Question

A chemist fills a reaction vessel with 0.980 g calcium phosphate (Ca3(PO4)2) solid, 0.212 M calcium (Ca+2) aqueous solution and 0.119 M phosphate (PO4−3) aqueous solution at a temperature of 25.0°C.

Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:

Ca3(PO4)2(s) 3Ca+2(aq)+2PO4−3(aq)

Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer 1

+2 I cas (Poul 3 ca cao) + 2 Pou Dag Standard Gibbs free energy of formatics so = Eproduct & React ant = 3+1 Cath) + 2 mol x Alpays – 1 mol Gaga ca espont - 3884.82 163/mol (standan Gibbs Free Energy taken from og = 38084-3884.82 kJ/mol Book) AG = DG - RT dm keal link= 2.303 log kear R= 8.314 T= 250 = 273 +25= 298k. Kes = [ca42] 3 [ Pouz] keqp = (0.21273 [0.199² Cappu). (ca (P4), ) -solid Concentration=1 Keal = 0.009528 x 0.014161 = 0.000 13 SG = 39 - 2:33 RT looked = -3884.82 - 2.3034 8.314x 298* log (6.00013] xioks! - -3884.82 - 2.303 x 8.314 x 298.6-3.8860) 19 = -3884.82 + 6 22172.92 x103 10 = 18 288 · 106 kJX103 139 = 18.288 ks.