The standard molar heats of formation of ethane, carbon dioxide, and liquid water are `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.

Options

A. `-372 kcal//mol`
B. `162 kcal//mol`
C. `-240 kcal//mol`
D. `183.5 kcal//mol`

Answer

To calculate the standard molar heat of combustion of ethane (C2H6), we will follow these steps:

### Step 1: Write the balanced combustion equation for ethane.
The combustion of ethane can be represented as:
\[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Balancing the equation, we get:
\[ \text{C}_2\text{H}_6 + 7/2 \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]

### Step 2: Write the formation equations for the reactants and products.
1. For ethane:
\[ 2 \text{C} + 3 \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \quad (\Delta H_f = -21.1 \text{ kcal}) \]

2. For carbon dioxide:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (\Delta H_f = -94.1 \text{ kcal}) \]

3. For water:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad (\Delta H_f = -68.3 \text{ kcal}) \]

### Step 3: Apply Hess's Law to find the heat of combustion.
According to Hess's Law, the heat of combustion can be calculated using the heats of formation of the products and reactants:
\[ \Delta H_c = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]

### Step 4: Calculate the heat of formation for products.
For the products (2 moles of CO2 and 3 moles of H2O):
\[ \Delta H_f (\text{products}) = 2(-94.1 \text{ kcal}) + 3(-68.3 \text{ kcal}) \]
\[ = -188.2 \text{ kcal} - 204.9 \text{ kcal} = -393.1 \text{ kcal} \]

### Step 5: Calculate the heat of formation for reactants.
For the reactants (1 mole of C2H6):
\[ \Delta H_f (\text{reactants}) = -21.1 \text{ kcal} + 0 \text{ kcal} \quad (\text{O}_2 \text{ has a heat of formation of } 0) \]
\[ = -21.1 \text{ kcal} \]

### Step 6: Calculate the heat of combustion.
Now substituting the values into the equation:
\[ \Delta H_c = -393.1 \text{ kcal} - (-21.1 \text{ kcal}) \]
\[ = -393.1 \text{ kcal} + 21.1 \text{ kcal} = -372.0 \text{ kcal} \]

### Step 7: Final result.
Since the combustion reaction is for 1 mole of ethane, the standard molar heat of combustion of ethane is:
\[ \Delta H_c = -372.0 \text{ kcal/mol} \]