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The lattice energy of an ionic solid such as NaCl is the enthalpy change ΔH° for the process in which the solid changes to ions. For example,
Assume that the ionization energy and electron a affinity are ΔH values for the processes defined by those terms. The ionization energy of Na is 496 kJ/mol. Use this, the
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General Chemistry - Standalone book (MindTap Course List)
- Of the five elements Sn, Si, Sb, O, Te, which has the most endothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? E(g)E+(g)+earrow_forwardWhat is the electron configuration of the Ba3+ ion? Suggest a reason why this ion is not normally found in nature.arrow_forwardGiven: Enthalpy of atomisation of calcium =+ 178 kJ First ionisation energy of calcium =+590 kJ Second ionisation energy of calcium = +1145 kJ Enthalpy of atomisation of chlorine =+ 121 kJ Electron affinity of chlorine Lattice energy of calcium chloride =- 2258 kJ = - 346 kJ Construct a Born-Haber cycle for calcium chloride, CaCl2 by using the data given above. Hence, calculate the enthalpy of formation of calcium chloride. b. The enthalpy of solution for calcium chloride crystal is -81.3 kJ mol'. Based on the data from the above Born-Haber cycle, calculate the enthalpy change for the reaction below: Ca" (g) + 2CI (g)–→ Ca* (aq) + 2CI¯ (aq)arrow_forward
- Consider an ionic compound, MXMX, composed of generic metal MM and generic, gaseous halogen XX. The enthalpy of formation of MXMX is Δ?∘f=−411ΔHf∘=−411 kJ/mol. The enthalpy of sublimation of MM is Δ?sub=101ΔHsub=101 kJ/mol. The ionization energy of MM is IE=461IE=461 kJ/mol. The electron affinity of XX is Δ?EA=−325ΔHEA=−325 kJ/mol. (Refer to the hint). The bond energy of X2X2 is BE=189BE=189 kJ/mol. Determine the lattice energy of MXMX.arrow_forwardThe lattice energy of potassium iodide is the energy required for the following reaction. KI(s) → K+(g) + I−(g) ΔHrxn = ΔHlattice Use the Born-Haber cycle to calculate ΔHlattice for KI(s) from the information given below. Equation 1: 2 K(s) + I2(g) → 2 KI(s) ΔH1 = −655 kJ/mol Equation 2: K(s) → K(g) ΔH2 = 89 kJ/mol Equation 3: I2(g) → 2 I(g) ΔH3 = 214 kJ/mol Equation 4: K(g) → K+(g) + e− ΔH4 = 419 kJ/mol Equation 5: I(g) + e− → I−(g) ΔH5 = −294 kJ/molarrow_forwardConsider an ionic compound, MXMX, composed of generic metal MM and generic, gaseous halogen XX. The enthalpy of formation of MXMX is Δ?∘f=−553ΔHf∘=−553 kJ/mol. The enthalpy of sublimation of MM is Δ?sub=129ΔHsub=129 kJ/mol. The ionization energy of MM is IE=491IE=491 kJ/mol. The electron affinity of XX is Δ?EA=−325ΔHEA=−325 kJ/mol. (Refer to the hint). The bond energy of X2X2 is BE=219BE=219 kJ/mol. Determine the lattice energy of MXMX. Δ?lattice=ΔHlattice= kJ/molarrow_forward
- 1) 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_forwardThe lattice energy of MgO is 3890 kJ/mol. The first and the second ionization energies (IE1 and IE2) of Mg are 738 kJ/mol and 1450.6 kJ/mol, respectively. The first ionization energy of O is 1314 kJ/mol. The first electron affinity (EA1) of O is +141 kJ/mol. Using these data, as well as data from a table of thermodynamic data at 1 atm and 25°C, determine the second electron affinity for oxygen, EA2(O). kJ/molarrow_forwardUse Born-Mayer equation to calculate the lattice energy for PbS (it crystallizes in theNaCl structure). Then, use the Born–Haber cycle to obtain the value of lattice energy for PbS.You will need the following data following data : ΔH Pb(g) = 196 kJ/mol; ΔHf PbS = –98kJ/mol; electron affinities for S(g)→S- (g) is -201 kJ/mol; ) S- (g) →S2-(g) is 640kJ/mol. Ionizationenergies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formationare calculated beginning with the elements in their standard states (S8 for sulfur). Diatomicsulfur, S2, is formed from S8 (ΔHf: S2 (g) = 535 kJ/mol. Can you just do the Born-Haber part?arrow_forward
- Calculate the lattice energy for LiBr(s) given the following: sublimation energy for Li(s) +166 kJ/mol ΔHf for Br(g) +97 kJ/mol first ionization energy of Li(g) +520. kJ/mol electron affinity of Br(g) –325 kJ/mol enthalpy of formation of LiBr(s) –351 kJ/molarrow_forwardConsider an ionic compound, MX2MX2, composed of generic metal MM and generic, gaseous halogen XX. The enthalpy of formation of MX2MX2 is Δ?∘f=−923ΔHf∘=−923 kJ/mol. The enthalpy of sublimation of MM is Δ?sub=131ΔHsub=131 kJ/mol. The first and second ionization energies of MM are IE1=755IE1=755 kJ/mol and IE2=1364IE2=1364 kJ/mol. The electron affinity of XX is Δ?EA=−329ΔHEA=−329 kJ/mol. (Refer to the hint). The bond energy of X2X2 is BE=153BE=153 kJ/mol. Determine the lattice energy of MX2MX2. Δ?lattice=ΔHlattice= kJ/molarrow_forwardThe lattice energy of cesium iodide is the energy required for the following reaction. CsI(s) → Cs+(g) + I−(g) ΔHrxn = ΔHlattice Use the Born-Haber cycle to calculate ΔHlatticefor CsI(s) from the information given below. Equation 1: 2 Cs(s) + I2(g) → 2 CsI(s) ΔH1 = −692 kJ/mol Equation 2: Cs(s) → Cs(g) ΔH2 = 77 kJ/mol Equation 3: I2(g) → 2 I(g) ΔH3 = 214 kJ/mol Equation 4: Cs(g) → Cs+(g) + e− ΔH4 = 376 kJ/mol Equation 5: I(g) + e− → I−(g) ΔH5 = −294 kJ/molarrow_forward
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