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? Use . 60.00 kJ 61.05 kJ 63.09 kJ 68.60 kJ

Answer 1

63.09 kJ is the quantity of heat that is released when 27.9 g of water condenses.

Step-by-step explanation:

`H_2O(l)\rightarrow H_2O(g),\Delta H_(vap)=40.7 kJ/mol`

Latent heat of vaporization =
`\Delta H_(vap)=40.7 kJ/mol`

Amount of heat required to condense 1 mole of water = 40.7 kJ

`H_2O(g)\rightarrow H_2O(l),\Delta H_(vap)=-40.7 kJ/mol`

Mass of water given = 27.9 g

Moles of water :

`(27.9 g)/(18 g/mol)=1.55 mol`

Heat required to vaporize 1.55 moles of water:

`1.55 moles* (-40.7 kJ/mol)=-63.085 kJ\approx -63.09 kJ`

Negative sign indicates that energy will release.

Answer 2

The quantity of heat released is 63.09 kJ.

n = 27.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 1.548 mol H₂O

ΔH_cond = -ΔH_vap = -40.7 kJ·mol⁻¹

The negative sign shows that the water vapour is releasing energy as it condenses.

q = nΔH_cond = 1.548 mol × (-40.7 kJ·mol⁻¹) = -63.0 kJ