For his 18th birthday in February Peter plants to turn a hut in the garden of his parents into a swimming pool with an artifical beach. In order to estimate the consts for heating the water and the house , peter obtains the data for the natural gas combustion and its price.
Write down the chemical equations for the complete conbustion of the main components of natural gas, methane and ethane , given in table-1. Assume tht nitrogen is inert under the chosen conditions. Calculate the reaction enthalpy , the reaction entropy , and the Gibbs energy under standard conditons `(1.013 xx 10^(5) Pa , 25.0^(@)C)` for the combustion of methane and ethane according to the equations above assuming that all products are gaseous .
The thermodynamic properties and the composition of natural gas can be found in table 1.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the Balanced Chemical Equations

1. **For Methane (CH₄)**:
The balanced equation for the complete combustion of methane is:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]

2. **For Ethane (C₂H₆)**:
The balanced equation for the complete combustion of ethane is:
\[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]

### Step 2: Calculate the Reaction Enthalpy (ΔH°)

1. **For Methane**:
The standard enthalpy change for the reaction can be calculated using:
\[ \Delta H^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \]
Using the provided values:
- ΔH_f°(CO₂) = -393.5 kJ/mol
- ΔH_f°(H₂O) = -241.8 kJ/mol
- ΔH_f°(CH₄) = -74.6 kJ/mol

\[ \Delta H^\circ = [(-393.5) + 2 \times (-241.8)] - [(-74.6) + 2 \times 0] \]
\[ \Delta H^\circ = (-393.5 - 483.6 + 74.6) = -802.5 \text{ kJ/mol} \]

2. **For Ethane**:
Similarly, for ethane:
- ΔH_f°(C₂H₆) = -84.0 kJ/mol

\[ \Delta H^\circ = [4 \times (-393.5) + 6 \times (-241.8)] - [2 \times (-84.0) + 7 \times 0] \]
\[ \Delta H^\circ = (-1574 - 1450.8 + 168) = -2856.8 \text{ kJ/mol} \]

### Step 3: Calculate the Reaction Entropy (ΔS°)

1. **For Methane**:
Using the standard entropy values:
- S°(CO₂) = 213.8 J/mol·K
- S°(H₂O) = 188.8 J/mol·K
- S°(CH₄) = 186.3 J/mol·K
- S°(O₂) = 205.2 J/mol·K

\[ \Delta S^\circ = [S^\circ(\text{CO}_2) + 2 \times S^\circ(\text{H}_2\text{O})] - [S^\circ(\text{CH}_4) + 2 \times S^\circ(\text{O}_2)] \]
\[ \Delta S^\circ = [213.8 + 2 \times 188.8] - [186.3 + 2 \times 205.2] \]
\[ \Delta S^\circ = (213.8 + 377.6) - (186.3 + 410.4) = -5.3 \text{ J/mol·K} \]

2. **For Ethane**:
\[ \Delta S^\circ = [4 \times S^\circ(\text{CO}_2) + 6 \times S^\circ(\text{H}_2\text{O})] - [2 \times S^\circ(\text{C}_2\text{H}_6) + 7 \times S^\circ(\text{O}_2)] \]
\[ \Delta S^\circ = [4 \times 213.8 + 6 \times 188.8] - [2 \times 229.2 + 7 \times 205.2] \]
\[ \Delta S^\circ = (855.2 + 1132.8) - (458.4 + 1436.4) = 33.2 \text{ J/mol·K} \]

### Step 4: Calculate the Gibbs Free Energy (ΔG°)

1. **For Methane**:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
\[ \Delta G^\circ = -802.5 \text{ kJ/mol} - (298.15 \text{ K} \times -0.0053 \text{ kJ/mol·K}) \]
\[ \Delta G^\circ = -802.5 + 1.58 = -800.9 \text{ kJ/mol} \]

2. **For Ethane**:
\[ \Delta G^\circ = -2856.8 \text{ kJ/mol} - (298.15 \text{ K} \times 0.0332 \text{ kJ/mol·K}) \]
\[ \Delta G^\circ = -2856.8 - 9.89 = -2866.7 \text{ kJ/mol} \]

### Summary of Results
- For Methane:
- ΔH° = -802.5 kJ/mol
- ΔS° = -5.3 J/mol·K
- ΔG° = -800.9 kJ/mol

- For Ethane:
- ΔH° = -2856.8 kJ/mol
- ΔS° = 33.2 J/mol·K
- ΔG° = -2866.7 kJ/mol