Question

\\(\text{h}_2\text{o}\\) has a \\(\delta\text{h}_{\text{vap}} = 40.7\text{ kj/mol}\\). what is the quantity of heat that is released when \\(27.9\text{ g}\\) of \\(\text{h}_2\text{o}\\) condenses?
use \\(q = n\delta\text{h}\\).

options:

• \\(63.09\text{ kj}\\)
• \\(60.00\text{ kj}\\)
• \\(61.05\text{ kj}\\)
• \\(68.60\text{ kj}\\)

Explanation:

Calculate moles of water

Find the molar mass of \( \text{H}_2\text{O} \).

$$ M_{\text{H}_2\text{O}} \approx 18.02\text{ g/mol} $$

Calculate moles \( n \) from the given mass.

$$ n = \frac{27.9\text{ g}}{18.02\text{ g/mol}} \approx 1.548\text{ mol} $$

Determine heat released during condensation

Using the Enthalpy of Reaction concept, relate phase change enthalpy to heat.
Since condensation is the reverse of vaporization:

$$ \Delta H_{\text{cond}} = -\Delta H_{\text{vap}} = -40.7\text{ kJ/mol} $$

Calculate the heat \( q \) released:

$$ q = n \cdot \Delta H_{\text{cond}} = 1.548\text{ mol} \cdot (-40.7\text{ kJ/mol}) \approx -63.01\text{ kJ} $$

The magnitude of heat released is approximately \( 63.01\text{ kJ} \) (matching \( 63.09\text{ kJ} \) using \( M_{\text{H}_2\text{O}} \approx 18.0\text{ g/mol} \)).

Verify with standard high school molar mass

Using \( M_{\text{H}_2\text{O}} = 18.0\text{ g/mol} \):

$$ n = \frac{27.9\text{ g}}{18.0\text{ g/mol}} = 1.55\text{ mol} $$

$$ |q| = 1.55\text{ mol} \times 40.7\text{ kJ/mol} = 63.085\text{ kJ} \approx 63.09\text{ kJ} $$

Answer:

• (A) 63.09 kJ (Correct answer)
• (B) 60.00 kJ
• (C) 61.05 kJ
• (D) 68.60 kJ