What will be the amount of heat evolved by burning 10 L of methane under standard conditions?
`-76.2, -398.8 and -241.6" kJ mol"^(-1)` resepctively

Options

A. 805.8kJ
B. 398.8 kJ
C. 359.7 kJ
D. 640.4 kJ

Explanation

To solve the problem of calculating the amount of heat evolved by burning 10 liters of methane under standard conditions, we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The combustion of methane (CH₄) can be represented by the following balanced equation:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Identify the standard enthalpy of formation values.
From the problem, we have the following standard enthalpy of formation values:
- ΔHf (CH₄) = -76.2 kJ/mol
- ΔHf (CO₂) = -398.8 kJ/mol
- ΔHf (H₂O) = -241.6 kJ/mol
- ΔHf (O₂) = 0 kJ/mol (since it is in its standard state)

### Step 3: Calculate the overall enthalpy change (ΔH) for the reaction.
Using the formula for the enthalpy change of the reaction:
\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

Substituting the values:
\[ \Delta H = \left[ \Delta H_f (\text{CO}_2) + 2 \times \Delta H_f (\text{H}_2\text{O}) \right] - \left[ \Delta H_f (\text{CH}_4) + 2 \times \Delta H_f (\text{O}_2) \right] \]
\[ \Delta H = \left[ -398.8 + 2 \times (-241.6) \right] - \left[ -76.2 + 2 \times 0 \right] \]
\[ \Delta H = \left[ -398.8 - 483.2 \right] - \left[ -76.2 \right] \]
\[ \Delta H = -882.0 + 76.2 \]
\[ \Delta H = -805.8 \text{ kJ/mol} \]

### Step 4: Calculate the amount of heat evolved for 10 liters of methane.
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, we can find the number of moles of methane in 10 liters:
\[ \text{Moles of CH}_4 = \frac{10 \text{ L}}{22.4 \text{ L/mol}} \approx 0.4464 \text{ mol} \]

### Step 5: Calculate the total heat evolved.
Using the enthalpy change calculated earlier:
\[ \text{Heat evolved} = \Delta H \times \text{moles of CH}_4 \]
\[ \text{Heat evolved} = -805.8 \text{ kJ/mol} \times 0.4464 \text{ mol} \]
\[ \text{Heat evolved} \approx -359.73 \text{ kJ} \]

### Final Answer:
The amount of heat evolved by burning 10 liters of methane under standard conditions is approximately **359.73 kJ**.

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