Given the following information:Li(s) → Li(g)Li(s) = 161 kJ/molHeat of sublimation ofHCI(g) → H(g) + Cl(g)Bond energy ofHCI = 427 kJ/molLi(g) → Li+(g) + e-lonization energy ofLi(g) = 520. kJ/molCI(g) + e- → CI-(g)Electron affinity ofCl(g) = -349 kJ/molLi+(g) + Cl-(g) →→ LiC(s)Lattice energy ofLİCI(s) = -829 kJ/molH2(g) → 2H(g)Bond energy ofH2 = 432 kJ/molcalculate the net change in energy for thereaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g)-179 kJ362 kJ-70 kJ-572 kJNoneese

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Given the following information: Li(s) → Li(g) Li(s) = 161 kJ/mol Heat of sublimation of HCI(g) → H(g) + CI(g) Bond energy of HCI = 427 kJ/mol %3D Li(g) → Li+(g) + e- lonization energy of Li(g) = 520. kJ/mol Cl(g) + e- → Cl-(g) CI(g) = -349 kJ/mol Electron affinity of %3D Li+(g) + Cl-(g) –→ LİCI(s) Lattice energy of LİCI(s) = -829 kJ/mol %3D H2(g) → 2H(g) Bond energy of H2 = 432 kJ/mol calculate the net change in energy for the reaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g)

Given the following information: Li(s) → Li(g) Li(s) = 161 kJ/mol Heat of sublimation of HCI(g) → H(g) + CI(g) Bond energy of HCI = 427 kJ/mol %3D Li(g) → Li+(g) + e- lonization energy of Li(g) = 520. kJ/mol Cl(g) + e- → Cl-(g) CI(g) = -349 kJ/mol Electron affinity of %3D Li+(g) + Cl-(g) –→ LİCI(s) Lattice energy of LİCI(s) = -829 kJ/mol %3D H2(g) → 2H(g) Bond energy of H2 = 432 kJ/mol calculate the net change in energy for the reaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g)

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter5: Electron Configurations And The Periodic Table

Section5.8: Ion Electron Configuration

Problem 5.16CE: Fluoride ion, F, has no unpaired electrons. Vanadium forms four binary fluoridesVF2, VF3, VF4, and...

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