Question

A metal salt with the formula MSO4 crystallizes from water to form a solid with the composition MSO4⋅5H2O. At 298 K, the equilibrium vapor pressure of water above this solid is 15.7 Torr.

MSO4⋅5H2O(s)↽−−⇀MSO4(s)+5H2O(g)


What is the value of ΔG for the reaction when the vapor pressure of water is 15.7 Torr?


Δ= ? J/mol


What is the value of Δ for the reaction when the vapor pressure of water is 571 Torr?


ΔG= ? J/mol

Answer 1

The value of Δ for the reaction when the vapor pressure of water is 571 Torr is as follows

Step-by-step explanation:

At 298 K, the equilibrium vapor pressure of water above this solid is 28.5 Torr.

MSO4⋅5H2O(s)<====> MSO4(s)+5H2O(g)

1 atm = 760 torr

thus, 28.5 torr= 0.0375 atm

Kp = [p(H2O)]5 = ( 0.0375)5 = 7.42 into 10-8

What is the value of Δ for the reaction when the vapor pressure of water is 28.5 Torr?

we have, Δ = - RT ln K

in this case K = Kp

Δ = - (8.314 J by K-mol into 298) ln 7.42 into 10-8 = + 40.673 kJ/mol

What is the value of Δ for the reaction at this temperature if the vapor pressure of water is 533 Torr?

533 torr= 0.701 atm

Kp = [p(H2O)]5 = ( 0.701)5 = 0.1697

Δ = - (8.314 J by K-mol into 298) ln 0.1697 = + 4.395 kJ/mo