Chapter 4, Problem 89QRT

Interpretation Introduction

Interpretation:

The amount of thermal energies produced from one gram of $C{H}_{3}OH$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ has to be compared.

Answer to Problem 89QRT

The enthalpy evolved from one gram of octane is more than enthalpy evolved from one gram of methanol. Methanol is oxidized more than octane hence its combustion will produce a smaller amount of energy.

Explanation of Solution

The equations for combustion of one mole of $C{H}_{3}OH$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ are as follows,

    $\begin{array}{l}C{H}_{3}OH\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(g\right)\\ \\ C_8{H}_{18}\left(l\right)+\frac{25}{2}{O}_2\left(g\right)\to 8C{O}_2\left(g\right)+9H_{2}O\left(g\right)\end{array}$

The combustion enthalpy is as follows,

    $\begin{array}{c}{\Delta }_c{H}^{\circ }{}_{C{H}_{3}OH}\text{ = }\left[{\Delta }_f{H}^{\circ }C{O}_2\left(g\right)+2{\Delta }_f{H}^{\circ }\left(H_{2}O\left(g\right)\right)\right]-\left[{\Delta }_f{H}^{\circ }C{H}_{3}OH\left(l\right)-\frac{3}{2}{\Delta }_f{H}^{\circ }\left(O_2\left(g\right)\right)\right]\\ \\ \text{= }\left(\text{-393}\text{.509 kJ/mol}\right)\text{+2}\left(\text{-241}\text{.818 kJ/mol}\right)\text{-}\left(\text{-238}\text{.66 kJ/mol}\right)\text{-}\frac{3}{2}\left(\text{0 kJ/mol}\right)\\ \\ =\text{ -638}\text{.49 kJ/mol}\end{array}$

$\begin{array}{c}{\Delta }_c{H}^{\circ }{}_{C_8{H}_{18}}\text{ = }\left[8{\Delta }_f{H}^{\circ }C{O}_2\left(g\right)+9{\Delta }_f{H}^{\circ }\left(H_{2}O\left(g\right)\right)\right]-\left[{\Delta }_f{H}^{\circ }{C}_8{H}_{18}-\frac{25}{2}{\Delta }_f{H}^{\circ }\left(O_2\left(g\right)\right)\right]\\ \\ \text{= }\left(\text{8}\times \text{-393}\text{.509 kJ/mol}\right)\text{+9}\left(\text{-241}\text{.818 kJ/mol}\right)\text{-}\left(\text{-250}\text{.1 kJ/mol}\right)\text{-}\frac{25}{2}\left(\text{0 kJ/mol}\right)\\ \\ =\text{ -5074}\text{.3 kJ/mol}\end{array}$

The enthalpy evolved for one gram of the two liquids are calculated as shown below,

    $\begin{array}{l}For{\text{ CH}}_{\text{3}}\text{OH }\Rightarrow \frac{638.49\text{ kJ produced}}{1{\text{ mole CH}}_{\text{3}}\text{OH}}\times \frac{1{\text{ mole CH}}_{\text{3}}\text{OH}}{32.0417{\text{ g CH}}_{\text{3}}\text{OH}}\\ \\ =\text{ 19}\text{.927 }\frac{kJ\text{ produced}}{g{\text{ CH}}_{\text{3}}\text{OH}}\left[{\text{Molar mass of CH}}_{\text{3}}\text{OH=32}\text{.0417g/mol}\right]\\ \\ For{\text{ C}}_{\text{8}}{\text{H}}_{\text{18}}\Rightarrow \frac{5074.3\text{ kJ produced}}{1{\text{ mole C}}_{\text{8}}{\text{H}}_{\text{18}}}\times \frac{1{\text{ mole C}}_{\text{8}}{\text{H}}_{\text{18}}}{114.2278{\text{ g C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ \\ =\text{ 44}\text{.422 }\frac{kJ\text{ produced}}{g{\text{ C}}_{\text{8}}{\text{H}}_{\text{18}}}\left[{\text{Molar mass of C}}_{\text{8}}{\text{H}}_{\text{18}}\text{=}114.2278\text{g/mol}\right]\end{array}$

From above calculations it is clear that enthalpy evolved from one gram of octane is more than enthalpy evolved from one gram of methanol. Methanol is oxidized more than octane hence its combustion will produce a smaller amount of energy.