Chapter 19, Problem 19.69QA

Interpretation Introduction

To find:

a) The fuel values of gaseous benzene and ethylene gas based on the thermochemical data in Appendix 4.

b) Does one mole of benzene have higher or lower fuel value than three moles of ethylene.

Answer to Problem 19.69QA

Solution:

a) Fuel value of benzene is $42.26 kJ/g,$ and fuel value of 3 moles of ethylene is $50.30 kJ/g$.

b) One mole of benzenehas alower fuel value than three mole ethylene.

Explanation of Solution

1) Concept:

To calculate the fuel value of the compound, we need to use heat of combustion of fuel and molar mass. First we will write the combustion reaction of gaseous benzene and ethylene gas. 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.

Fuel value is the quantity of energy released during the complete combustion of 1 g of a substance.

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) $Benzene$,   $∆H_f^0 = 82.9 kJ/mol$$$$$$$

ii) $Benzene$,   $molar mass = 78.114 g/mol$$$$$$$

iii) $Ethylene$,   $∆H_f^0 = 52.4 kJ/mol$$$$$$$

iv) $Ethylene$,   $molar mass =28.054 g/mol$

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 benzene.

$C_6{H}_6\left(g\right)+\frac{15}{2}{O}_2\left(g\right) \to 6 C{O}_2\left(g\right)+3 H_{2}O \left(l\right)$

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

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

$∆H_{comb}^0 = \left[6 \times \left(-393.5\frac{kJ}{mol}\right)+3 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[1 \times \left(82.9 \frac{kJ}{mol}\right)+\frac{15}{2} \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-3218.4\right) kJ-\left(82.9\right) kJ= -3301.3 kJ$

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

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

$Fuel Value= \frac{3301.3 kJ}{78.114 \frac{g}{mol} \times \left(1 mol\right)}= 42.26 kJ /g$

The fuel value of benzeneis $42.26 kJ/g$.

Write the balanced combustion reaction of ethylene.

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

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

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

$∆H_{comb}^0 = \left[6 \times \left(-393.5\frac{kJ}{mol}\right)+6 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[3 \times \left(52.4\frac{kJ}{mol}\right)+9 \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-4075.8\right) kJ-\left(157.2\right) kJ= -4233.0 kJ$

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

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

$Fuel Value= \frac{4233.0 kJ}{28.054 \frac{g}{mol} \times \left(3 mol\right)}= 50.30 kJ /g$

The fuel value of ethyleneis $50.30 kJ/g$.

The fuel value of one mole of benzene is lower than 3 mole of ethylene.

Conclusion:

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