6.58 The common lead-acid car battery produces a large burst ofcurrent, even at low temperatures, and is rechargeable. The reac-tion that occurs while recharging a “dead" battery is2PBSO,(s) + 2H,O(1) →(a) Use AH values from Appendix B to calculate AH(b) Use the following equations to check your answer to part (a):Pb(s) + PbO2(s) + 2H,SO,(!)rxn(1) Pb(s) + PbO,(s) + 2SO;(g)2PBSO,(s)= -768 kJAHAH = -132 kJrxn(2) SO;(g) + H,O(I)H,SO,(1)

∆Hrxn is calculated by using the standard enthalpy of formation of each molecule involved in the…

6.58 The common lead-acid car battery produces a large burst of current, even at low temperatures, and is rechargeable. The reac- tion that occurs while recharging a "dead" battery is 2PBSO,(s) + 2H,O(1) Pb(s) + PbO,(s) + 2H,SO,(1) (a) Use AH values from Appendix B to calculate AH (b) Use the following equations to check your answer to part (a): rxn (1) Pb(s) + PbO,(s) + 2SO;(g) 2PBSO,(s) AH = -768 kJ rxn (2) SO;(g) + H,O(1) H,SO,(1) AH = -132 kJ rxn

6.58 The common lead-acid car battery produces a large burst of current, even at low temperatures, and is rechargeable. The reac- tion that occurs while recharging a "dead" battery is 2PBSO,(s) + 2H,O(1) Pb(s) + PbO,(s) + 2H,SO,(1) (a) Use AH values from Appendix B to calculate AH (b) Use the following equations to check your answer to part (a): rxn (1) Pb(s) + PbO,(s) + 2SO;(g) 2PBSO,(s) AH = -768 kJ rxn (2) SO;(g) + H,O(1) H,SO,(1) AH = -132 kJ rxn

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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Question

6.58 The common lead-acid car battery produces a large burst of
current, even at low temperatures, and is rechargeable. The reac-
tion that occurs while recharging a “dead" battery is
2PBSO,(s) + 2H,O(1) →
(a) Use AH values from Appendix B to calculate AH
(b) Use the following equations to check your answer to part (a):
Pb(s) + PbO2(s) + 2H,SO,(!)
rxn
(1) Pb(s) + PbO,(s) + 2SO;(g)
2PBSO,(s)
= -768 kJ
AH
AH = -132 kJ
rxn
(2) SO;(g) + H,O(I)
H,SO,(1)

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