How can I show the relevant steps to calculate Deltaf H degrees of MgCl(s) given the following data: Enthalpy of sublimation for Mg(s) = + 146 kJ/mol Enthalpy of dissociation of Cl2 (g)= +244 kJ/mol EA (electron affinity ) of Cl- = -349 kJ/mol First ionization energy of magnesium Mg (g) = 738 kJ/mol Lattice energy of MgCl(s) = -676 kJ/mol
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How can I show the relevant steps to calculate Deltaf H degrees of MgCl(s) given the following data:
Enthalpy of sublimation for Mg(s) = + 146 kJ/mol
Enthalpy of dissociation of Cl2 (g)= +244 kJ/mol
EA (
First ionization energy of magnesium Mg (g) = 738 kJ/mol
Lattice energy of MgCl(s) = -676 kJ/mol
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Solved in 2 steps
- Given the following thermodynamic data, calculate the lattice energy of LiCl:ΔH°f[LiCl(s)] = -409 kJ/molΔH°sublimation [Li] = 161 kJ/molBond energy [Cl-Cl] = 243 kJ/molIE1 (Li) = 520 kJ/molEA1 (Cl) = -349 kJ/mol -1682 kJ/mol -984 kJ/mol -1560 kJ/mol -862 kJ/mol -1213 kJ/molDetermine the second ionization energy of calcium (in kJ mol-1) from the given data (all in kJ mol-1): AH°[CaCl,(s)] = -796 AH¡[Ca(g)] = 178 AH°[CI(g)] = 122 First ionization energy of Ca(g) = 590 Electron affinity of Cl(g) = -349 Lattice enthalpy of CaCl,(s) = -2260 А 1150 В 1235 C 1093 D 1210Calculate the enthalpies of formation, AH;, of group 1 chloride compounds from their elements by using the Born- Haber cycle. Process ΔΗ, k/mol Sublimation of Li(s) 161 Sublimation of Cs(s) 78 Dissociation of Cl, (g) 244 Ionization energy of Li(g) 513 Ionization energy of Cs(g) 376 Electron affinity of Cl(g) -349 Lattice enthalpy of LiCl(s) 852 Lattice enthalpy of CsCl(s) 676 AH; of LiCl = kJ/mol AH; of CsCl = kJ/mol
- The lattice energy of magnesium sulfide is the energy change accompanying the process Mg2*(g) + + S2-(g) → MgS(s) Calculate the lattice energy of MgS using the following data: Mg(s) → Mg(g) AH° = 148 kJ/mol Mg(g) → Mg2*(g) + 2e- AH° = 2186 kJ/mol Sg(s) → 8S(g) AH° = 2232 kJ/mol S(g) + 2e-- s2-(g) AH° = 450 kJ/mol 8Mg(s) + Sg(s) → 8MGS(s) AH° = -2744 kJ/mol Mg2*(g) + S2-(g)→ MgS(s) AH°lattice = ?Calculate the standard enthalpy of formation of the M20(s) metal oxide (AH in kJ/mol) using the following data: Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 124 kJ/mol First ionization energy of M = + 372 kJ/mol Lattice enthalpy of M20(s) = -2115 kJ/mol Refer to the textbook for definitions of ionization energy and electron affinity. Do not use scientific notation for your answer. Do not enter units. Your Answer: AnswerCalculate the lattice enthalpy for RbC1. You will need the following information: Species AfH°, kJ/mol Rb(g) RbCl(s) Cl(g) 80.9 - 435.4 121.3 Enthalpy of ionization for Rb(g) is 403.0 kJ/mol; electron attachment enthalpy for Cl(g) is −349.0 kJ/mol. Lattice enthalpy = kJ/mol
- Which member of each pair is more metallic?(a) Na or Cs(b) Mg or Rb(c) As or NCalculate the standard enthalpy of formation of the M20(s) metal oxide (AHe in kJ/mol) using the following data: Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 107 kJ/mol First ionization energy of M = + 488 kJ/mol Lattice enthalpy of M20(s) = -2108 kJ/molN Calculate the lattice enthalpy for RbCl. You will need the following information: Species A,H, kJ/mol Rb(e) RbCl() CI(g) Enthalpy of ionization for Rb(g) is 403.0 kl/mol; electron attachment enthalpy for Cl(g) is-349.0 kJ/mol Lattice enthalpy kJ/mol Submit Answer *** 1 43 E D 80.9 -435.4 121.3 Try Another Version 80 54 R F di a 25 2 item attempts remaining T G ‹6 P FA Y & 29 7 H 8 #7 U 8 Dil FA - 6 8 K
- 4. The common oxidation number for an alkaline earth metal is +2. (a) Using the Born-Mayer equation (for determining the lattice enthalpy) and a Born-Haber cycle (draw it), show that CaCl is an exothermic compound (negative AHf). Make a reasonable prediction to estimate the ionic radius of Ca (explain your reasoning). The sublimation (atomization) enthalpy for Ca(s) is 178 kJ/mol. (b) Show that an explanation for the non-existence of CaCl can be found in the enthalpy change for the reaction below. The AHf for CaCl2(s) is -190.2 kcal/mol. 2 CaCl(s) → Ca(s) + CaCl2(s)Calculate the standard enthalpy of formation of the M2O(s) metal oxide (AH+ in kJ/mol) using the following data: Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M + 116 kJ/mol First ionization energy of M = + 463 kJ/mol Lattice enthalpy of M₂O(s) = -2248 kJ/mol Refer to sections 9.4 and 9.5 in the textbook for definitions of ionization energy and electron affinity. Do not use scientific notation for your answer. Do not enter units. Your Answer:4G VOLTE O 76 14 4:58 7 February 2021 4:57 PM (ii) Calculate the lattice energy of Silica oxide (SiO2) from the data given below: Si (g) - Si (s) AH° = -454 kJ mol-1 S1A+ (E) + 4e" Si (3) AH° = -9949 kJ mol-1 20 (g) – 02 (3) AH° = -498 kJ mol-1 02 (3) 0 (g) + 2e AH° = -737 kj mol-1 Si (s) + 02 (3) → SiO2 (s) s14+ (6) - 202- (3) – SiO2 (s) AHf (SIO2)= -910.9 kj mol1 AH°lattice = ? Share Favourite Edit Delete More