

$\text{ΔH}=\text{-ve}$

Also, it is important to state the physical state of water in any reaction because while breaking the bonds in any reaction, the amount of heat requires depends on the physical state. The change in enthalpy for water in liquid and vapor state is different.