Question

For the following reaction dH^o_rxn= 63.11 kJ/mol and dS^o_rxn= 148 J/mol K.

BaCl₂H₂O(s) <---> BaCl_2(s) + H₂O_(g)

a) Write out the equilibrium constant for the reaction and use it to calculate the vapor pressure of the gaseous water (P_H2O) avoe the BaCl₂ H₂O at 298K.

b) Assuming dH^o_rxn and dS^o_rxn are temperature independent estimate the temperature at which the quilibrium constant (and the P_H2O) = 1bar.

Answer 1

a. dG^o_rxn = dH^o_rxn - T.dS^o_rxn

= 63110 J/mol - 298 K.148 J/mol K

= 19006 J/mol

dG^o_rxn = -RT lnK_eq

or, lnK_eq = dG^o_rxn / -RT = (19006 J/mol)/ -(8.314 J.mol^-1.K^­-1)(298K)

or, K_eq = e^{(19006 J/mol)/ -(8.314 J.mol-1.K­-1)(298K)} = 4.66*10^-4

BaCl₂H₂O(s) <---> BaCl_2(s) + H₂O_(g)

K_eq = a _BaCl2 * a_H2O/a _BaCl2H2O           where a = activity. as activity of solid substance is assumed to be 1, so we can write

K_eq = a_H2O = p_H2O = 4.66*10^-4 atm

b. when K_eq = P_H2O = 1

dG^o_rxn = -RT lnK_eq = 0

dG^o_rxn = dH^o_rxn - T.dS^o_rxn

or, 0 = dH^o_rxn - T.dS^o_rxn

or, T = dH^o_rxn/ dS^o_rxn = 63110/148 = 426.4 K