Chapter 6, Problem 6.125QP

Interpretation Introduction

Interpretation:

The standard enthalpies of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{and}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ and change in enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ have to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.125QP

Standard enthalpy of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{=226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{C}}_6{\text{H}}_6\text{=49}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

Change in enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$= $\text{-630}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

The chemical equations can be given,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{6(l)}}\text{+}\frac{\text{15}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{6CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-3267}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+}\frac{\text{5}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{2CO}}_{\text{2(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-1299}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Using the values of standard enthalpies of formation,

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$= $\text{-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Water = $\text{-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

The equations can be given as,

$\begin{array}{l}{\text{C}}_{\left(\text{Graphite}\right)}{\text{+O}}_{\text{2}}\to {\text{CO}}_{\text{2(g)}}\text{ΔH}°\text{=-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The equations are summed up to get the standard enthalpy of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$

$\begin{array}{l}{\text{C}}_{\left(\text{Graphite}\right)}{\text{+O}}_{\text{2}}\to {\text{CO}}_{\text{2(g)}}\text{ΔH}°\text{=-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$${\text{C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+}\frac{\text{5}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{2CO}}_{\text{2(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-1299}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

${\text{2C}}_{\left(\text{Graphite}\right)}{\text{+H}}_{\text{2(g)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{ΔH°=+226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$

Therefore, standard enthalpy of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ = $\text{226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$

Similarly, the equations are summed up to get the standard enthalpy of formation of ${\text{C}}_6{\text{H}}_6$

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{6(l)}}\text{+}\frac{\text{15}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{6CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-3267}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\\ \end{array}$

$\begin{array}{l}{\text{C}}_{\left(\text{Graphite}\right)}{\text{+O}}_{\text{2}}\to {\text{CO}}_{\text{2(g)}}\text{ΔH}°\text{=-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

${\text{6C}}_{\left(\text{graphite}\right)}{\text{+3H}}_{\text{2(g)}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{6(l)}}\text{ΔH°=+49}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the change in enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$

The equation can be given as,

${\text{3C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{6(g)}}$

$\begin{array}{l}{\text{ΔH}}_{\text{reaction}}\text{=(1)(49}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-(3)(226}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{ΔH}}_{\text{reaction}}\text{=-630}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Change in enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$= $\text{-630}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Conclusion

The change in enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ was calculated using the values of Standard enthalpy of formation of products and reactants.  The standard enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{from}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ was found to be $\text{-630}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$.