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

Step 1: Understanding the Concept:
Enthalpy of combustion (\( \Delta H_c \)) is the heat released per mole of substance. We first calculate the molar enthalpy of combustion for ethanol and then use Hess's Law (\( \Delta H_{\text{rxn}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \)) to find the enthalpy of formation.

Step 2: Key Formula or Approach:
Molar Mass of \( \text{C}_2\text{H}_5\text{OH} = 46 \, \text{g/mol} \).
\[ \Delta H_c (\text{molar}) = \frac{\text{Heat liberated}}{\text{moles burnt}} \] Combustion equation: \( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O} \).

Step 3: Detailed Explanation:
1. Find moles of ethanol: \( n = \frac{3.395}{46} \approx 0.0738 \, \text{mol} \).
2. Calculate molar \( \Delta H_c \):
\[ \Delta H_c = -\frac{101.11}{0.0738} \approx -1370 \, \text{kJ/mol} \] 3. Use Hess's Law:
\[ \Delta H_c = [2 \Delta H_f(\text{CO}_2) + 3 \Delta H_f(\text{H}_2\text{O})] - [\Delta H_f(\text{C}_2\text{H}_5\text{OH}) + 0] \] Given \( \Delta H_f(\text{CO}_2) = \Delta H_c(\text{graphite}) = -393.5 \).
\[ -1370 = [2(-393.5) + 3(-285.8)] - \Delta H_f(\text{C}_2\text{H}_5\text{OH}) \] \[ -1370 = [-787 - 857.4] - \Delta H_f(\text{C}_2\text{H}_5\text{OH}) \] \[ -1370 = -1644.4 - \Delta H_f(\text{C}_2\text{H}_5\text{OH}) \] \[ \Delta H_f(\text{C}_2\text{H}_5\text{OH}) = -1644.4 + 1370 = -274.4 \, \text{kJ/mol} \]

Step 4: Final Answer:
The enthalpy of formation of \( \text{C}_2\text{H}_5\text{OH(l)} \) is \( -274.43 \, \text{kJ/mol} \).