Reactivity 1.2.5—A Born–Haber cycle is an application of Hess’s law, used to show energy changes in the formation of an ionic compound. Interpret and determine values from a Born–Haber cycle for compounds composed of univalent and divalent ions.
Description
[N/A]Directly related questions
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21N.1A.HL.TZ0.16:
Consider the Born–Haber cycle for the formation of sodium oxide:
What is the lattice enthalpy, in kJ mol−1, of sodium oxide?
A. 414 + 2(108) + 249 + 2(496) − 141 + 790B. 414 + 2(108) + 249 + 2(496) + 141 + 790
C. −414 + 2(108) + 249 + 2(496) − 141 + 790
D. −414 − 2(108) − 249 − 2(496) + 141 − 790
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21N.1A.HL.TZ0.16:
Consider the Born–Haber cycle for the formation of sodium oxide:
What is the lattice enthalpy, in kJ mol−1, of sodium oxide?
A. 414 + 2(108) + 249 + 2(496) − 141 + 790B. 414 + 2(108) + 249 + 2(496) + 141 + 790
C. −414 + 2(108) + 249 + 2(496) − 141 + 790
D. −414 − 2(108) − 249 − 2(496) + 141 − 790
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21N.1A.HL.TZ0.16:
Consider the Born–Haber cycle for the formation of sodium oxide:
What is the lattice enthalpy, in kJ mol−1, of sodium oxide?
A. 414 + 2(108) + 249 + 2(496) − 141 + 790B. 414 + 2(108) + 249 + 2(496) + 141 + 790
C. −414 + 2(108) + 249 + 2(496) − 141 + 790
D. −414 − 2(108) − 249 − 2(496) + 141 − 790
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21N.1A.HL.TZ0.16:
Consider the Born–Haber cycle for the formation of sodium oxide:
What is the lattice enthalpy, in kJ mol−1, of sodium oxide?
A. 414 + 2(108) + 249 + 2(496) − 141 + 790B. 414 + 2(108) + 249 + 2(496) + 141 + 790
C. −414 + 2(108) + 249 + 2(496) − 141 + 790
D. −414 − 2(108) − 249 − 2(496) + 141 − 790
- 22M.1A.HL.TZ1.16: Which compound has the largest value of lattice enthalpy? A. Na2O B. K2O C. Na2S D. K2S
- 22M.1A.HL.TZ1.16: Which compound has the largest value of lattice enthalpy? A. Na2O B. K2O C. Na2S D. K2S
- 22M.1A.HL.TZ1.16: Which compound has the largest value of lattice enthalpy? A. Na2O B. K2O C. Na2S D. K2S
- 22M.1A.HL.TZ1.16: Which compound has the largest value of lattice enthalpy? A. Na2O B. K2O C. Na2S D. K2S