What is the value of `DeltaH-DeltaU` for the combustion of Heptane `(l)` ?

Options

A. `-4RT`
B. `-3RT`
C. `4RT`
D. `3RT`

Answer

To find the value of \( \Delta H - \Delta U \) for the combustion of heptane (\( C_7H_{16} \)), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of heptane.

The combustion of heptane can be represented by the following equation:

\[ C_7H_{16}(l) + O_2(g) \rightarrow CO_2(g) + H_2O(l) \]

### Step 2: Balance the chemical equation.

1. **Carbon atoms**: There are 7 carbon atoms in heptane, so we need 7 \( CO_2 \):
\[ C_7H_{16}(l) + O_2(g) \rightarrow 7 CO_2(g) + H_2O(l) \]

2. **Hydrogen atoms**: There are 16 hydrogen atoms in heptane, which means we need 8 \( H_2O \):
\[ C_7H_{16}(l) + O_2(g) \rightarrow 7 CO_2(g) + 8 H_2O(l) \]

3. **Oxygen atoms**: Now we count the oxygen atoms. The products have:
- \( 7 \times 2 = 14 \) from \( CO_2 \)
- \( 8 \times 1 = 8 \) from \( H_2O \)
- Total = \( 14 + 8 = 22 \) oxygen atoms.
- Since \( O_2 \) has 2 oxygen atoms, we need \( \frac{22}{2} = 11 \) \( O_2 \) molecules.

Thus, the balanced equation is:
\[ C_7H_{16}(l) + 11 O_2(g) \rightarrow 7 CO_2(g) + 8 H_2O(l) \]

### Step 3: Calculate \( \Delta N_G \).

\( \Delta N_G \) is defined as the change in the number of moles of gas, which is calculated as:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

- **Moles of gaseous products**: 7 (from \( 7 CO_2 \))
- **Moles of gaseous reactants**: 11 (from \( 11 O_2 \))

Thus,
\[ \Delta N_G = 7 - 11 = -4 \]

### Step 4: Use the relationship between \( \Delta H \) and \( \Delta U \).

The relationship between \( \Delta H \) and \( \Delta U \) is given by the equation:
\[ \Delta H = \Delta U + \Delta N_G RT \]

Rearranging this gives:
\[ \Delta H - \Delta U = \Delta N_G RT \]

### Step 5: Substitute \( \Delta N_G \) into the equation.

Substituting \( \Delta N_G = -4 \) into the equation:
\[ \Delta H - \Delta U = -4 RT \]

### Final Answer

Thus, the value of \( \Delta H - \Delta U \) for the combustion of heptane is:
\[ \Delta H - \Delta U = -4 RT \]