The molar enthalpies of formation for H₂O(l)and H₂O(g)are −285.8 kJ and −241.8 kJ,respectively.How much heat is released when 25 g of water condenses from the gas to the liquid phase?

Question

Quizplus · Accepted Answer

The enthalpy change for the condensation of water can be found using the formula:

ΔH = H(vapor) - H(liquid)

We can use the enthalpies of formation given in the problem to find the enthalpies of the two phases:

H(vapor) = -241.8 kJ/mol
H(liquid) = -285.8 kJ/mol

To convert from moles to grams, we need to use the molar mass of water:

18.015 g/mol

So the enthalpy change for the condensation of 1 mole of water is:

ΔH = -285.8 kJ/mol - (-241.8 kJ/mol) = -44 kJ/mol

To find the enthalpy change for 25 g of water, we can use a conversion factor:

ΔH = -44 kJ/mol x (25 g / 18.015 g/mol) = -61 kJ

Therefore, the answer is choice C, -61 kJ.

The Molar Enthalpies of Formation for H2O(l)and H2O(g)are −285

The Molar Enthalpies of Formation for H₂O(l)and H₂O(g)are −285