The fat, `C_(57)H_(104)O_(6)(s)`, is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.0 g of this fat reacts.
`C_(57)H_(104)O_(6)(s)+80 O_(2)(g)rarr57CO_(2)(g)+52 H_(2)O(l)`
`Delta_(f)H^(@)(C_(57)H_(104)O_(6),s)=-70870" kJ"//"mol, "Delta_(f)H^(@)(H_(2)O,l)=-285.8 " kJ"//"mol"`,
`Delta_(f)H^(@)(CO_(2),g)=-393.5 " kJ"//"mol"`

Options

A. 37.98
B. 40.4
C. 33.4
D. 30.2

Explanation

To calculate the energy liberated when 1.0 g of the fat \( C_{57}H_{104}O_{6} \) reacts, we will follow these steps:

### Step 1: Write the reaction and identify the enthalpy of formation values.
The reaction is:
\[ C_{57}H_{104}O_{6}(s) + 80 O_{2}(g) \rightarrow 57 CO_{2}(g) + 52 H_{2}O(l) \]

Given enthalpies of formation:
- \( \Delta_f H^\circ (C_{57}H_{104}O_{6}, s) = -70870 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ (H_{2}O, l) = -285.8 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ (CO_{2}, g) = -393.5 \, \text{kJ/mol} \)

### Step 2: Calculate the total enthalpy of formation for products and reactants.
**Products:**
- For \( 57 \, CO_{2} \):
\[ \text{Total for } CO_{2} = 57 \times (-393.5) = -22469.5 \, \text{kJ} \]
- For \( 52 \, H_{2}O \):
\[ \text{Total for } H_{2}O = 52 \times (-285.8) = -14881.6 \, \text{kJ} \]

**Total enthalpy of formation for products:**
\[ \text{Total Products} = -22469.5 + (-14881.6) = -37351.1 \, \text{kJ} \]

**Reactants:**
- For \( C_{57}H_{104}O_{6} \):
\[ \text{Total for } C_{57}H_{104}O_{6} = -70870 \, \text{kJ} \]
- For \( O_{2} \):
\[ \text{Total for } O_{2} = 0 \, \text{kJ} \, (\text{elemental form}) \]

**Total enthalpy of formation for reactants:**
\[ \text{Total Reactants} = -70870 + 0 = -70870 \, \text{kJ} \]

### Step 3: Calculate the enthalpy change for the reaction.
\[ \Delta H = \text{Total Products} - \text{Total Reactants} \]
\[ \Delta H = -37351.1 - (-70870) = 33518.9 \, \text{kJ} \]

### Step 4: Calculate the molar mass of \( C_{57}H_{104}O_{6} \).
\[ \text{Molar mass} = (57 \times 12) + (104 \times 1) + (6 \times 16) = 684 + 104 + 96 = 884 \, \text{g/mol} \]

### Step 5: Calculate the number of moles in 1.0 g of \( C_{57}H_{104}O_{6} \).
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.0 \, \text{g}}{884 \, \text{g/mol}} = 0.00113 \, \text{mol} \]

### Step 6: Calculate the energy released for 1.0 g of fat.
Using the enthalpy change calculated for 1 mole:
\[ \text{Energy released} = \Delta H \times \text{number of moles} \]
\[ \text{Energy released} = -33518.9 \, \text{kJ/mol} \times 0.00113 \, \text{mol} = -37.9 \, \text{kJ} \]

### Final Answer:
The energy liberated when 1.0 g of the fat reacts is approximately **-37.9 kJ**.

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