Standard enthalpy of combustion of cyclopropane is `-2091 kJ//"mole"` at `25^(@)C` then calculated the enthalpy formation of cyclopropane. If `DeltaH_(f)^(o)(CO_(2))= -393.5 kJ//"mole"` and `Delta H_(f)^(o)(H_(2)O)= -285.8 kJ//"mole"`.

Answer

To calculate the enthalpy of formation of cyclopropane (C3H6) using the given data, we will follow these steps:

### Step 1: Write the combustion reaction of cyclopropane
The combustion reaction of cyclopropane can be represented as:
\[ \text{C}_3\text{H}_6 + \frac{9}{2} \text{O}_2 \rightarrow 3 \text{CO}_2 + 3 \text{H}_2\text{O} \]
The standard enthalpy of combustion (\( \Delta H_c^\circ \)) is given as \(-2091 \, \text{kJ/mol}\).

### Step 2: Write the formation reactions for CO2 and H2O
The formation reactions for carbon dioxide and water are:
1. For CO2:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (\Delta H_f^\circ = -393.5 \, \text{kJ/mol}) \]
2. For H2O:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad (\Delta H_f^\circ = -285.8 \, \text{kJ/mol}) \]

### Step 3: Write the formation reaction of cyclopropane
The formation reaction of cyclopropane can be represented as:
\[ 3 \text{C} + 3 \text{H}_2 + \frac{9}{2} \text{O}_2 \rightarrow \text{C}_3\text{H}_6 \]
This is the reverse of the combustion reaction.

### Step 4: Apply Hess's Law
According to Hess's Law, the enthalpy change for the formation of cyclopropane can be calculated using the enthalpies of formation of the products and the enthalpy of combustion. The equation is:
\[ \Delta H_f^\circ (\text{C}_3\text{H}_6) = [3 \Delta H_f^\circ (\text{CO}_2) + 3 \Delta H_f^\circ (\text{H}_2\text{O})] - \Delta H_c^\circ \]

### Step 5: Substitute the values
Substituting the values into the equation:
\[ \Delta H_f^\circ (\text{C}_3\text{H}_6) = [3(-393.5) + 3(-285.8)] - (-2091) \]
Calculating the values:
\[ = [(-1180.5) + (-857.4)] + 2091 \]
\[ = -2037.9 + 2091 \]
\[ = 53.1 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of formation of cyclopropane is:
\[ \Delta H_f^\circ (\text{C}_3\text{H}_6) = 53.1 \, \text{kJ/mol} \]

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