The energy evolved when one mole of the crystalline substance is formed from its constituent gaseous ions” is known as lattice energy (U) of the crystal.
Draw the Born-Haber Cycle with these values and calculate lattice energy. Problem 1: Label each reaction...
16. What is the lattice energy of CaO? A. -3414 kJ B -2144 C.-2667 kg D4139741 -E 73028 RJ Cas-Calg) Calg) - Ca' (g) + Ca'(g) - Ca?(g) + 2 O(g)- 0(g) O(g) + O(g) O(g) + -02 (8) CaS) +12 0:) - CAO3 AH(kJ) +193 +590 +1010 498 -141 +878 -635
Construct a Born-Haber cycle and calculate the lattice energy of CaC2 (s). Note that this solid contains the diatomic ion C22–.Useful Information:?H°f (CaC2(s)) ?Hsub (Ca (s)) ?Hsub (C (s)) Bond dissociation energy of C2 (g) = +614 kJ/molFirst ionization energy of Ca (g) = +590 kJ/mol Second ionization energy of Ca (g) = +1143 kJ/mol First electron affinity of C2 (g) = –315 kJ/mol Second electronaffinity of C2 (g) = +410 kJ/mol= –60 kJ/mol = +178 kJ/mol = +717 kJ/mol
Part A Use the Born-Haber cycle, data from Appendix llBand the following table to calculate the lattice energy of Cao. (AsubH for calcium is 178 kJ/mol: EA 141 kJ/mol, EA 744 kJ/mol: IEC 590 kJ/mol. IE2 1145 kJ/mol. TABLE 9.1 Average Bond Energies Bond Energy Bond Energy Bond Energy Bond kJ mo 1) Bond (kJ mol 1) Bond (kJ mol 237 414 418 218 389 946 193 464 208 H 0 N-0 222 590 565 272 51 200 H-Br 364...
Given the following information, construct a Born-Haber cycle to calculate the lattice energy of CaC2(s): Net energy change for the formation of CaC2(s)=−60kJ/mol Heat of sublimation for Ca(s)=+178kJ/mol Ei1 for Ca(g)=+590kJ/mol Ei2 for Ca(g)=+1145kJ/mol Heat of sublimation for C(s)=+717kJ/mol Bond dissociation energy for C2(g)=+614kJ/mol Eea1 for C2(g)=−315kJ/mol Eea2 for C2(g)=+410kJ/mol Express your answer using four sig figs
Question 4 4 pts Use the Born-Haber Cycle to calculate the lattice energy for the formation of X2Y. Input your answer in units of kJ/mole with the correct sign. Process Enthalpy (kJ/mol). X(s)--> X(g) 115 X(g) -->X*(8) + le 499 Y2 (8) --> 2Y (8) 264 -295 Y (8) + 1e.-->Y (8) Y (8) + 1e' --> Y2 () 115 2X(s) +% Y2 (8)--> X2Y(s) -549
7) For the ionic solid AlzOs a) Determine its lattice energy using the appropriate Born-Haber cycle and the following values. All values in kJ/mol: IEi (A)-557.5:IE2 (A)-1817; IEs (A)-2745; IE(Al) 11580 E (0)-1314; IE2 (0) 3388; IEs (O)-5300 ΔΗ"a (O) =-141 (first electron affinity) ; ΔΗ'EA AH (Al) 330; AHa (O)-249;AH (Al Os)--1669.8 (o)- 798 (second electron affinity) b) Al:O, crystallizes in a corundum structure. How does the above lattice energy compare to the lattice energy determined by an electrostatic...
Use the Born Haber cycle (see equations and enthalpy values below) to determine the lattice energy for BeI2 (s) (∆H LE (BeI2 (s))= ?) Show your work. Box your final answer. A. Be(g)→Be1+ (g) + 1 e–∆H = + 899.5kJ B. Be1+ (g) →Be2+ (g) + 1 e–∆H = +1757 kJ C. Be(s)→Be(g)∆H= +302kJ D. I2(s)→I2(g)∆H= + 62.4kJ E. I(g) + e–→I–(g)∆H= –295kJ F. I2(g)→2I(g)∆H= + 151 kJ G. Be(s) + I2(s) →BeI2(s)∆H= –208 kJ
Draw a Born Haber cycle for gallium(I) oxide and calculate the crystal lattice energy for Gallium(I) oxide, given: ΔH°sub (Ga) = 277 kJ/mol E.A.1 (O) = –133 kJ/mol I.E.1(Ga) = 578.84 kJ/mol E.A.2 (O) = 247 kJ/mol I.E.2(Ga) = 1979.4 kJ/mol B.D.E.(O2) = 495 kJ/mol I.E.3(Ga) = 2964.5 kJ/mol ΔHf° (Ga2O) = – 349.8 kJ/mol ANS: -2779 kJ/mol (It was in the Answer Key)
4) Calculate the lattice enthalpy for calcium fluoride using the Born-Haber cycle method, using the provided table. (Show all your work; 2 points) Enthalpies, AH/(kJ mol) +192 Process Sublimation of Ca(s) Ionization of Ca(g) Dissociation of F2(g) Electron gain by F(g) Formation of CaF (s) +1735 to Ca(ag +157 -328 -1220
Use the Born-Haber Cycle and information given below to determine the lattice energy of LiF. Li(s) → Li(g) +159.3 kJ Li(g) → Li+(g) + e– +500.9 kJ F2(g) → 2 F(g) +158.8 kJ F(g) + e– → F–(g) –332.6 kJ Li+(g) + F–(g) → LiF(s) ? --------------------------------------------------- Li(s) + ½ F2(g) → LiF(s) – 616.0 kJ Group of answer choices +209.0 kJ –1023 kJ +1023 kJ –209.0 kJ –1102.4 kJ