Electron affinity = -442.9 kJ mol-1
Review Problem 8.065 Construct an enthalpy diagram with the following data to calculate the electron affinity...
17. What is the first ionization energy (IE) of sodium (Na) based on the following information? A. +399.4 kJ B. +464.4 kJ C. +495.4 kJ D. +526.4 kJ E. +591.4 kJ Brig) + e + Br (g) Na(s) + / Bra(l) + NaBr(s) Na (g) + Br-(g) NaBr(s) Br2(g) + 2 Brig) Br(1) Br (g) Na(s) + Nag) AH (kJ mol-') -331.4 -360.0 -743.3 +192.0 +31.0 +107.8
Physical Chemistry: Use a Born-Haber cycle to find an experimentally based value for the lattice enthalpy of sodium bromide (NaBr(s)). The lattice enthalpy corresponds to the enthalpy change for the process NaBr(s) rightarrow Na^+(g) + Br^-(g) Use the following information in doing this problem. delta H degree_f(Na(g)) = 107.32 kJ/mol delta H degree_IE1(Na(g)) = 495.8 kJ/mol delta H degree_f(Br(g)) = 111.88 kJ/mol delta H degree_EA(Br(g)) = -324.6 kJ/mol delta H degree_f(NaBr(s)) = -361.06 kJ/mol The ionization enthalpy (IE_1) and electron...
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
Knowing the enthalpy of formation of calcium fluoride is -1228 KJ / mol, the enthalpy of the sublimation of calcium is 168kj/mole, the bond enthalpy of fluorine is 155kj/mole, the electron affinity of fluorine is -328 kj/mole, the first ionization enthalpy of calcium is 590kj/mole and the second ionization enthaply is 1145kj/mole, calculate the lattice enthalpy of calcium fluoride.
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.
Given the following information, calculate the lattice energy of CaF2 The enthalpy of formation of CaF2 -1228 kJ/mol Heat of sublimation of Ca 177.8 kJ/mol Bond dissociation energy of F2 159 kJ/mol First ionization energy of Ca 589.8 kJ/mol Second ionization energy of Ca 1145.4 kJ/mol . Electron affinity of F -328 kJ/mot [ Answer : -2644 KJİ I
Calculate the enthalpies of formation, ΔHfo, of the following group 1 fluoride compounds from their elements using the Born–Haber cycle. NaF RbF Number Number kJ ol kJ mol AHO, kJ/mol Sublimation of Na(s) 108 86 Sublimation of Rb(s) 158 Dissociation of F2(g Ionization energy of Na(g) 496 ionization energy of Rb(g) 403 Electron affinity of F(g) -322 Lattice enthalpy of NaF(s) 926 Lattice enthalpy of RbF(s) 789
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
A. Calculate the lattice energy of NaI(s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea. Na(s) ΔHsublimation = 88 kJ/mol Na(g) Ionization energy = 476 kJ/mol I-I(g) Bond energy = 131 kJ/mol I(g) Electron affinity = -315 kJ/mol NaI(s) ΔH°f = -308 kJ/mol kJ/mol Do you expect this value to be larger or smaller than the lattice energy of...
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