A metal salt with the formula MSO4 crystallizes from water to form a solid with the composition MSO4⋅3H2O. At 298 K, the equilibrium vapor pressure of water above this solid is 14.7 Torr. MSO4⋅3H2O(s)↽−−⇀MSO4(s)+3H2O(g) What is the value of ΔG for the reaction when the vapor pressure of water is 14.7 Torr?

ΔG =  6259.11 kJ/mol

Explanation:

At 298 K, the equilibrium vapor pressure of water above this solid is 14.7 Torr.

MSO4⋅5H2O(s)⇔ MSO4(s)+5H2O(g)

→ 1 atm = 760 torr

So, 14.7 torr= 0.0193 atm = 0.0193 / 760

Kp = [p(H2O)]3  

= ( 0.0193/760)³

= 1.637 x 10^-14

as we know that, by Kp and Kc reaction

Kp = Kc (RT)^Δng

Δng = 3 -0  = 3

Kc = Kp / (RT)³

= 1.637 x 10^-14 / (0.082 x 298)³

=  1.637 x 10^-14  / 14,591.177

1.12237 x 10^-18

What is the value of Δ for the reaction when the vapor pressure of water is 14.7 Torr?

we have, Δ = - RT ln K

Δ = - (8.314 J/K-mol × 298) ln 1.12237 x 10^-18

= 2477.57 x  ln 1.12237 x 10^-18

= 2477.57 x 2.5263

= 6259.11 kJ/mol

the answer is na (sodium)