Lattice energy for CuBr
dHf = dHsub[Cu] + dH[IE Cu] + dH[BE Br2] + dH[EA Br] + lattice energy
feeding values from above,
-125 = 318 + 725 + 173 - 345 + lattice energy
lattice energy of CuBr = -996 kJ/mol
For CuI, we expect the lattice energy to be lower than CuBr as the electronegativity difference between Cu and Br is lower. Less attraction between the two ions gives lower lattice energy.
Use the References to access important values if needed for this question. Calculate the lattice energy...
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...
Calculate the lattice energy of LiBr(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. Li(s) AHgublimation 139 kJ/mol Li(g Ionization energy-500 kJ/mol Br-Brig Bond energy 173 kJ/mol Br(g) Electron affinity 345 kJ/mol LiBr(s) AH°f-371 kJ/mol kJ/mol Submit Answer Retry Entire Group 8 more group attempts remaining Calculate the lattice energy of LiBr(s) using the following thermodynamic data (all data...
Use the ces to ac importan Tor this Calculate the lattice energy of AgCI(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. AHsublimation 265 kJ/mol Ag(s) Ag(g) lonization energy 711 kJ/mol CI-CI(g) Bond energy 223 kJ/mol -369 kJ/mol CI(g) Electron affinity AgCl(s) AH-147 kJ/mol kJ/mol Do you expect this value to be larger or smaller than the lattice energy...
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 of RbH(s) using the following thermodynamic data (all data is in kJ/mol). Rb(s) AHŞublimation = 61 kJ/mol Rb(g) Ionization energy = 383 kJ/mol H-H(g) Bond energy = 416 kJ/mol Hg) Electron affinity = -93 kJ/mol RbH) AHºr= -72 kJ/mol kJ/mol
[References) Use the References to access important values if needed for this question. The free energy change for the following reaction at 25 °C, when Cu2+] = 1.16 M and Co?'] = 3.17X10M, is -134 KJ: Cu2+(1.16 M) + Co(s) Cu(s) + Co2+(3.17X10M) AG = -134 kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction? Hot Visited Submit Answer Retry Entire Group...
Part I. Use a Born-Haber cycle to calculate the lattice energy of KCl from the following data. (5 marks) Ionization energy of K(g) = 444.0 kJ mol-1 Electron Affinity of Cl(g) = -381.0 kJ mol-1 Energy to Sublime K(s) = 152.0 kJ mol-1 Bond energy of Cl2 = 201.0 kJ mol-1 ∆rH for K(s) + 1/2 Cl2(g) ↔ KCl(s) = -480.0 kJ mol-1 art II. Using the lattice energy calculated in part I determine the enthalpy of solution potassium chloride...
Question 10 1 pts Given the following thermodynamic data, calculate the lattice energy of CaBrz(s). Term Value (kJ/mol) AH°formation[CaBr2(s)] -675 AH sublimation Ca(g)] 178 AH Sublimation[Brz(s)] 31 AH°bond energy[Br2(g)] 194 IE(Ca) IE2(Ca) Ea(Br) 590. 1145 -325 Enter your answer in units of kJ.
Calculate the lattice enthalpy of AgCl (s) using the following thermodynamic data. Note that the data given has been perturbed, so looking up the answer is probably not a good idea. Cl - Cl (g) Enthalpy of dissociation = 223 kJ/mol Ag (g) Enthalpy of formation = 265 kJ/mol Cl (g) Electron attachment enthalpy = -369 kJ/mol Ag (g) Enthalpy of ionization = 711 kJ/mol AgCl (s) Enthalpy of formation = -147 kJ/mol kJ/mol
Given the following thermodynamic data, calculate the lattice energy of CaBr2(s).Term Value (kJ/mol)ΔH∘f[CaBr2(s)] -675ΔH∘f[Ca(g)] 178ΔH∘f[Br(g)] 112I1(Ca) 590.I2(Ca) 1145EA(Br) -325Express your answer to four significant figures, and include the appropriate units.