Chapter 11, Problem 11.48QP

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

$\begin{array}{l}{\text{ΔH}}_{\text{r}}{}^{\text{o}}\text{=}{\text{ΣnΔH}}^{\text{o}}\left(\text{products}\right)\text{-}{\text{ΣnΔH}}^{\text{o}}\left(\text{reactants}\right)\\ \text{where,}\\ \text{n}\text{is}\text{Stoichiometric}\text{coefficient}\end{array}$

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

$2{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)+3O_2\left(g\right){\text{ΔH}}^{\text{o}}=+1411kJ/mol$

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

${\text{C}}_{\text{2}}{\text{H}}_2\left(\text{g}\right)+\frac{5}{2}{O}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(\text{l}\right){\text{ΔH}}^{\text{o}}=-1299.5kJ/mol$

Equating the all the given equations as,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right)+3O_2\left(g\right){\text{ΔH}}^{\text{o}}=+1411kJ/mol\\ {\text{C}}_{\text{2}}{\text{H}}_2\left(\text{g}\right)+\frac{5}{2}{O}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(\text{l}\right){\text{ΔH}}^{\text{o}}=-1299.5kJ/mol\\ {\text{H}}_2\left(\text{g}\right)+\frac{1}{2}{O}_2\left(g\right)\to {\text{H}}_2\text{O}\left(\text{l}\right){\text{ΔH}}^{\text{o}}=-285.8kJ/mol\end{array}}\\ overall:{\text{C}}_{\text{2}}{\text{H}}_2\left(\text{g}\right)+{\text{H}}_2\left(\text{g}\right)\to {\text{C}}_{\text{2}}{\text{H}}_4\left(\text{g}\right){\text{ΔH}}^{\text{o}}=-174.3kJ/mol\end{array}}$

The standard enthalpy change in kilojoules per mole for the given reaction is $-174.3kJ/mol$