Answer to Question #133621 in General Chemistry for Rethabile Ndlovu

Question

0.5 mol of a compound is converted from 250 °C to 38 °C at constant pressure. Calculate the total heat in J for this process. The heat of vaporization is 38.6 kJ mol-1. The boiling point is 78.5 °C. The molar mass of the compound is 46.07 g mol-1. The specific heat capacities are Cgas: 1.43 J g-1 °C-1 and Cliquid: 2.45 J g-1 °C-1

Answer 1

The given process includes:

1) cooling of the gaseous compound from 250 to 78.5°C;

2) condensation of it at 78.5°C;

3) cooling of the liquid compound from 78.5 to 38°C.

1)Q_gas=C_gas×m×∆T

m=n×M=0.5×46.07=23.035(g)

∆T=78.5-250=-171.5°C

Q_gas=1.43×23.035×(-171.5)=5649.218(J);

2)Q_condensation=-Q_vaporization=-38.6kJ/mol=-38600J/mol

And for 0.5 mol:

Q_cond=-38699×0.5=-19300(J);

3)Q_liquid=C_l×m×∆T

∆T=38-78.5=-40.5°C

Q_liquid=2.45×23.035×(-40.5)=-2268.717(J)

Q_total=-5649.219-19300-2268.717=-27217.936(J).