Chapter 17.4, Problem 41LC

Interpretation Introduction

Interpretation: The value of change in enthalpy of the given reaction needs to be calculated using Hess’s law and two thermochemical equations.

Concept Introduction: According to Hess’s law, the total enthalpy change for a complete reaction is equal to the sum of the enthalpy change of all the steps involved in the reaction. This is because enthalpy is a state function.

Explanation of Solution

The given reaction for which change in enthalpy needs to be calculated is as follows:

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

The two thermochemical equations are as follows:

The following reaction shows the production of hydrogen after the reaction of methane and water vapor.

  ${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CH}}_{\text{4}}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\Delta \text{H}=+\text{206 kJ}$   ....... (1)

Now, when carbon monoxide reacts with the added stream, more hydrogen is produced as follows:

  $\text{CO}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\Delta \text{H}=-41\text{ kJ}$   ........ (2)

The given reaction can be obtained by adding reactions (1) and (2) as follows:

  $\begin{array}{l}{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CH}}_{\text{4}}\left(g\right)\to \cancel{\text{CO}\left(g\right)}+3{\text{H}}_2\left(g\right)\Delta \text{H}=+\text{206 kJ}\\ \cancel{\text{CO}\left(g\right)}+{\text{H}}_{\text{2}}\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\Delta \text{H}=-41\text{ kJ}\\ \overline{\underset{¯}{{\text{2H}}_{\text{2}}\text{O}\left(g\right)+{\text{CH}}_{\text{4}}\left(g\right)\to {\text{CO}}_2\left(g\right)+4{\text{H}}_2\left(g\right)\Delta \text{H}=+\text{206 kJ}-41\text{ kJ}=+165\text{ kJ }}}\end{array}$

Therefore, the change in enthalpy for the given reaction is $+165\text{ kJ}$ .