Step 1: Understanding the Concept:
Enthalpy of vaporization (\( \Delta H_{\text{vap}} \)) is an intensive property defined as the amount of heat energy required to vaporize one mole of a substance at constant pressure and temperature.
The problem gives us the heat required for a specific mass, so we need to convert that mass to moles to find the molar enthalpy.
Step 2: Key Formula or Approach:
The formula relating heat, moles, and molar enthalpy of vaporization is:
\[ q = n \times \Delta H_{\text{vap}} \]
Therefore, \( \Delta H_{\text{vap}} = \frac{q}{n} \)
Where:
\( q \) = heat supplied (in kJ)
\( n \) = number of moles of the substance
Moles (\( n \)) can be calculated as:
\[ n = \frac{\text{Mass (m)}}{\text{Molar Mass (M)}} \]
Step 3: Detailed Explanation:
First, determine the molar mass (\( M \)) of benzene (\( \text{C}_6\text{H}_6 \)).
Atomic mass of \( \text{C} = 12 \text{ g mol}^{-1} \)
Atomic mass of \( \text{H} = 1 \text{ g mol}^{-1} \)
\[ M = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g mol}^{-1} \]
Next, calculate the number of moles (\( n \)) in 13 grams of benzene:
\[ n = \frac{m}{M} = \frac{13 \text{ g}}{78 \text{ g mol}^{-1}} \]
\[ n = \frac{1}{6} \text{ mol} \]
We are given that the heat supplied (\( q \)) is \( 5.1 \text{ kJ} \).
Now, substitute these values into the enthalpy of vaporization formula:
\[ \Delta H_{\text{vap}} = \frac{q}{n} = \frac{5.1 \text{ kJ}}{\frac{1}{6} \text{ mol}} \]
\[ \Delta H_{\text{vap}} = 5.1 \times 6 \text{ kJ mol}^{-1} \]
\[ \Delta H_{\text{vap}} = 30.6 \text{ kJ mol}^{-1} \]
Step 4: Final Answer:
The enthalpy change of vaporisation is \( 30.6 \text{ kJ mol}^{-1} \).