Concept explainers
Interpretation:
Born-Haber cycle for formation of
Concept introduction:
The pictorial representation of formation of ionic solids from its constituent elements is known as Born-Haber cycle. Following are the steps required to draw Born-Haber cycle of any ionic compound:
Step 1: Solid metal is converted into gaseous isolated atoms. It takes place by process called sublimation.
Step 2: Gaseous molecules are broken down into separate atoms. Energy is supplied to break molecules apart and this is called
Step 3: Isolated metal atoms are converted into respective cations with the help of ionization energy.
Step 4: Anions are formed from gaseous atoms with the help of
Step 5: Ionic compound is formed by the combination of cation and anion. Energy is released in this process.
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CHEMISTRY-TEXT
- Given the following information, construct a Born-Haber cycle to calculate the lattice energy of CrCl₂I(s): Net energy change for the formation of CrCl₂I(s) = -420 kJ/mol Bond dissociation energy for I2(g) = +243 kJ/mol Bond dissociation energy for Cl2(g) for Cl2(g) for 12(g) = +151 kJ/mol Heat of sublimation for I2(s) = +62 kJ/mol Heat of sublimation for Cr(s) = +397 kJ/mol = E₁₁ for Cr(g) = 652 kJ/mol E₁₂ for Cr(g) == 1588 kJ/mol E₁3 for Cr(g) = 2882 kJ/mol Eea for Cl(g)=-349 kJ/mol === Eea for I(g) = -295 kJ/molarrow_forwardUse 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_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
- Consider the following data for tungsten: atomic mass electronegativity electron affinity ionization energy heat of fusion 183.84 2.36 78.6 (2) W (g) + e 770. 1 35. g mol kJ mol You may find additional useful data in the ALEKS Data tab. kJ mol Does the following reaction absorb or release energy? 2+ (1) W²(g) + e w (g) kJ mol Is it possible to calculate the amount of energy absorbed or released by reaction (1) using only the data above? w (g) If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (1): Does the following reaction absorb or release energy? Is it possible to calculate the amount of energy absorbed or released by reaction (2) using only the data above? If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (2): release absorb Can't be decided with the data given. O yes no kJ/mol release absorb Can't be decided with the data given. yes no 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_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
- 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_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_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=105ΔHsub=105 kJ/mol. The ionization energy of MM is IE=483IE=483 kJ/mol. The electron affinity of XX is Δ?EA=−307ΔHEA=−307 kJ/mol. (Refer to the hint). The bond energy of X2X2 is BE=213BE=213 kJ/mol. Determine the lattice energy of MXMX. Δ?lattice=ΔHlattice= kJ/molarrow_forward
- Estimate the change in enthalpy (delta H) for the dissociation for HCl(g) into H+(g) and Cl-(g) Ionization energy for H 1312 kJ/mol Electron affinity for H (-72.8 kJ/mol) Ionization energy for Cl 1251.2 kJ/mol Electron affinity for Cl (-349 kJ/mol)arrow_forwardChoose the related energy for the following reaction: 2 Cs* (g) + O2- (g) → Cs20 (s) electron affinity ionization energy heat of formation lattice energyarrow_forwardCalculate the lattice energy for NaCI(s) using a Born-Haber cycle and the following information: 3. NaCI(s) → Na*(g) + Cl'(g) ? Na(s) + 1/2 CI2(g) NaCI(s) -411.0 kJ/mol Na(s) Na(g) +107.3 kJ/mol Na*(g) + e¨ +495.8 kJ/mol 1/2 Cl2(g) → CI(g) +121.7 kJ/mol Na(g) CI(g) + e → Cl'(g) -348.6 kJ/mol A) +690.3 kJ/mol B) +787.2 kJ/mol C) +34.8 kJ/mol O D) +1512 kJ/mol 身arrow_forward
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