Calculate the standard enthalpy of combustion of ` CH_4 (g) `if ` Delta_f H^(0) (CH_4)=-74.8 Kj mol^(-1)`
` Delta_f H^(0) (CO_2) =- 395 .5 KJ mol^(-1)` and ` Delta f H^(0 ) (H_2 O) =- 285.8 KJ mol^(-1)`

Answer

To calculate the standard enthalpy of combustion of methane (\( CH_4(g) \)), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The combustion of methane can be represented by the following balanced equation:

\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \]

### Step 2: Identify the Standard Enthalpy of Formation Values
We are given the following standard enthalpy of formation values:
- \( \Delta_f H^{\circ} (CH_4) = -74.8 \, \text{kJ/mol} \)
- \( \Delta_f H^{\circ} (CO_2) = -395.5 \, \text{kJ/mol} \)
- \( \Delta_f H^{\circ} (H_2O) = -285.8 \, \text{kJ/mol} \)

### Step 3: Apply the Formula for Standard Enthalpy of Combustion
The standard enthalpy of combustion (\( \Delta_r H^{\circ} \)) can be calculated using the formula:

\[ \Delta_r H^{\circ} = \sum (\Delta_f H^{\circ} \text{ of products}) - \sum (\Delta_f H^{\circ} \text{ of reactants}) \]

### Step 4: Calculate the Enthalpy of Products
For the products:
- For \( CO_2 \): \( 1 \times \Delta_f H^{\circ} (CO_2) = 1 \times (-395.5) = -395.5 \, \text{kJ} \)
- For \( H_2O \): \( 2 \times \Delta_f H^{\circ} (H_2O) = 2 \times (-285.8) = -571.6 \, \text{kJ} \)

Total for products:
\[ \text{Total products} = -395.5 + (-571.6) = -967.1 \, \text{kJ} \]

### Step 5: Calculate the Enthalpy of Reactants
For the reactants:
- For \( CH_4 \): \( 1 \times \Delta_f H^{\circ} (CH_4) = 1 \times (-74.8) = -74.8 \, \text{kJ} \)
- For \( O_2 \): The standard enthalpy of formation of elements in their standard state (like \( O_2 \)) is \( 0 \, \text{kJ/mol} \).

Total for reactants:
\[ \text{Total reactants} = -74.8 + 0 = -74.8 \, \text{kJ} \]

### Step 6: Substitute Values into the Formula
Now we substitute the total enthalpy values into the formula:

\[ \Delta_r H^{\circ} = (-967.1) - (-74.8) \]

### Step 7: Calculate the Final Value
Calculating the above expression gives:

\[ \Delta_r H^{\circ} = -967.1 + 74.8 = -892.3 \, \text{kJ} \]

### Final Answer
The standard enthalpy of combustion of \( CH_4(g) \) is:

\[ \Delta_r H^{\circ} = -892.3 \, \text{kJ/mol} \]

---