Chapter 19, Problem 19.90QA

Interpretation Introduction

To find:

Which has the higher fuel value, propanol or isopropanol?

Answer to Problem 19.90QA

Solution:

Propanol has the higher fuel value than isopropanol.

Explanation of Solution

1) Concept:

To calculate the fuel value of a compound, we need to use heat of combustion of fuel and molar mass. First we will write the combustion reaction of propanol and isopropanol. 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. From the fuel values, we will check which has the higher 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) $Propanol$, $∆H_f^0=-302.6 kJ/mol$

ii) $Propanol$, $molar mass=60.095 g/mol$

iii) $Isopropanol$, $∆H_f^0=-318.1 kJ/mol$

iv) $Isopropanol$, $molar mass=60.095 g/mol$

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 propanol.

$2 C{H}_{3}C{H}_{2}C{H}_2-OH \left(l\right)+9 O_2\left(g\right) \to 6 C{O}_2\left(g\right)+8 H_{2}O \left(l\right)$

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

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

$∆H_{comb}^0 = \left[6 \times \left(-393.5\frac{kJ}{mol}\right)+8 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[2 \times \left(-302.6\frac{kJ}{mol}\right)+9 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-4647.4\right) kJ-\left(-605.2\right) kJ= -4042.2 kJ$

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

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

$Fuel Value= \frac{4042.2 kJ}{60.095 \frac{g}{mol} \times \left(2 mol\right)}= 33.63 kJ /g$

The fuel value of propanol is $33.63 kJ/g$.

Write the balanced combustion reaction of isopropanol.

$2 C{H}_{3}CHOHC{H}_3 \left(l\right)+9 O_2\left(g\right) \to 6 C{O}_2\left(g\right)+8 H_{2}O \left(l\right)$

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

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

$∆H_{comb}^0 = \left[6 \times \left(-393.5\frac{kJ}{mol}\right)+8 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[2 \times \left(-318.1\frac{kJ}{mol}\right)+9 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-4647.4\right) kJ-\left(-636.2\right) kJ= -4011.2 kJ$

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

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

$Fuel Value= \frac{4011.2 kJ}{60.095 \frac{g}{mol} \times \left(2 mol\right)}= 33.37 kJ /g$

The fuel value of isopropanol is $33.37 kJ/g$.

The fuel value of propanol is higher than isopropanol.

Conclusion:

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