The difference between `Delta H` and `DeltaU(DeltaH - DeltaU)`, when the combustion of one mole of heptane (l) is carried out at a temperature T, is equal to:

Options

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

Explanation

To solve the question regarding the difference between ΔH (enthalpy change) and ΔU (internal energy change) during the combustion of one mole of heptane at temperature T, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Chemical Reaction**:
The combustion of heptane (C7H16) can be represented as:
\[ C_7H_{16} + O_2 \rightarrow CO_2 + H_2O \]

2. **Balance the Reaction**:
Balancing the combustion reaction:
\[ C_7H_{16} + 11 O_2 \rightarrow 7 CO_2 + 8 H_2O \]
Here, we have 7 moles of carbon dioxide and 8 moles of water produced.

3. **Determine the Change in Moles of Gas (ΔNG)**:
- **Products**: 7 moles of CO2 + 8 moles of H2O = 15 moles of gas
- **Reactants**: 11 moles of O2
- Thus, ΔNG (change in moles of gas) = moles of products - moles of reactants = 15 - 11 = 4.

4. **Relate ΔH and ΔU**:
The relationship between enthalpy change (ΔH) and internal energy change (ΔU) is given by the equation:
\[ ΔH = ΔU + ΔNG \cdot R \cdot T \]
Rearranging this gives:
\[ ΔH - ΔU = ΔNG \cdot R \cdot T \]

5. **Substitute ΔNG**:
Now, substituting ΔNG = 4 into the equation:
\[ ΔH - ΔU = 4RT \]

### Final Answer:
Thus, the difference between ΔH and ΔU when the combustion of one mole of heptane is carried out at temperature T is:
\[ ΔH - ΔU = 4RT \]

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