Chapter 13, Problem 13.69QP

Interpretation Introduction

Interpretation: The fuel values of gaseous benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ and ethylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ are to be calculated. The compound, benzene with $1$ mole whether has a higher fuel value or a lower fuel value as compared to $3$ moles of ethylene gas is to be predicted.

Concept introduction: The amount of energy that is produced by the complete combustion of a specific mass of any fuel is termed as fuel value.

The heat of combustion of any fuel is the total amount of heat that is released at the time of combustion of a particular amount of that fuel.

To determine: The fuel values of gaseous benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ and ethylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ and if $1$ mole of benzene has a higher fuel value or a lower fuel value as compared to $3$ moles of ethylene gas.

Answer to Problem 13.69QP

The fuel value of gaseous benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is $\text{0}\text{.531}\text{kJ/g}$ and the fuel value of ethylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ is $1.\text{865}\text{kJ/g}$.

The fuel value of $\text{1}\text{mole}$ of benzene is higher than $3\text{moles}$ of ethylene.

Explanation of Solution

The balanced chemical equation for the combustion of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ in air is,

$2{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(g\right)+15{\text{O}}_{\text{2}}\left(g\right)\to 12{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The standard heat of combustion for gaseous benzene is $82.9\text{kJ/mol}$.

The molar mass of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is $78.11\text{g/mol}$.

There are $2$ moles of benzene consumed during the combustion. Thus, the total mass of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is,

$2\times 78.11\text{g/mol}=156.22\text{g/mol}$

The fuel value is calculated by the formula,

$\text{Fuel}\text{value}=\frac{\text{Heat}\text{of}\text{combustion}}{\text{Total}\text{mass}\text{of}\text{the}\text{substance}}$ (1)

Substitute the value of heat of combustion and total mass of benzene in equation (1).

$\begin{array}{c}\text{Fuel}\text{value}\text{of}\text{benzene}=\frac{82.9\text{kJ/mol}}{156.22\text{g/mol}}\\ =\text{0}\text{.531}\text{kJ/g}\end{array}$

Thus, the fuel value of benzene is $\text{0}\text{.531}\text{kJ/g}$.

The balanced chemical equation for the combustion of ethylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ in air is,

${\text{C}}_{\text{2}}{\text{H}}_4\left(g\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The standard heat of combustion for ethylene gas is $52.3\text{kJ/mol}$.

The molar mass of ethylene $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ is $28.05\text{g/mol}$.

Substitute the value of heat of combustion and total mass of ethylene in equation (1).

$\begin{array}{c}\text{Fuel}\text{value}\text{of}\text{ethylene}=\frac{52.3\text{kJ/mol}}{28.05\text{g/mol}}\\ =1.\text{865}\text{kJ/g}\end{array}$

Thus, the fuel value of ethylene gas is $1.\text{865}\text{kJ/g}$.

The fuel value of $\text{1}\text{mole}$ of benzene is higher than $3\text{moles}$ of ethylene.

The molar mass of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is $78.11\text{g/mol}$.

Substitute the value of heat of combustion and mass of $\text{1}\text{mole}$ benzene in equation (1).

$\begin{array}{c}\text{Fuel}\text{value}\text{of}\text{benzene}=\frac{82.9\text{kJ/mol}}{78.11\text{g/mol}}\\ =1.\text{061}\text{kJ/g}\end{array}$

Thus, the fuel value of $\text{1}\text{mole}$ benzene is $1.\text{061}\text{kJ/g}$.

The molar mass of ethylene $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ is $28.05\text{g/mol}$.

If there are $\text{3}\text{moles}$ of ethylene $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ consumed during the combustion. Thus, the total mass of ethylene $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ is,

$3\times 28.05\text{g/mol}=84.15\text{g/mol}$

Substitute the value of heat of combustion and total mass of ethylene in equation (1).

$\begin{array}{c}\text{Fuel}\text{value}\text{of}\text{ethylene}=\frac{52.3\text{kJ/mol}}{84.15\text{g/mol}}\\ =0.\text{622}\text{kJ/g}\end{array}$

Therefore, it is proven from the above calculation that the fuel value of $\text{1}\text{mole}$ of benzene is higher than $3\text{moles}$ of ethylene.

Conclusion

The fuel value of gaseous benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ is $\text{0}\text{.531}\text{kJ/g}$ and the fuel value of ethylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ is $1.\text{865}\text{kJ/g}$.

The fuel value of $\text{1}\text{mole}$ of benzene is higher than $3\text{moles}$ of ethylene.