The lattice energy of NaCl is -786 kJ/mol, and the enthalpy of hydration of one mole of gaseous Na+ and one mole of gaseous Cl- ions is -783 kJ/mol. Calculate the enthalpy of solution per mole of solid NaCl.
____ kJ/mol NaCl
The lattice energy of NaCl is -786 kJ/mol, and the enthalpy of hydration of one mole...
Consider the following information. The lattice energy of NaCl is ΔH lattice=−788 kJ/mol The enthalpy of sublimation of Na is ΔHsub=107.5 kJ/mol The first ionization energy of Na is IE1=496 kJ/mol. The electron affinity of Cl is ΔHEA=−349 kJ/mol. The bond energy of Cl2 is BE=243 kJ/mol. Determine the enthalpy of formation, ΔHf, for NaCl(s). ΔHf= kJ/mol
Consider the following information. • The lattice energy of NaCl is AHlattice = –788 kJ/mol. • The enthalpy of sublimation of Na is AHsub = 107.5 kJ/mol. • The first ionization energy of Na is IE1 = 496 kJ/mol. • The electron affinity of Cl is AHEA = -349 kJ/mol. • The bond energy of Cl, is BE = 243 kJ/mol. Determine the enthalpy of formation, AHf, for NaCl(s). AH= kJ/mol
calculate the lattice energy of NaCl based on the given information : ΔH°f[NaCl(s)] = -411 kJ/mol ΔH°f [Clg] = 121.5 kJ/mol ΔH°sublimation [Na] = 109 kJ/mol IE1 (Na) = 496 kJ/mol EA1 (Cl) = -349 kJ/mol
2. Use the following data to calculate the lattice energy (U) of NaCl(s) from sodium me chlorine: Enthalpy of formation (4H) for NaCl(s) - -411 kJ/mol Enthalpy of sublimation (4Hub) of Na 107.3 kJ/mol The first ionization energy of Na (E,)-495.8 kJ/mol The bond dissociation energy (D) of Clh- 243 kJ/mol The electron affinity of Cl (Eea)- 348.6 kJ/mol.
calculate the electron affinity of Cl from the following data for NaCl. Bond energy of Cl2= 239 kJ/mol; ΔHsub(Na)=109kJ/mol; IE(Na)=496kJ/mol; lattice enthalpy(Nacl)=-789kJ/mol and ΔHf, NaCl=-413kJ/mol.
Consider the following information. The lattice energy of LiCl is ΔH lattice = −834 kJ/mol. The enthalpy of sublimation of Li is ΔH sub = 159.3 kJ/mol. The first ionization energy of Li is IE 1 = 520 kJ/mol. The electron affinity of Cl is ΔH EA = -349 kJ/mol. The bond energy of Cl2 is BE = 243 kJ/mol. Determine the enthalpy of formation, ΔHf, for LiCl(s).
Consider the following information. • The lattice energy of KCl is AHlattice = -701 kJ/mol. • The enthalpy of sublimation of K is AHsub = 89.0 kJ/mol. • The first ionization energy of K is IE1 = 419 kJ/mol. • The electron affinity of Cl is AHEA = -349 kJ/mol. • The bond energy of Cl, is BE = 243 kJ/mol. Determine the enthalpy of formation, AHf, for KCl(s). AHư= kJ/mol
Part II: lonic Bonds Earlier in the semester you used solubility rules to Now you can explore why NaOH is soluble and all Lattice Energy bility rules to decide it a substance would dissolve in solution soluble and Al(OH)) is insoluble. The key to this concept is parate a mole of a solid into its gaseous ions. lue of AH for the following reaction. e stability of the ionic solid. The higher the lattice Lattice Energy - the enthalpy required...
Sodium hydroxide (NaOH) has a lattice energy of -887 kJ/mol and a heat of hydration of -932 kJ/mol. Part A How much solution could be heated to boiling by the heat evolved by the dissolution of 22.0 g of NaOH? (For the solution, assume a heat capacity of 4.0 J/g⋅∘C, an initial temperature of 25.0 ∘C, a boiling point of 100.0 ∘C, and a density of 1.05 g/mL.) Express your answer using two significant figures. V = mL
Part II: lonic Bonds Earlier in the semester you used solubility rules to decide if a substance would dissolve in solution. Now you can explore why NaOH is soluble and Al(OH), is insoluble. The key to this concept is Lattice Energy Lattice Energy - the enthalpy required to separate a mole of a solid into its gaseous fons. Or in terms of a chemical equation, it is the value of All for the following reaction. NaCl(s) - Na" (s) Cl...