Use the following information plus the data given in Tables6.2 and 6.3 to calculate the second
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- Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? (a) MgO (b) SrO (c) KF (d) CsF (e) MgF2arrow_forwardCalculate the lattice energy of potassium fluoride, KF, using the BornHaber cycle. Use thermodynamic data from Appendix C to obtain the enthalpy changes for each step. (Note: You will obtain a slightly different answer if you use values given in Chapter 8 for the ionization energy and electron affinity, which are energy values at 0 K rather than the enthalpy changes at 298 K.)arrow_forwardUsing the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2.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_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_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
- The standard heat of formation of BaBr2BaBr2 is −−764 kJ/molkJ/mol. The first ionization energy of BaBa is 503 kJ/molkJ/mol and its second ionization energy is 965 kJ/molkJ/mol. The heat of sublimation of Ba[Ba(s)→Ba(g)]Ba[Ba(s)→Ba(g)] is 175 kJ/molkJ/mol. The bond energy of Br2Br2 is 193 kJ/molkJ/mol, the heat of vaporization of Br2(l)Br2(l) is 31 kJ/molkJ/mol, and the electron affinity of BrBr is −−325 kJ/molkJ/mol. Calculate the lattice energy of BaBr2BaBr2.arrow_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 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_forward
- 8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of thesearrow_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_forwardThe bond energy of N2 is 160 kJ/mol. What is the wavelength of a photon that can cause the dissociation of a N2 molecule? (h = 6.63 × 10–34J s; c = 3.0 × 108m/s and Avogadro's number = 6.022 × 1023 molecules/mole)arrow_forward
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