Change in enthalpy and change in internal energy are state functions. The value of `DeltaH, DeltaU` can be determined by using Kirchoff's equation.
Calculate the heat of formation of methane, given that heat of formation of water `= -286kJ mol^(-1)`, heat of combustion of methane `= -890kJ mol^(-1)` heat of combustion of carbon `= -393.5 kJ mol^(-1)`

Options

A. `90.5 kJ mol^(-1)`
B. `-240 kJ mol^(-1)`
C. `-75.5kJ mol^(-1)`
D. `95.6 kJmol^(-1)`

Answer

To calculate the heat of formation of methane (ΔH_f(CH₄)), we can use the provided heats of formation and combustion along with Hess's law. Here’s a step-by-step breakdown of the calculation:

### Step 1: Write the relevant reactions
We need to consider the following reactions:

1. **Combustion of methane (CH₄)**:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta H = -890 \text{ kJ/mol} \]

2. **Formation of water (H₂O)**:
\[ \text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H = -286 \text{ kJ/mol} \]

3. **Combustion of carbon (C)**:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H = -393.5 \text{ kJ/mol} \]

### Step 2: Write the formation reaction for methane
The formation reaction for methane is:
\[ \text{C} + 2\text{H}_2 \rightarrow \text{CH}_4 \quad \Delta H_f = ? \]

### Step 3: Apply Hess's law
To find ΔH_f(CH₄), we can manipulate the above reactions. We will reverse the combustion of methane and the formation of water to align with the formation of methane.

1. Reverse the combustion of methane:
\[ \text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_4 + 2\text{O}_2 \quad \Delta H = +890 \text{ kJ/mol} \]

2. Use the formation of water (multiply by 2 since we need 2 moles of water):
\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad \Delta H = 2 \times (-286) = -572 \text{ kJ/mol} \]

3. Use the combustion of carbon:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H = -393.5 \text{ kJ/mol} \]

### Step 4: Combine the reactions
Now we can add the modified reactions together:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (-393.5 \text{ kJ}) \]
\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad (-572 \text{ kJ}) \]
\[ \text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_4 + 2\text{O}_2 \quad (+890 \text{ kJ}) \]

### Step 5: Calculate ΔH_f(CH₄)
Now, we sum the enthalpy changes:
\[ \Delta H_f(CH₄) = -393.5 - 572 + 890 \]
\[ \Delta H_f(CH₄) = -393.5 - 572 + 890 = -75.5 \text{ kJ/mol} \]

### Final Answer
The heat of formation of methane (ΔH_f(CH₄)) is:
\[ \Delta H_f(CH₄) = -75.5 \text{ kJ/mol} \]