The heat of combustion of ethylene at `18^(@)C` and at constant volume is `-330.0 kcal` when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at `18^(@)C`?

Options

A. `-333.3 Kcal`
B. `-331.03 Kcal`
C. 0
D. None of these

Answer

To calculate the heat of combustion of ethylene at constant pressure from the given heat of combustion at constant volume, we will use the following steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Heat of combustion at constant volume (ΔU) = -330.0 kcal
- Temperature (T) = 18°C = 291 K (after converting to Kelvin)
- Gas constant (R) = 2 cal/(mol·K)

2. **Write the Reaction:**
The combustion of ethylene (C₂H₄) can be represented as:
\[ C_2H_4(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2CO_2(g) \]

3. **Determine Change in Moles of Gas (ΔNG):**
- Moles of gaseous products = 2 (from 2 CO₂)
- Moles of gaseous reactants = 4 (1 C₂H₄ + 3 O₂)
- Therefore, ΔNG = Moles of gaseous products - Moles of gaseous reactants
\[ ΔNG = 2 - 4 = -2 \]

4. **Use the Relationship Between ΔH and ΔU:**
The relationship between the heat of combustion at constant pressure (ΔH) and at constant volume (ΔU) is given by:
\[ ΔH = ΔU + ΔNG \cdot R \cdot T \]

5. **Substitute Values into the Equation:**
Now substituting the known values into the equation:
\[ ΔH = -330.0 \, \text{kcal} + (-2) \cdot (2 \, \text{cal/(mol·K)}) \cdot (291 \, \text{K}) \]
First, calculate the term involving ΔNG:
\[ -2 \cdot 2 \cdot 291 = -1164 \, \text{cal} = -1.164 \, \text{kcal} \]

6. **Calculate ΔH:**
Now substitute this back into the equation:
\[ ΔH = -330.0 \, \text{kcal} - 1.164 \, \text{kcal} = -331.164 \, \text{kcal} \]

7. **Final Answer:**
The heat of combustion at constant pressure is approximately:
\[ ΔH \approx -331.16 \, \text{kcal} \]