Exercise 8.122 Trends in First lonization Energy He 2372 2081 1312 168 Electron

Question

Exercise 8.122 Trends in First lonization Energy He 2372 2081 1312 168 Electron Affinities (kJ/mol) 402 314 Ar Be B 1086 CI 152 801 1A. 8A 125 520 Si 1012 1000 Kr Mg He Br 1351 Na 738 2500 496 2A 3A 5A 7A. As Se 1140 Ge 947 941 Xe Ga 762 Li Be B C N O F Ne 2000 590 419 579 Sb 1008 170 In 209 834 869 Rn 558 60 >0 27 122 141 328 >0 1500 Sr Rb 59 1000 403 Na Mig Al Si P S Cl Ar Pb -53 -43 -134 -72 -200 -349 >0 716 703 812 589 503 500- 376 K Ca Ga Ge As Se Br Kr -48 -2 -30 -119 -78 -195 -325 >0 Rb Sr In Sn Sb Te I Xe 1A 4A. 7A -47 -5 30 107 103 190 -295 Increasing ionization energy Part A Using the data in the figures above calculate AE for the reaction Nalg) Cl(g) Na (g) Cl (g) AE kJ/mol

Explanation / Answer

Solution :-

Reaction equation

Na(g) + Cl(g) ----- > Na+(g) + Cl-(g)

First ionization energy of the Na   ---- > Na+ = 496 kJ/mol

Electron affinity of the Cl --- > Cl-          = -349 kJ/ mol

Delta E = ?

Formula to calculate the delta E is as follows

Delta E = IE + EA

Lets put the values in the formula.

Delta E = 496 kJ per mol + (-349 kJ per mol )

             = 147 kJ/ mol

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