Question

Calculate the temperature at which each of the following reactions becomes spontaneous. CaO(s) + H2O(l) --> <--- (equillibrium arrows) Ca(OH)2(s).

I got it now. The answer was 2611 K. I still don't understand why it was spontaneous below though. The K wasn't negative.

Given values:

ΔH_f⁰ (kJ/mol)

CaO (s) = -634.9 kJ/mol

H₂O (l) = -285.8 kJ/mol

Ca(OH)₂ (s) = -985.2 kJ/mol

ΔS^⁰_f (J/mol x K)

CaO (s) = 38.1

H₂O (l) = 70

Ca(OH)₂ (s) = 83.4

Formula to find K:

T = ΔH/ΔS

Answer 1

   CaO(s) + H2O(l) <---->    Ca(OH)2(s)

DHrxn = DH0fCa(OH)2 - DH0fCaO + DH0fH2O

      = (-985.2)-(-285.8-634.9)

      = -64.5 kj

DSrxn = DS0fCa(OH)2 - DS0fCaO + DS0fH2O

      = (83.4)-(38.1+70)

      = -24.7 j/mol.k

DG = DH - TDS

the process to be the spontaneous, at least DG = 0 or less than 0.

   0 = -64.5*10^3 - (T*-24.7)

T = 2611 K

so that, if DH = -ve , DS = -ve

process is spontaneous, if T < DH/DS because at this condition DG = -ve.

process is non-spontaneous, if T> DH/DS because at this condition DG =+ve.