If the lattice energy for a compound is 666 kJ/mol and the ΔH(soln) is 12.9 kJ/mol, what is the Δ(hydr) in kJ/mol to the nearest integer?
ΔH(soln) = ΔH(hydr) - lattice energy -------------------> formula
12.9 = ΔH(hydr) - (-666)
ΔH(hydr) = -653
hydration energy = -653 kJ / mol
If the lattice energy for a compound is 666 kJ/mol and the ΔH(soln) is 12.9 kJ/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 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
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
Determine the lattice energy of MX2. Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX2 is ΔH∘f=−681 kJ/mol. The enthalpy of sublimation of M is ΔHsub=145 kJ/mol. The first and second ionization energies of M are IE1=607 kJ/mol and IE2=1444 kJ/mol. The electron affinity of X is ΔHEA=−315 kJ/mol. The bond energy of X2 is BE=247 kJ/mol. Determine the lattice energy of MX2. I've been stuck on this question...
Given the information below, determine the lattice energy for the fictitious ionic compound ZX 3, where Z is a metal with a change of +3 and X is a nonmetal with a charge of -1. ΔH sublimation(Z) = 195 kJ/mol BE (X2) = 320 kJ/mol IE 1 (Z) =433 kJ/mol EA1(x) = -298 kJ/mol IE2 (Z) = 678 kJ/mol EA 2(X) = -591 Kj/mol IE3 (Z) =1291 kJ/mol ΔH ° (ZX3) = 661 kJ/mol
The lattice energy of LiBr is -807 kJ/mol. If 150g of LiBr is dissolved in water, 83.9 kJ of heat is released. What is the heat of hydration of LiBr? Give your answer in units of kJ/mol to the nearest kJ/mol (don't forget to use the correct sign of + or -)
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
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
Calculate the lattice energy of AgF(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. Ag(s) Asublimation -265 kJ/mol Ag(g) Ionization energy-711 kJ/mol F-F(g) Bond energy- 138 kJ/mol F(g) Electron affinity348 kJ/mol AgF(s) AHor-225 kJ/mol kJ/mol Do you expect this value to be larger or smaller than the lattice energy of AgCI(s)?
Calculate the lattice energy for LiF(s) given the following: sublimation energy for Li(s) = +166 KJ/mol delta Hf for F(g) = +77 KJ/mol first ionization energy of Li(g) = +520 KJ/mol electron affinity of F(g) = -328 KJ/mol enthalpy of formation of LiF(s) = -617 KJ/mol