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:

\[ \text{CH}_4 + \text{C}_2\text{H}_4 \rightarrow \text{C}_3\text{H}_8 \]

we will use the standard heat of formation values given for each compound:

- ΔH° for CH₄ = -17.9 kcal/mol
- ΔH° for C₂H₄ = 12.5 kcal/mol
- ΔH° for C₃H₈ = -24.8 kcal/mol

### Step 1: Write the formula for ΔH
The change in enthalpy for the reaction can be calculated using the following formula:

\[ \Delta H = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]

### Step 2: Identify the products and reactants
In our reaction:
- Products: C₃H₈
- Reactants: CH₄ and C₂H₄

### Step 3: Substitute the values into the formula
Now we substitute the values of ΔH° for the products and reactants into the formula:

\[ \Delta H = \Delta H_{\text{C}_3\text{H}_8} - (\Delta H_{\text{CH}_4} + \Delta H_{\text{C}_2\text{H}_4}) \]

Substituting the values:

\[ \Delta H = (-24.8 \text{ kcal/mol}) - [(-17.9 \text{ kcal/mol}) + (12.5 \text{ kcal/mol})] \]

### Step 4: Calculate the sum of the reactants
First, calculate the sum of the enthalpy of the reactants:

\[ -17.9 + 12.5 = -5.4 \text{ kcal/mol} \]

### Step 5: Substitute back into the ΔH equation
Now substitute this value back into the ΔH equation:

\[ \Delta H = -24.8 - (-5.4) \]

### Step 6: Simplify the equation
This simplifies to:

\[ \Delta H = -24.8 + 5.4 = -19.4 \text{ kcal/mol} \]

### Final Answer
Thus, the change in enthalpy (ΔH) for the reaction is:

\[ \Delta H = -19.4 \text{ kcal/mol} \]

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