Chapter 19, Problem 19.87QA

Interpretation Introduction

To find:

a) The fuel values of diethyl ether and butanol based on the thermochemical data in Appendix 4.

b) Which has the higher fuel values?

Answer to Problem 19.87QA

Solution:

a) Fuel value of diethyl ether is $36.74\frac{kJ}{g},$ and fuel value of butanol is $36.10 kJ/g.$

b) Diethyl ether has a higher fuel value than butanol.

Explanation of Solution

1) Concept:

To calculate the fuel value of the compound, we need to use the heat of combustion of fuel and molar mass. First we will write the combustion reaction of diethyl ether and butanol. Then we will calculate the $∆H_{comb}^0$ from the thermodynamic value, $∆H_f^0$, of compounds present in the reaction. By using $∆H_{comb}^0$ and the molar mass of the fuel, we will calculate the fuel value.

2) Formula:

i) $∆H_{comb}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

ii) $Fuel Value= \frac{∆H_{comb}^0}{molar mass \times number of moles of fuel}$

3) Given:

i) $Butanol$, $∆H_f^0=-327.3 kJ/mol$

ii) $Butanol$, $molar mass=74.122 g/mol$

iii) $Diethyl ether$, $∆H_f^0=-279.6 kJ/mol$

iv) $Diethyl ether$, $molar mass=74.122 g/mol$

v) $O_2$, $∆H_f^0 = 0.0 kJ/mol$

vi) ${CO}_2 \left(g\right)$, $∆H_f^0=-393.5 kJ/mol$

vii) $H_{2}O \left(l\right)$, $∆H_f^0=-285.8 kJ/mol$

4) Calculations:

Write the balanced combustion reaction of diethyl ether.

$C{H}_{3}C{H}_2-O-C{H}_{2}C{H}_3 \left(l\right)+6 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+5 H_{2}O \left(l\right)$

Calculate the $∆H_{comb}^0$ of diethyl ether by using the thermodynamic value given from Appendix 4.

$∆H_{comb}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

$∆H_{comb}^0 = \left[4 \times \left(-393.5\frac{kJ}{mol}\right)+5 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[1 \times \left(-279.6\frac{kJ}{mol}\right)+6 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-3003.0\right) kJ-\left(-279.6\right) kJ= -2723.4 kJ$

If we divide the absolute value of $∆H_{comb}^0$ by the molar mass of diethyl ether, we will get the fuel value.

$Fuel Value= \frac{∆H_{comb}^0}{molar mass \times number of moles of fuel}$

$Fuel Value= \frac{2723.4 kJ}{74.122\frac{g}{mol} \times \left(1 mol\right)}= 36.74 kJ /g$

The fuel value of diethyl ether is $36.74 kJ/g$.

Write the balanced combustion reaction of butanol.

$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_2-OH \left(l\right)+6 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+5 H_{2}O \left(l\right)$

Calculate the $∆H_{comb}^0$ of butanol by using the thermodynamic value given from Appendix 4.

$∆H_{comb}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

$∆H_{comb}^0 = \left[4 \times \left(-393.5\frac{kJ}{mol}\right)+5 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[1 \times \left(-327.3\frac{kJ}{mol}\right)+6 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-3003.0\right) kJ-\left(-327.3\right) kJ= -2675.7 kJ$

If we divide the absolute value of $∆H_{comb}^0$ by the molar mass of butanol, we will get the fuel value.

$Fuel Value= \frac{∆H_{comb}^0}{molar mass \times number of moles of fuel}$

$Fuel Value= \frac{2675.7 kJ}{74.122\frac{g}{mol} \times \left(1 mol\right)}= 36.10 kJ /g$

The fuel value of butanol is $36.10 kJ/g$.

The fuel value of diethyl ether is higher than butanol.

Conclusion:

The fuel value is calculated from the thermodynamic data, combustion reaction, and molar mass.