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[tex] H_2O [/tex] has a [tex] \Delta H_{\text {vap}} = 40.7 \, \text{kJ/mol} [/tex]. What is the quantity of heat that is released when [tex] 27.9 \, \text{g} [/tex] of [tex] H_2O [/tex] condenses? Use [tex] q = n \Delta H [/tex].

A. [tex] 60.00 \, \text{kJ} [/tex]
B. [tex] 61.05 \, \text{kJ} [/tex]
C. [tex] 63.09 \, \text{kJ} [/tex]
D. [tex] 68.60 \, \text{kJ} [/tex]



Answer :

GinnyAnswer
To determine the quantity of heat released when 27.9 g of H2O condenses, we'll follow these steps:

1. Identify given data and constants:
- The enthalpy of vaporization ([tex]\(\Delta H_{\text{vap}}\)[/tex]) of water is 40.7 kJ/mol.
- The mass of water (H2O) is 27.9 g.
- The molar mass of water (H2O) is approximately 18.015 g/mol.

2. Calculate the number of moles of H2O:
Using the molar mass of H2O, we can convert the mass of water to moles. The formula to find the number of moles ([tex]\(n\)[/tex]) is:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Substituting the given values:
[tex]\[ n = \frac{27.9 \text{ g}}{18.015 \text{ g/mol}} \approx 1.5487 \text{ mol} \][/tex]

3. Calculate the quantity of heat released:
The quantity of heat ([tex]\(q\)[/tex]) released when water condenses can be found using the formula:
[tex]\[ q = n \, \Delta H_{\text{vap}} \][/tex]
Substituting the values for [tex]\(n\)[/tex] and [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[ q = 1.5487 \text{ mol} \times 40.7 \text{ kJ/mol} \][/tex]
[tex]\[ q \approx 63.03 \text{ kJ} \][/tex]

Thus, the quantity of heat released when 27.9 g of H2O condenses is approximately [tex]\(63.03 \text{ kJ}\)[/tex].

Among the given options:
- [tex]\(60.00 \text{ kJ}\)[/tex]
- [tex]\(61.05 \text{ kJ}\)[/tex]
- [tex]\(63.09 \text{ kJ}\)[/tex]
- [tex]\(68.60 \text{ kJ}\)[/tex]

The closest option to our calculated value (63.03 kJ) is [tex]\(63.09 \text{ kJ}\)[/tex].

Therefore, the correct answer is [tex]\(63.09 \text{ kJ}\)[/tex].

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