Question

CuSO4 · 5H2O (s) ⇌ CuSO4 (s) + 5 H2O (g)
a) Calculate ΔH ° and ΔS ° for the reaction at 25 ° C. (3p)
b) Calculate ΔG ° for the reaction at 150 ° C. (2p)
c) What will be the equilibrium constant of the reaction at 150 ° C? (2p)
d) In what direction is the reaction spontaneous at 150 ° C and the pressure 1 bar? What temperature would be required to change the reaction direction? .

I especially need help with d. The answer to d is T = 398 K

Answer 1

(a) We know according of Thermodynamical laws,

Enthalpy change in the reaction (∆H) = H_product - H_reactant

now, for reaction CuSO4 · 5H2O (s) ⇌ CuSO4 (s) + 5 H2O (g)

∆H_solutionof CuSO4(s) = –66.5 kJ mol^–1

∆H_solutionof CuSO4.5H2O(s) = +11.7 kJ mol^–1

The reaction involved are:

(1) CuSO4(s) → Cu²⁺(aq) + SO4 ^2- (aq) ∆H° _CuSO4 = -66.5 kJ mol^-1

(2) Cu²⁺(aq) + SO4 ^2- (aq) + 5H2O(l) → CuSO₄.5H2O(s) ∆H°= -11.7 kJ mol-1

Net reaction involved is:

CuSO4(s) + 5H2O(l) → CuSO4.5H2O(s)

∆H°_CuSO4.5H2O  = ∆H° _CuSO4 + ∆H°= -66.5 kJ mol^-1 + (-11.7 kJ mol^-1) i.e the reaction is spontaneous.

∆H°_CuSO4.5H2O = -78.2 kJ mol^-1

∆S°_CuSO4.5H2O = ∆H°_CuSO4.5H2O/ T = 78.2 kJ mol^-1/ 298K = 0.2624 kJ mol^-1K^-1

(b) ∆G°_CuSO4.5H2O = ∆H°_CuSO4.5H2O - T∆S°_CuSO4.5H2O

= -78.2 kJ mol^-1 - 298K * 0.2624 kJ mol^-1K^-1

= -78.2 kJ mol^-1 -78.2 kJ mol^-1 = -156.4 kJ mol^-1

As ∆G = is negative hence, the reaction is spontaneous.

(c) ∆G° = -RT ln K  

T = 448K (150^oC)

ln K = ∆G°/ -RT

= -156.4 kJ mol^-1 / - 8.314 kJ mol^-1 * 448K  

ln K = 0.042

K = e^0.042 = 1.04289

(d) at T = 448K (150^oC)

∆G°_CuSO4.5H2O = ∆H°_CuSO4.5H2O - T∆S°_CuSO4.5H2O

= -78.2 kJ mol^-1 - 448K * 0.2624 kJ mol^-1K^-1 = - 195.7552 kJ mol^-1

the reaction is spontaneous at 150 °C as we have obtained ∆G = negative.

And the reaction will shift towards right.

hope this helps!