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 for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter6: The Periodic Table And Atomic Structure
Section: Chapter Questions
Problem 6.107PAE
Related questions
Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fefb8e691-6f9b-4b15-9ff4-9dbd730b8fe0%2F4f7b770f-bdb6-4ae5-9577-00093af0b7e3%2Fzmg23l.png&w=3840&q=75)
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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