Chapter 5, Problem 85GQ

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

                       ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

Answer to Problem 85GQ

The enthalpy change of combustion per gram of isooctane, is $\text{-47}\text{.82}\text{J/gK}$ and the enthalpy change of combustion per gram of methanol is $\text{-22}\text{.68J/gK}$

Explanation of Solution

For isooctane the balanced reaction is

  ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{+ 25/2 O}}_{\text{2}}\to {\text{8 CO}}_{\text{2}}{\text{+ 9 H}}_{\text{2}}\text{O}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\left[\left(\text{8mol×-393}\text{.5kJ/mol}\right)\text{+}\left(\text{9mol×-285}\text{.83kJ/mol}\right)\right]$ -$\left[\left(\text{1mol×-259}\text{.2kJ/mol}\text{+0}\right)\right]$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-5461}\text{.3kJ}$

The change in enthalpy is $\text{-5461}\text{.3kJ}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-5461}\text{.3kJ}$$\text{/mol}\text{×}\frac{\text{1mol}}{\text{114}\text{.2g}}$=$\text{-47}\text{.82}\text{J/gK}$

So, the enthalpy of isooctane is $\text{-47}\text{.82}\text{J/gK}$

For methanol,

  ${\text{2 CH}}_{\text{3}}{\text{OH + 3 O}}_{\text{2}}\to {\text{2CO}}_{\text{2}}{\text{+ 4 H}}_{\text{2}}\text{O}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\left[\left(\text{2mol×-393}\text{.5kJ/mol}\right)\text{+}\left(\text{4mol×-285}\text{.83kJ/mol}\right)\right]$-$\left[\left(\text{2mol×-238}\text{.4kJ/mol}\text{+0}\right)\right]$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-1453}\text{.5kJ}$

The change in enthalpy is $\text{-5461}\text{.3kJ}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\frac{\text{-1453}\text{.5kJ}}{\text{2mol}}\text{×}\frac{\text{1mol}}{\text{32}\text{.04g}}$=$\text{-22}\text{.68J/gK}$

So, the enthalpy of methanol is $\text{-22}\text{.68J/gK}$

Conclusion

Enthalpy change of combustion per gram of methanol and isooctane was determined.