Chapter 19, Problem 19.89QA

Interpretation Introduction

To find:

a) The fuel values of methanol and ethanol based on the thermochemical data in Appendix 4.

b) Does your answer support the prediction made in problem 19.80?

Answer to Problem 19.89QA

Solution:

a) Fuel value of methanol is $22.67 kJ/g,$ and fuel value of ethanol is $29.67 kJ/g$.

b) Yes; the answer supports the prediction made in problem 19.80.

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 methanol and ethanol. 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. From the fuel values and the prediction made in problem 19.80, we can check if it supports

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) $Methanol$,   $∆H_f^0 = -238.7 kJ/mol$$$$$$$

ii) $Methanol$,   $molar mass = 32.042 g/mol$$$$$$$

iii) $Ethanol$,   $∆H_f^0 = -277.7kJ/mol$$$$$$$

iv) $Ethanol$,   $molar mass = 46.069 g/mol$

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 methanol.

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

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

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

$∆H_{comb}^0 = \left[2 \times \left(-393.5\frac{kJ}{mol}\right)+4 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[2 \times \left(-238.7\frac{kJ}{mol}\right)+3 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-1930.2\right) kJ-\left(-477.7\right) kJ= -1452.8 kJ$

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

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

$Fuel Value= \frac{1452.8 kJ}{32.042\frac{g}{mol} \times \left(2 mol\right)}= 22.67 kJ /g$

The fuel value of methanol is $22.67 kJ/g$.

Write the balanced combustion reaction of ethanol.

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

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

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

$∆H_{comb}^0 = \left[2 \times \left(-393.5\frac{kJ}{mol}\right)+3 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[1 \times \left(-277.7\frac{kJ}{mol}\right)+3 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-1644.4\right) kJ-\left(-277.7\right) kJ= -1366.7 kJ$

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

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

$Fuel Value= \frac{1366.7 kJ}{46.069\frac{g}{mol} \times \left(1 mol\right)}= 29.67 kJ /g$

The fuel value of ethanol is $29.67 kJ/g$.

The fuel value of ethanol is higher than methanol. As the number of carbon atoms in the alcohol increases, the fuel value of alcohols increases.

Thus, the answer supports the prediction made in problem 19.80.

Conclusion:

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