Answer letter B only 1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following:а.AH2660 EmoleBe(g)ALLAĦ+107 kJ/moleBeo 2 IT(9Bel:(s)AHBets) L(s)208 KJ/mole Beb,Mn(g) → Mn“(g) + 2 eMn(s) → Mn(g)S(s) → S(g)S(g) +2 e→ s(g)Mn* (g) + s (g) → MnS(s)Mn(s) + S(s) →> MnS(s)b.AH= +2230 kJ/moleAH=AH=+264 kJ/moleAH=+246 kJ/moleAH=-3176 kJ/moleAH=-212 kJ/mole

Mn (g) →Mn 2+ (g) + 2e- +2230 kJ/mole Ionization energy (IE) Mn (s) → Mn (g) ………

Answered: Mn(g) → Mn“(g)+2 e¯ Mn(s) → Mn(g) S(s)…

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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Answer letter B only

1. Use a Born-Haber Cycle to calculate the missing AH (bold reaction) of the following:
а.
AH2660 Emole
Be(g)
ALL
AĦ
+107 kJ/mole
Beo 2 IT(9
Bel:(s)
AH
Bets) L(s)
208 KJ/mole Beb,
Mn(g) → Mn“(g) + 2 e
Mn(s) → Mn(g)
S(s) → S(g)
S(g) +2 e→ s(g)
Mn* (g) + s (g) → MnS(s)
Mn(s) + S(s) →> MnS(s)
b.
AH= +2230 kJ/mole
AH=
AH=+264 kJ/mole
AH=+246 kJ/mole
AH=-3176 kJ/mole
AH=-212 kJ/mole

Expert Solution

Step 1

Mn (g) $\to$ Mn ²⁺ (g) + 2e^-	+2230 kJ/mole	Ionization energy (IE)
Mn (s) $\to$ Mn (g)	$\dots \dots$	$∆$ H₁_sublimation(of Mn)
S (s) $\to$ S (g)	+264 kJ/mole	$∆$ H₂_sublimation (of S)
S(g) + 2e^- $\to$ S^2-(g)	+246 kJ/mole	electron affinity (EA)
Mn^2-(g) + S^2-(g) $\to$ MnS (s)	-3176 kJ/mole	Lattice energy (U)
Mn(s) + S(s) $\to$ MnS (s)	-212 kJ/mole	$∆$ H_formation

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