Chapter 10, Problem 10.62PAE

Recall that incomplete combustion of fossil fuels occurs when too little oxygen is present and results in the production of carbon monoxide rather than carbon dioxide. Water is the other product in each case.

(a) Write balanced chemical equations for the complete and incomplete combustion of propane.

(b) Using these equations, predict which will have the larger change in entropy.

(c) Use tabulated thermodynamic data to calculate $\Delta \text{G}°$ for each reaction.

(d) Based on these results, predict the sign and value of $\Delta \text{G}°$ for the combustion of carbon monoxide to form carbon dioxide.

Interpretation Introduction

Interpretation:

The incomplete combustion of fossil fuels occurs when too little oxygen is present and result in the production of carbon monoxide rather than carbon dioxide. Water is the other product in each case.

1. Balanced equation for complete and incomplete combustion of propane needs to be written.
2. The reaction with larger change in entropy needs to be identified.

The value of $\Delta {\text{G}}^0$ for each reaction needs to be calculated.

The sign and value of $\Delta {\text{G}}^0$

3. for the combustion of carbon monoxide to form carbon dioxide needs to be determined.

Concept introduction:

The combustion of hydrocarbon results in the production of carbon dioxide gas and water. Below are the two types of combustion: -

1. Complete Combustion: - In this process the burning of the hydrocarbon completely occurs and resulting in the formation of CO₂ and H₂O. In complete combustion blue or orange fame is produced

2. Incomplete Combustion: - In this process the burning of the hydrocarbon is not completebecause of insufficient oxygen. This results in the formation of CO and H₂O. In incomplete combustion yellow fame is produced.

Answer to Problem 10.62PAE

Solution:

1. ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{3CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O }\left(\text{g}\right)$ (Complete combustion

   ${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 7O}}_{\text{2}}\left(\text{g}\right)\to \text{6CO}\left(\text{g}\right){\text{ + 8H}}_{\text{2}}\text{O}\left(\text{g}\right)$ ( Incomplete combustion

   • Complete combustion has larger entropy.

   • $\Delta {\text{G}}^0$ for complete combustion is -2108.11 kJ

   • $\Delta {\text{G}}^0$ for incomplete combustion is -2649.53 kJ

   • $\Delta {\text{G}}^0$= -772.57 kJ

(a)

Explanation of Solution

The balanced chemical reaction for complete combustion of propane is as follows:

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{3CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O }\left(\text{g}\right)$

The incomplete combustion of propane takes place in the presence of low oxygen and results in the formation of carbon monoxide instead of carbon dioxide.

The balanced chemical reaction for incomplete combustion of propane is as follows:

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 3/2O}}_{\text{2}}\left(\text{g}\right)\to 3\text{CO}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Or,

${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 7O}}_{\text{2}}\left(\text{g}\right)\to \text{6CO}\left(\text{g}\right){\text{ + 8H}}_{\text{2}}\text{O}\left(\text{g}\right)$

(b)

In complete combustion, the entropy will be higher becausemore number of gaseous molecules are formed during complete combustion resulting increase in degree of randomness and increase in entropy.

(c)

The change in Gibbs free energy of a reaction is calculated as follows:

${\text{ΔG}}^0$ = sum of ${\text{ΔG}}^0$ (product) - sum of ${\text{ΔG}}^0$ (reactant)

Balanced equation for complete combustion of propane is as follows:

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{3CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O}$

The change in Gibbs free energy of reaction is calculated as follows:

$\Delta {\text{G}}_{\text{rxn}}^0=3\times \Delta {\text{G}}^0\left({\text{CO}}_2\left(\text{g}\right)\right)+4\times \Delta {\text{G}}^0\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)-\left[\begin{array}{l}1\times \Delta {\text{G}}^0\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)\right)\\ +5\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]$

Thus,

$\begin{array}{c}\Delta {\text{G}}_{\text{rxn}}^0\text{=}\left[\begin{array}{l}\text{3 mol}\left(-\text{394}\text{.36 kJ/mol}\right)\\ \text{+ 4 mol}\left(-\text{237}\text{.13kJ/mol}\right)\end{array}\right]-\left[\begin{array}{l}\text{1 mol}\left(-\text{23}\text{.49 kJ/mol}\right)\\ \text{+5 mol}\left(\text{0 kJ/mol}\right)\end{array}\right]\\ =-2108.11\text{ kJ}\end{array}$

The balance equation for incomplete combustion of propane

${\text{2 C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 7O}}_{\text{2}}\left(\text{g}\right)\to \text{6CO}\left(\text{g}\right){\text{ + 8H}}_{\text{2}}\text{O}$

The change in Gibbs free energy of reaction is calculated as follows:

$\Delta {\text{G}}_{\text{rxn}}^0=6\times \Delta {\text{G}}^0\left(\text{CO}\left(\text{g}\right)\right)+8\times \Delta {\text{G}}^0\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)-\left[\begin{array}{l}2\times \Delta {\text{G}}^0\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)\right)\\ +7\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]$

Thus,

$\begin{array}{c}\Delta {\text{G}}_{\text{rxn}}^0\text{=}\left[\begin{array}{l}\text{6 mol}\left(-\text{137}\text{.16 kJ/mol}\right)\\ \text{+ 8 mol}\left(-\text{237}\text{.13kJ/mol}\right)\end{array}\right]-\left[\begin{array}{l}\text{3 mol}\left(-\text{23}\text{.49 kJ/mol}\right)\\ \text{+7 mol}\left(\text{0 kJ/mol}\right)\end{array}\right]\\ =-2649.53\text{ kJ}\end{array}$

As per the Hess’s law method, the two reactions i.e. complete combustion and incomplete combustion can be summed together for the ΔG° of this process.

So, here first step is to be taken to divide the carbon monoxide i.e. incomplete combustion by 2. And after that when two below reactions are added we will get the final reaction: -

Reaction A: ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{3CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O }\left(\text{g}\right)$ (Complete combustion

Reaction B: ${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 7O}}_{\text{2}}\left(\text{g}\right)\to \text{6CO}\left(\text{g}\right){\text{ + 8H}}_{\text{2}}\text{O}\left(\text{g}\right)$ ( Incomplete combustion

Now dividing the reaction (B) by 2 and adding them we will get the followingreaction:

$\text{3CO +}\frac{3}{2}\text{ O2 }\to \text{ 3CO2}$

Now calculating $\Delta {\text{G}}^0$ by using Hess’s Law: -

$\Delta {\text{G}}_{\text{rxn}}^0=3\times \Delta {\text{G}}^0\left({\text{CO}}_2\left(\text{g}\right)\right)-\left[\begin{array}{l}3\times \Delta {\text{G}}^0\left(\text{CO}\left(\text{g}\right)\right)\\ +\frac{3}{2}\times \Delta {\text{G}}^0\left({\text{O}}_2\left(\text{g}\right)\right)\end{array}\right]$

Thus,

$\begin{array}{c}\Delta {\text{G}}_{\text{rxn}}^0\text{=}\left[\text{3 mol}\left(-394.359\text{ kJ/mol}\right)\right]-\left[\begin{array}{l}\text{3 mol}\left(-\text{137}\text{.168 kJ/mol}\right)\\ \text{+}\frac{3}{2}\text{ mol}\times \text{0}\end{array}\right]\\ =-772.57\text{ kJ}\end{array}$

Conclusion

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer. Therefore, based on this reaction the values are:

1. the balance equation for complete and incomplete combustion of propane

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{3CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O }\left(\text{g}\right)$

• The balance equation for incomplete combustion of propane

  ${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right){\text{ + 7O}}_{\text{2}}\left(\text{g}\right)\to \text{6CO}\left(\text{g}\right){\text{ + 8H}}_{\text{2}}\text{O}\left(\text{g}\right)$

  • In complete combustion, it has higher entropy. Because during complete combustion all the molecules are disordered, and hence entropy increase.

  • $\Delta {\text{G}}^{\text{0}}$ for complete combustion is -2108.11 kJ

    $\Delta {\text{G}}^{\text{0}}$ for incomplete combustion is -2649.53 kJ

  • $\Delta {\text{G}}^{\text{0}}$ for overall reaction is -772.57 kJ