Chapter 6, Problem 145MP

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapour. Given the following data, determine $\Delta H_f^{°}$ for the hydrocarbon:

$\begin{array}{ll}\Delta H_{reaction}^{°}\hfill & =-2044.5\text{KJ/mol hydrocarbon}\hfill \\ \Delta H_f^{°}\left({\text{CO}}_2\right)\hfill & =-393.5\text{KJ/mol}\hfill \\ \Delta H_f^{°}\left({\text{H}}_{2}O\right)\hfill & =-242\text{KJ/mol}\hfill \end{array}$

Density of CO₂ and H₂O product mixture at 1.00 atm, 200.°c = 0.751g/L.

The density of the hydrocarbon is less than the density of Kr at the same conditions.

Interpretation Introduction

Interpretation: The standard enthalpies of formation of given hydrocarbon should be determined.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, with all substances  in its standard states is called as a standard enthalpy of formation.
• Formula:

               ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

Answer to Problem 145MP

Answer

• The standard enthalpies of formation of given hydrocarbon $\text{(C}\text{H}{}_{\text{8(g)}}\text{)}$ is $\text{-104}\text{kJ/mol}$.

Explanation of Solution

Explanation

To calculate the average molar mass of ${\text{CO}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ mixture.

given:

                 $\begin{array}{l}\text{Enthalpy}\text{change}{\text{ΔΗ}}_{\text{reaction}}\text{ = - 2044}\text{.5 kJ/mol hydrocarbon}\\ \text{Standard enthalpies of formation}{\text{CO}}_{\text{2}}\text{is}\text{-393}\text{.5}\text{kJ/mol}\\ \text{Standard enthalpies of formation}{\text{H}}_{\text{2}}\text{O}\text{is}\text{-242}\text{kJ/mol}\\ {\text{Density of CO}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}\text{mixture}\text{is}\text{0}\text{.751g/L}\\ \text{Pressure}\text{is}\text{1}\text{.00}\text{atm}\\ \text{Temperature}\text{is}\text{200°C}\end{array}$

                   $\begin{array}{l}\text{Density}\text{}\text{of}\text{gas}\text{=}\frac{\text{P}\text{.MM}}{\text{RT}}\\ \text{P}\text{is}\text{pressure}\\ \text{MM}\text{is}\text{the}\text{molar}\text{mass}\\ \text{R}\text{is}\text{gas}\text{constant}\\ \text{T}\text{is}\text{temperature}\\ \\ \text{0}\text{.751g/L}\text{=}\frac{\text{1}\text{.00}\text{atm×}\text{MM}}{\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K}\text{mol}}\text{×473}\text{K}}\\ \text{MM}\text{=}\text{29}\text{.1}\text{g/mol}\end{array}$

• The given values are plugging in to above equation and do sum rearrangement to give  the molar mass of ${\text{CO}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ mixture.
• The molar mass of ${\text{CO}}_{\text{2}}{\text{/H}}_{\text{2}}\text{O}$ mixture is $\text{29}\text{.1}\text{g/mol}$.

To calculate the mole of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ .

Let a is the mole of ${\text{CO}}_{\text{2}}$ and b is the mole of ${\text{H}}_{\text{2}}\text{O}$

Molecular weight of ${\text{CO}}_{\text{2}}$ is $\text{44}\text{.01g}$

Molecular weight of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.02}\text{g}$

From the assumption moles of mixture is $\text{1}\text{.00}$

                                   $\text{29}\text{.1=(44}\text{.01)a+(1}\text{.00}\text{-a)×18}\text{.02}$

$\text{solve}\text{the}\text{above}\text{equation,}$

                                  $\begin{array}{l}\text{a}\text{=}\text{0}\text{.426}\text{mole}{\text{CO}}_{\text{2}}\\ \text{b=}\text{0}\text{.574}\text{mole}{\text{H}}_{\text{2}}\text{O}\end{array}$

• The Molecular weight of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ values are plugging in to equation to give mole of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ .
• The mole of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.426}\text{mole}$ and ${\text{H}}_{\text{2}}\text{O}$ $\text{0}\text{.574}$ is .

To determine the given hydrocarbon ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ .

The general representation of hydrocarbon is $\text{CnH2n+2}$

                                              $\text{nC=}\frac{\text{nH-2}}{\text{2}}$

                            $\text{CxHy+}\left(\frac{\text{2x+y/2}}{\text{2}}\right){\text{O}}_{\text{2}}\to {\text{xCO}}_{\text{2}}{\text{+y/2H}}_{\text{2}}\text{O}$

The combustion of hydrocarbon produce x of carbon and y/2 of hydrogen so

                                            $\begin{array}{l}\text{=}\frac{\text{0}\text{.574}}{\text{0}\text{.426}}\text{=}\frac{\text{y/2}}{\text{x}}\\ \text{2}\text{.69=}\frac{\text{y}}{\text{x}}\\ \text{y=}\text{(2}\text{.69)x}\end{array}$

From the result the hydrocarbon is,

Density of the hydrocarbon is less than the density of Kr so

                                     ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$= ${\text{C}}_3{\text{H}}_8$

• The calculated mole of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ where do some mathematical modification to give the number of Carbon and Hydrogen.
• The number of Carbon is 3 number of Hydrogen is 8 so the hydrocarbon is propane.

To calculate standard enthalpies of formation propane.

Standard enthalpies of formation of Carbon is $\text{-393}\text{.5}\text{kJ/mol}$

Standard enthalpies of formation of Hydrogen is $\text{-121}\text{kJ/mol}$

                                    $\begin{array}{l}\text{ΔH°=}{\text{3ΔH}}_{\text{carbon}}{\text{-8ΔH}}_{\text{hydrogen}}{\text{-ΔH}}_{\text{propane}}\\ \text{-2044}\text{.5}\text{=}\text{-3(393}{\text{.5)-8(121)-ΔH}}_{\text{propane}}\\ {\text{ΔH}}_{\text{propane}}\text{=}{\text{ΔH°-3ΔH}}_{\text{carbon}}{\text{-8ΔH}}_{\text{hydrogen}}\\ {\text{ΔH}}_{\text{propane}}\text{=}\text{2044}\text{.5-3(393}\text{.5)-8(121)}\\ \text{=}\text{-104}\text{kJ/mol}\end{array}$

• The standard enthalpies of formation values taking form standard data and they are plugging in equation (1) and do sum rearrangement to give standard enthalpies of formation ${\text{C}}_{\text{3}}{\text{H}}_8\text{(g)}$.
• Standard enthalpies of formation of ${\text{C}}_{\text{3}}{\text{H}}_8\text{(g)}$ is $\text{-104}\text{kJ/mol}$.

Conclusion

Conclusion

The standard enthalpies of formation ${\text{C}}_{\text{3}}{\text{H}}_8\text{(g)}$ was calculated by using Standard enthalpies of formation values and the standard enthalpies of formation of ${\text{C}}_{\text{3}}{\text{H}}_8\text{(g)}$ was found to be $\text{-104}\text{kJ/mol}$.