The Change in Enthalpy of Water Vapor Condensation

What is the change in enthalpy when 180 g of water vapor condenses at 100°C? (ΔHv = 40.67 kJ/mol)

a. 565 kJ

b. -565 kJ

c. -407 kJ

d. 407 kJ

Answer:

Q = -407 kJ

Hello there!

In this case, considering that the heat has two forms, sensible (variable temperature) and latent (constant temperature), we can notice that phase changes account for latent heat as the temperature remains the same. In such a way, given the enthalpy of vaporization of water, 40.67 kJ/mol, the enthalpy of condensation (reverse process) is the negative value, -40.67 kJ/mol; therefore, the associated latent heat would be:

Q = 180 g * (1 mol / 18.02 g) * -40.67 kJ/mol

Q = -407 kJ

Best regards!

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