The combustion enthalpies of carbon, hydrogen and methane are `-395.5 "kJ" mol^-1, -285.8 "kJ" mol^-1 "and" -890.4 kJ mol^-1 "respectively at " 25^@C`. The value of standard formation enthalpy of methane at that temperature is ____________ .

Options

A. `890.4 kJ mol^-1`
B. `-298.8 kJ mol^-1`
C. `-76.1 kJ mol^-1`
D. `-107.7 kJ mol^-1`

Answer

To find the standard formation enthalpy of methane (CH₄) at 25°C, we can use the given combustion enthalpies of carbon, hydrogen, and methane. The combustion reactions and their respective enthalpies are as follows:

1. **Combustion of Carbon**:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -395.5 \, \text{kJ/mol} \]

2. **Combustion of Hydrogen**:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H_2 = -285.8 \, \text{kJ/mol} \]

3. **Combustion of Methane**:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_3 = -890.4 \, \text{kJ/mol} \]

Next, we need to write the formation reaction for methane:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta H_f = ? \]

### Step 1: Reverse the Combustion of Methane
To find the formation enthalpy, we can reverse the combustion reaction of methane:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \]
When we reverse a reaction, the sign of the enthalpy change also reverses:
\[ \Delta H_5 = +890.4 \, \text{kJ/mol} \]

### Step 2: Adjust the Combustion of Hydrogen
Next, we need to adjust the combustion of hydrogen to match the stoichiometry of our target equation. Since we need 2 moles of hydrogen, we can multiply the combustion reaction of hydrogen by 2:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H_6 = 2 \times (-285.8) = -571.6 \, \text{kJ/mol} \]

### Step 3: Combine the Reactions
Now we can combine the reactions to find the formation enthalpy of methane:
1. From the reversed combustion of methane:
\[ CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H_5 = +890.4 \, \text{kJ/mol} \]
2. From the adjusted combustion of hydrogen:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H_6 = -571.6 \, \text{kJ/mol} \]
3. The combustion of carbon remains the same:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -395.5 \, \text{kJ/mol} \]

### Step 4: Write the Overall Enthalpy Change
The overall enthalpy change for the formation of methane can be calculated as:
\[ \Delta H_f = \Delta H_1 + \Delta H_6 + \Delta H_5 \]
Substituting the values:
\[ \Delta H_f = (-395.5) + (-571.6) + (890.4) \]
Calculating this gives:
\[ \Delta H_f = -395.5 - 571.6 + 890.4 = -76.7 \, \text{kJ/mol} \]

### Final Answer
The standard formation enthalpy of methane at 25°C is:
\[ \Delta H_f = -76.7 \, \text{kJ/mol} \]