Chapter 19, Problem 19.88QA

Interpretation Introduction

Interpretation & Concept Introduction:

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

b) Which has higher fuel values?

Answer to Problem 19.88QA

Solution:

a) Fuel value of liquid diethyl ether is $36.74\frac{kJ}{g},$ and fuel value of methyl propyl ether  is $36.93 kJ/g$.

b) Methyl propyl ether has the higher fuel value than diethyl ether.

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 liquid diethyl ether and methyl propyl ether. 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 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) $\mathrm{M}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}$, $∆H_f^0=-266.0 kJ/mol$

ii) $\mathrm{M}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}$, $molar mass=74.122 g/mol$

iii) $Diethyl ether \left(l\right)$, $∆H_f^0=-279.6 kJ/mol$

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 liquid 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 liquid 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 liquid diethyl ether is $36.74 kJ/g$.

Write the balanced combustion reaction of methyl propyl ether.

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

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

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

$∆H_{comb}^0 = \left[4 \times \left(-393.5\frac{kJ}{mol}\right)+5 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[1 \times \left(-266.0\frac{kJ}{mol}\right)+6 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-3003.0\right) kJ-\left(-266.0\right) kJ= -2737.0 kJ$

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

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

$Fuel Value= \frac{2737.0 kJ}{74.122\frac{g}{mol} \times \left(1 mol\right)}= 36..93 kJ /g$

The fuel value of methyl propyl ether is $36.93 kJ/g$.

The fuel value of methyl propyl ether is higher than liquid diethyl ether.

Conclusion

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