Question

calculate the enthalpy change for converting 1.00 mol of liquid water at 100 C to water vapor at 145 degree celsius for water Δ H vaporization=40.7kj/mol and the specfic heat of water vapor is 1.84j/g-K

Answer 1

q1 = mCpΔT

= 18.016g × 1.84J/g.K × (418.15-373.15)

= 1491.72 J

q2 = n×ΔH vap

= 1mol ×44.0kJ/mol

= 44KJ

∴ qtotal = q1+ q2

= 1.498kJ + 44.0kJ

= 45.498KJ

Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.