Question

When 3.395 gm \(C_2H_5OH_{(l)}\) is burnt completely, it liberates 101.11 kJ of heat, then find the enthalpy of formation of \(C_2H_5OH_{(l)}\). Given: \(\Delta H_f^\circ\) of \(H_2O_{(l)} = -285.8 \, \text{kJ/mol}\) \(\Delta H_c^\circ\) of graphite \(= -393.5 \, \text{kJ/mol}\)

• A. \(-274.43 \, \text{kJ/mol}\)
• B. \(-548.86 \, \text{kJ/mol}\)
• C. \(-683.5 \, \text{kJ/mol}\)
• D. \(-872.1 \, \text{kJ/mol}\)

Answer 1

The Correct Option is A

Concept: Using Hess's law, the enthalpy of combustion can be related to enthalpies of formation of products and reactants. \[ \Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \]
Step 1: Write the combustion reaction \[ C_2H_5OH_{(l)} + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \]
Step 2: Calculate moles of ethanol \[ \text{Moles of } C_2H_5OH = \frac{3.395}{46} = 0.0738 \]
Step 3: Heat released for 1 mole ethanol \[ \frac{101.11}{0.0738} = 1369.97 \, \text{kJ/mol} \] Since heat is released, \[ \Delta H_c^\circ = -1369.97 \, \text{kJ/mol} \]
Step 4: Apply Hess law \[ \Delta H^\circ = 2\Delta H_f^\circ(CO_2) + 3\Delta H_f^\circ(H_2O) - \Delta H_f^\circ(C_2H_5OH) \] \[ -1369.97 = 2(-393.5) + 3(-285.8) - \Delta H_f^\circ(C_2H_5OH) \] \[ -1369.97 = -787 - 857.4 - \Delta H_f^\circ(C_2H_5OH) \] \[ -1369.97 = -1644.4 - \Delta H_f^\circ(C_2H_5OH) \] \[ \Delta H_f^\circ(C_2H_5OH) = -274.43 \, \text{kJ/mol} \]