Question

Determine deltaE for the process H2O(l) --> H2O(g) at 25 degrees Celsius and 1 atm. H2O(l) deltaHf = -286 H2O(g) deltaHf = -242

Answer 1

H^o = H^o_f products - H^o_f reactants

H^o = H^o_f H₂O (g) - H^o_f H₂O (l)

H^o = (-242 kJ/mol) - (286 kJ/mol)

H^o = -242 kJ/mol + 286 kJ/mol

H^o = 44 kJ/mol

Expansion work, w_ex = (n_g) * (R) * (T)

where n_g = change in the number of gaseous moles = 1

R = constant = 8.314 J/mol-K

T = temperature = 25 ^oC = 298 K

w_ex = (1) * (8.314 J/mol-K) * (298 K)

w_ex = 2500 J/mol

w_ex = 2.5 kJ/mol

H = E + w_ex

E = H - w_ex

E = (44 kJ/mol) - (2.5 kJ/mol)

E = 41.5 kJ/mol