Calculate the overall energy change in kilojoules per mole for the formation of CaCl(s) from the elements. The following data are needed:
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- Given the following information: Li(s) → Li(g) Li(s) = 161 kJ/mol Heat of sublimation of HCI(g) → H(g) + Cl(g) Bond energy of HCI = 427 kJ/mol Li(g) → Li+(g) + e- lonization energy of Li(g) = 520. kJ/mol CI(g) + e- → CI-(g) Electron affinity of Cl(g) = -349 kJ/mol Li+(g) + Cl-(g) →→ LiC(s) Lattice energy of LİCI(s) = -829 kJ/mol H2(g) → 2H(g) Bond energy of H2 = 432 kJ/mol calculate the net change in energy for the reaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g) -179 kJ 362 kJ -70 kJ -572 kJ None esearrow_forwardConsider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X. • The enthalpy of formation of MX is AH; = -513 kJ/mol. • The enthalpy of sublimation of M is AHsub = 123 kJ/mol. • The ionization energy of M is IE = 465 kJ/mol. • The electron affinity of X is AHEA = -325 kJ/mol. (Refer to the hint). • The bond energy of X, is BE = 187 kJ/mol. Determine the lattice energy of MX. AHjattice -795.5 kJ/mol Incorrectarrow_forwardDraw the energy diagram of the reaction: Li(s) + ½F2 → LiF(s). Enthalpy (kJ/mol) Li(s) → Li(g) +155.2 Li(g) → Li+(g) + e +520 ½F2 →F(g) +75.3 F(g) + e → F-(g) -3arrow_forward
- Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is ΔHf∘=−965 kJ/mol. The enthalpy of sublimation of M is ΔHsub=123 kJ/mol. The first, second, and third ionization energies of M are IE1=557 kJ?mol, IE2=1751 kJ/mol, and IE3=2731 kJ/mol. The electron affinity of X is ΔHEA=−339 kJ/mol The bond energy of X2 is BE=235 kJ/mol. Determine the lattice energy of MX3.arrow_forwardConsider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X. • The enthalpy of formation of MX is AH; = -517 kJ/mol. • The enthalpy of sublimation of M is AHub 141 kJ/mol. • The ionization energy of M is IE = 449 kJ/mol. • The electron affinity of X is AHEA = -323 kJ/mol. (Refer to the hint). • The bond energy of X, is BE = 217 kJ/mol. Determine the lattice energy of MX. -33 AHjattice kJ/mol Incorrectarrow_forward1) Calculate the lattice energy for NaCl(s) using a Born-Haber cycle and the following information: NaCl(s) → Nat(g) + Cl-(g) Na(s) + 1/2 C12(g) → NaCl(s) Na(s) → Na(g) Na(g) → Na+(g) + e- 1/2 C12(g) → Cl(g) Cl(g) + e- → Cl-(g) ? -411.0 kJ/mol +107.3 kJ/mol +495.8 kJ/mol +121.7 kJ/mol -348.6 kJ/molarrow_forward
- Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than thelattice energy of NaCl? Explain.Electron affinity of Cl = –349 kJ/mol1st ionization energy of Rb = 403 kJ/molBond energy of Cl2 = 242 kJ/molSublimation energy of Rb = 86.5 kJ/molΔHf [RbCl (s)] = –430.5 kJ/molarrow_forwardI'm not sure how to solve this one.arrow_forward6- Draw Born – Haber Cycle and Calculate the lattice enthalpy for lithium fluoride, given the following information: • Enthalpy of sublimation for solid lithium = 151 kJ/mol • First ionization energy for lithium = 519 kJ/mol • F-F bond dissociation energy = 164 kJ/mol • Enthalpy of formation for F(g) = 82 kJ/mol • Electron affinity for fluorine = -348 kJ/mol Enthalpy of formation for solid lithium fluoride = -617 kJ/molarrow_forward
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