Enthalpy of combustion of methane and ethane are `-210` kcal/mol and `-368` kcal/mol respectively. The enthalpy of combustion of decane is

Options

A. `-1582` kcal
B. `-1632` kcal
C. `-1700` kcal
D. `-1480` kcal

Explanation

To find the enthalpy of combustion of decane (C10H22), we can use the enthalpy of combustion values for methane (CH4) and ethane (C2H6) provided in the question. The steps are as follows:

### Step 1: Write the combustion equations
The combustion of decane can be represented as:
\[ \text{C}_{10}\text{H}_{22} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Balance the combustion equation
To balance the combustion equation for decane:
1. **Carbon atoms**: 10 CO2 (since there are 10 carbon atoms in decane)
2. **Hydrogen atoms**: 11 H2O (since there are 22 hydrogen atoms in decane)
3. **Oxygen atoms**: Total oxygen needed = 10 CO2 + 11 H2O = 20 + 11 = 31 oxygen atoms. Therefore, we need \( \frac{31}{2} \) O2.

The balanced equation is:
\[ \text{C}_{10}\text{H}_{22} + \frac{31}{2} \text{O}_2 \rightarrow 10 \text{CO}_2 + 11 \text{H}_2\text{O} \]

### Step 3: Determine the enthalpy change
The enthalpy of combustion of an alkane can be estimated using the enthalpy of combustion of its smaller homologues (methane and ethane).

1. **Enthalpy of combustion of methane (CH4)**:
\[ \Delta H_{combustion} = -210 \text{ kcal/mol} \]

2. **Enthalpy of combustion of ethane (C2H6)**:
\[ \Delta H_{combustion} = -368 \text{ kcal/mol} \]

### Step 4: Calculate the enthalpy of combustion of CH2 unit
The difference in enthalpy between ethane and methane represents the enthalpy change for adding a CH2 unit:
\[ \Delta H_{C2H6} - \Delta H_{CH4} = -368 - (-210) = -158 \text{ kcal/mol} \]

This means that each additional CH2 unit contributes -158 kcal/mol to the enthalpy of combustion.

### Step 5: Calculate the enthalpy of combustion for decane
Decane has 9 CH2 units in addition to one CH4 unit:
\[ \Delta H_{C10H22} = \Delta H_{CH4} + 9 \times \Delta H_{CH2} \]
\[ \Delta H_{C10H22} = -210 + 9 \times (-158) \]
\[ \Delta H_{C10H22} = -210 - 1422 \]
\[ \Delta H_{C10H22} = -1632 \text{ kcal/mol} \]

### Final Answer
The enthalpy of combustion of decane (C10H22) is:
\[ \Delta H_{combustion} = -1632 \text{ kcal/mol} \]

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