Question

Calculate the enthalpy of vaporisation of ethanol if 11.5 g of ethanol is completely vaporised by supplying 11.8 kJ of heat.

• A. $21.7 \text{ kJ mol}^{-1}$
• B. $47.2 \text{ kJ mol}^{-1}$
• C. $65.1 \text{ kJ mol}^{-1}$
• D. $39.0 \text{ kJ mol}^{-1}$

Hint:

$\Delta H$ is always "per mole." Always find the moles of the substance first. Dividing by 0.25 is the same as multiplying by 4!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Enthalpy of vaporization ($\Delta H_{\text{vap}}$) is defined as the amount of heat energy required to completely vaporize exactly one mole of a liquid substance at constant temperature and pressure.
Step 2: Key Formula or Approach:
The formula connecting heat, moles, and enthalpy of vaporization is: \[ \Delta H_{\text{vap}} = \frac{q}{n} \] where $q$ is the total heat supplied and $n$ is the number of moles of the substance.
Step 3: Detailed Explanation:
1. First, we must calculate the molar mass of ethanol ($\text{C}_2\text{H}_5\text{OH}$). Using standard atomic masses: $\text{C} = 12 \text{ g/mol}$, $\text{H} = 1 \text{ g/mol}$, $\text{O} = 16 \text{ g/mol}$. Molar mass of $\text{C}_2\text{H}_5\text{OH} = (2 \times 12) + (6 \times 1) + 16 = 24 + 6 + 16 = 46 \text{ g/mol}$. 2. Next, calculate the number of moles ($n$) of ethanol present in the given sample. Given mass = $11.5 \text{ g}$ \[ n = \frac{\text{Mass}}{\text{Molar mass}} = \frac{11.5 \text{ g}}{46 \text{ g/mol}} = 0.25 \text{ mol} \] 3. Finally, calculate the enthalpy of vaporization per mole. Heat supplied ($q$) for $0.25 \text{ mol} = 11.8 \text{ kJ}$ \[ \Delta H_{\text{vap}} = \frac{11.8 \text{ kJ}}{0.25 \text{ mol}} \] \[ \Delta H_{\text{vap}} = 11.8 \times 4 \text{ kJ mol}^{-1} \] \[ \Delta H_{\text{vap}} = 47.2 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The calculated enthalpy of vaporization is $47.2 \text{ kJ mol}^{-1}$.