Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)H_(8)` is :

Options

A. `-55.2 " Kcal"`
B. `-30.2 " Kcal"`
C. `55.2 " Kcal"`
D. `-19.4 " Kcal"`

Answer

To calculate the change in enthalpy (ΔH) for the reaction \( CH_4 + C_2H_4 \rightarrow C_3H_8 \), we can use the standard heat of formation values provided for each compound.

### Step-by-Step Solution:

1. **Identify the standard heat of formation values**:
- For \( CH_4 \) (methane): \( \Delta H_f = -17.9 \, \text{kcal/mol} \)
- For \( C_2H_4 \) (ethylene): \( \Delta H_f = 12.5 \, \text{kcal/mol} \)
- For \( C_3H_8 \) (propane): \( \Delta H_f = -24.8 \, \text{kcal/mol} \)

2. **Write the formula for ΔH of the reaction**:
\[ \Delta H = \Delta H_f \text{(products)} - \Delta H_f \text{(reactants)} \]

3. **Substitute the values into the formula**:
- Products: \( C_3H_8 \)
- Reactants: \( CH_4 + C_2H_4 \)
\[ \Delta H = \Delta H_f(C_3H_8) - [\Delta H_f(CH_4) + \Delta H_f(C_2H_4)] \]
\[ \Delta H = (-24.8 \, \text{kcal/mol}) - [(-17.9 \, \text{kcal/mol}) + (12.5 \, \text{kcal/mol})] \]

4. **Calculate the sum of the reactants' enthalpy**:
\[ \Delta H_f(CH_4) + \Delta H_f(C_2H_4) = -17.9 + 12.5 = -5.4 \, \text{kcal/mol} \]

5. **Complete the calculation for ΔH**:
\[ \Delta H = -24.8 - (-5.4) = -24.8 + 5.4 = -19.4 \, \text{kcal/mol} \]

6. **Final result**:
\[ \Delta H = -19.4 \, \text{kcal/mol} \]

### Conclusion:
The change in enthalpy (ΔH) for the reaction \( CH_4 + C_2H_4 \rightarrow C_3H_8 \) is \( -19.4 \, \text{kcal/mol} \).