Calculate the standard enthalpy of formation of propane `(C_(3)H_(8))` if its enthalpy of combustion is `-2220.2 kJ "mol"^(-1)`.The enthalpies of formation of `CO_(2)(g)` and `H_(2)O(l)` are -393.5 and -285.8 kJ `"mol"^(-1)` respectively.

Explanation

To calculate the standard enthalpy of formation of propane (C₃H₈) using the given data, we can follow these steps:

### Step 1: Write the combustion reaction of propane
The combustion of propane can be represented by the following balanced equation:
\[ C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l) \]
The enthalpy change for this reaction (ΔH_combustion) is given as:
\[ \Delta H_{combustion} = -2220.2 \, \text{kJ/mol} \]

### Step 2: Write the formation reactions for CO₂ and H₂O
The formation reactions for carbon dioxide (CO₂) and water (H₂O) are:
1. For CO₂:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_f = -393.5 \, \text{kJ/mol} \]
2. For H₂O:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_f = -285.8 \, \text{kJ/mol} \]

### Step 3: Set up the enthalpy of formation equation for propane
The standard enthalpy of formation (ΔH_f) of propane can be calculated using Hess's law. According to Hess's law:
\[ \Delta H_{reaction} = \sum \Delta H_{f(products)} - \sum \Delta H_{f(reactants)} \]

For the combustion of propane, we can express it as:
\[ \Delta H_{combustion} = [3 \Delta H_f(CO_2) + 4 \Delta H_f(H_2O)] - \Delta H_f(C_3H_8) \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ -2220.2 = [3(-393.5) + 4(-285.8)] - \Delta H_f(C_3H_8) \]

### Step 5: Calculate the total enthalpy of products
Calculating the total enthalpy of the products:
1. For CO₂:
\[ 3 \times (-393.5) = -1180.5 \, \text{kJ} \]
2. For H₂O:
\[ 4 \times (-285.8) = -1143.2 \, \text{kJ} \]
Adding these together:
\[ -1180.5 + (-1143.2) = -2323.7 \, \text{kJ} \]

### Step 6: Rearrange the equation to find ΔH_f(C₃H₈)
Now we can rearrange the equation to solve for ΔH_f(C₃H₈):
\[ -2220.2 = -2323.7 - \Delta H_f(C_3H_8) \]
\[ \Delta H_f(C_3H_8) = -2323.7 + 2220.2 \]
\[ \Delta H_f(C_3H_8) = -103.5 \, \text{kJ/mol} \]

### Final Answer
The standard enthalpy of formation of propane (C₃H₈) is:
\[ \Delta H_f(C_3H_8) = -103.5 \, \text{kJ/mol} \]

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