Question

H2O has a Hvap = 40.7 kJ/mol. What is the quantity of heat that is released when 27.9 g of H2O condenses?

Answer 1

H=40.7 kJ/mol
m=27.9 g
M(H₂O)=18.0 g/mol

n=m/M(H₂O)

Q=Hn=Hm/M(H₂O)

Q=40.7*27.9/18.0=63.1 kJ

Answer 2

Quantity of heat released = -63.1 kJ

Step-by-step explanation:

Given:

Enthalpy of vaporization of water, ΔHvap = 40.7 kJ/mol

Mass of water, m = 27.9 g

To determine:

The amount of heat (Q) released

Step-by-step explanation:

The reaction is: H2O(g) ↔ H2O(l)

The amount of heat evolved during condensation which involves a phase transition from vapor to liquid is given as:

`Q = n*\-Delta Hvap`

n = moles of water

Since this is a condensation process: ΔHcond = -ΔHvap

`Q = (27.9 g)/(18 g/mol) * -40.7 kJ/mol = -63.1 kJ`