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8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these

8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these

Chemistry
Chemistry
9th Edition
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Steven S. Zumdahl
Chapter7: Atomic Structure And Periodicity
Section: Chapter Questions
Problem 126E: Using data from the text, determine the following values (justify your answer): a. the electron...
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8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these
Transcribed Image Text:8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these
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