How much energy is released when 6 mole of octane is burnt in air ? Given `DeltaH_(f)^(@)` for `CO_(2)(g),H_(2)O(g)` and `C_(8)H_(18)(l)` respectively are `-490,-240` and `+160KJ//mol`

Options

A. `-6.2KJ`
B. `-37.4KJ`
C. `-35.5KJ`
D. `-20.0KJ`

Explanation

To determine the energy released when 6 moles of octane (C8H18) are burnt in air, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of octane.
The combustion of octane can be represented by the following equation:
\[ C_8H_{18}(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g) \]
To balance this equation, we need to determine the coefficients for each compound. The balanced equation is:
\[ C_8H_{18}(l) + \frac{25}{2} O_2(g) \rightarrow 8 CO_2(g) + 9 H_2O(g) \]

### Step 2: Write down the enthalpy of formation values.
We are given the following enthalpy of formation values:
- \( \Delta H_f^\circ (CO_2(g)) = -490 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (H_2O(g)) = -240 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (C_8H_{18}(l)) = +160 \, \text{kJ/mol} \)

### Step 3: Calculate the total enthalpy of products and reactants.
Using the enthalpy of formation values, we can calculate the total enthalpy change for the combustion reaction.

**Products:**
- For \( 8 \, CO_2 \):
\[ 8 \times (-490) = -3920 \, \text{kJ} \]
- For \( 9 \, H_2O \):
\[ 9 \times (-240) = -2160 \, \text{kJ} \]

**Total for products:**
\[ \text{Total Products} = -3920 + (-2160) = -6080 \, \text{kJ} \]

**Reactants:**
- For \( C_8H_{18} \):
\[ 1 \times (+160) = +160 \, \text{kJ} \]
- For \( O_2 \) (elemental state):
\[ \Delta H_f^\circ (O_2) = 0 \, \text{kJ} \]

**Total for reactants:**
\[ \text{Total Reactants} = +160 + 0 = +160 \, \text{kJ} \]

### Step 4: Calculate the enthalpy change for the reaction.
Using the formula:
\[ \Delta H_{reaction} = \text{Total Products} - \text{Total Reactants} \]
Substituting the values:
\[ \Delta H_{reaction} = -6080 - 160 = -6240 \, \text{kJ/mol} \]

### Step 5: Calculate the energy released for 6 moles of octane.
Since the enthalpy change calculated is for 1 mole of octane, for 6 moles:
\[ \text{Energy released} = -6240 \times 6 = -37440 \, \text{kJ} \]

### Final Answer:
The energy released when 6 moles of octane are burnt in air is **-37440 kJ**.

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