Using the Born-Haber cycle, the AH ofKBr is equal toAs used here AHf [K(g)]=AHsub, andAH [Br(g)]= AH vap+1/2BE(Br₂), and E=EAO AH [K(g)] + AH [Br(g)] + 1₁(K) + E(Br)+ AH latticeAH [K(g)] + AH [Br(g)] - 1₁(K) - E(Br)+ AHlatticeO AH [K(g)] + AH, [Br(g)] + 1₁(K) + E(Br)AHlatticeO AH[K(g)] - AH, [Brg)] - 1₁(K)- E(Br) -AHlatticeAH, [K(g)] - AH-[Br(g)] + 1₁(K) - E(Br)+ AHlattice

From the Born-Haber cycle, one can calculate the lattice enthalpy of an ionic crystal. It involves a…

Using the Born-Haber cycle, the AHf of KBr is equal to - As used here AH₁°[K(g)]=AHsub, and AH [Br(g)]= AH vap+1/2BE(Br₂), and E =EA O AH [K(g)] + AH₁°[Br(g)] + 1₁(K) + E(Br) + AH lattice OAH [K(g)] + AH [Br(g)]- 1₁(K) - E(Br) + AHlattice OAH [K(g)] + AH [Br(g)] + 1₁(K) + E(Br) AHlattice OAH, K(g)] - AH [Brg)] - 1₁(K) - E(Br) - AHlattice OAH, K(g)] - AH-[Br(g)] + 1₁(K) - E(Br) + AHlattice

Using the Born-Haber cycle, the AHf of KBr is equal to - As used here AH₁°[K(g)]=AHsub, and AH [Br(g)]= AH vap+1/2BE(Br₂), and E =EA O AH [K(g)] + AH₁°[Br(g)] + 1₁(K) + E(Br) + AH lattice OAH [K(g)] + AH [Br(g)]- 1₁(K) - E(Br) + AHlattice OAH [K(g)] + AH [Br(g)] + 1₁(K) + E(Br) AHlattice OAH, K(g)] - AH [Brg)] - 1₁(K) - E(Br) - AHlattice OAH, K(g)] - AH-[Br(g)] + 1₁(K) - E(Br) + AHlattice

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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