Chapter 19, Problem 19.70QA

Interpretation Introduction

To find:

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

b) Whether one mole of benzene has higher or lower fuel value than three moles of acetylene

Answer to Problem 19.70QA

Solution:

a) Fuel value of benzene is $42.26\frac{kJ}{g},$ and fuel value of acetylene is $49.91 kJ/g$

b) One mole of benzene has a lower fuel value than three mole acetylene.

Explanation of Solution

1) Concept:

Fuel value is the quantity of energy released during the complete combustion of one gram of a substance.  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 gaseous benzene and acetylene gas. Then we will calculate the $∆H_{comb}^0$ from the thermodynamic value($∆H_f^0$) of compounds present in the reaction. By using 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) $Benzene$,   $∆H_f^0 = 82.9 kJ/mol$$$$$$$

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

iii) $Acethylene$,   $∆H_f^0 =226.7 kJ/mol$$$$$$$

iv) $Acethylene$,   $molar mass =26.038 g/mol$

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 $∆H_{comb}^0$ of benzene 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 twice 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 benzene is $42.26 kJ/g$.

Write the balanced combustion reaction of acetylene.

$3 HC\equiv CH \left(g\right)+\frac{15}{2}{O}_2\left(g\right) \to 6 C{O}_2\left(g\right)+3 H_{2}O \left(l\right)$

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

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

$∆H_{comb}^0 = \left[6 \times \left(-393.5\frac{kJ}{mol}\right)+3 \times \left(-285.8\frac{kJ}{mol}\right)\right]-\left[3 \times \left(226.7\frac{kJ}{mol}\right)+\frac{15}{2} \times \left(0.0\frac{kJ}{mol}\right)\right]$

$∆H_{comb}^0 =\left(-3218.4\right) kJ-\left(680.1\right) kJ= -3898.5 kJ$

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

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

$Fuel Value= \frac{3898.5 kJ}{26.038 \frac{g}{mol} \times \left(3 mol\right)}= 49.91 kJ /g$

The fuel value of acetyleneis $49.91 kJ/g$.

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

Conclusion:

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