(ii)The first and second electron affinities of the O atom are-141.8 and +844.4 kJ mol,respectively (iii)The enthalpy of sublimation of calcium metal is+192.5 kJ mol (iv)The first and second ionization potentials of calcium are+589.5 and+1145.2 kJmol,respectively (v)The enthalpy of formation of is-635.5 kJ mol Cas→Cag +192.5kJmo Cag→Cag+e +589.5 kJ mol Cag→Cag+e +1145 2 kI mol h02g→0g 2(+495.0)=+247.5 kJ mol1 Og+e→0g -141.8 kJ mol-l 0e+e→02g +844.4 kJ mol! Ca”g+02g→Ca0s Cas+⅓02g→Ca0s-635.5 kJmol Thus,-U=-635.5-(192.5+589.5+1145.2+247.5-141.8+844.4)=-3512.8kJmo Thus,lattice energy=U=+3512.8 kJ mol Part C.Answer any three of the six questions.If you answer more than three,the best three will be used to calculate your mark(20 marks each) 1.(a)Balance the following REDOX reaction in acidic solution: 2 2 (ii) The first and second electron affinities of the O atom are -141.8 and +844.4 kJ mol-1, respectively. (iii) The enthalpy of sublimation of calcium metal is +192.5 kJ mol-1 (iv) The first and second ionization potentials of calcium are +589.5 and +1145.2 kJ mol-1, respectively. (v) The enthalpy of formation of CaO(s) is -635.5 kJ mol-1 Ca(s) → Ca(g) +192.5 kJ mol-1 Ca(g) → Ca+ (g) + e- +589.5 kJ mol-1 Ca+ (g) → Ca+2 (g) + e- +1145.2 kJ mol-1 ½ O2(g) → O(g) ½ (+495.0) = +247.5 kJ mol-1 O(g) + e- → O- (g) -141.8 kJ mol-1 O- (g) + e- → O-2 (g) +844.4 kJ mol-1 Ca+2 (g) + O-2 (g) → CaO(s) -U _____________________________________________________ Ca(s) + ½ O2(g) → CaO(s) -635.5 kJ mol-1 Thus, -U = -635.5 – (192.5 + 589.5 + 1145.2 + 247.5 – 141.8 + 844.4) = -3512.8 kJ mol-1 Thus, lattice energy = U = +3512.8 kJ mol-1 Part C. Answer any three of the six questions. If you answer more than three, the best three will be used to calculate your mark (20 marks each). 1. (a) Balance the following REDOX reaction in acidic solution: