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66. Calculate AH°for the process Co304(s)3Co(8)+2O2(g) from the following information: Co(s) + O2 (g) → CoO(s) 3C00(s) + ½ O2 (g) → Co3O4 AH*298 =-237.9 kJ AH 298 -177.5 kJ

66. Calculate AH°for the process Co304(s)3Co(8)+2O2(g) from the following information: Co(s) + O2 (g) → CoO(s) 3C00(s) + ½ O2 (g) → Co3O4 AH*298 =-237.9 kJ AH 298 -177.5 kJ

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Chem Hw need help with question 66 and question 68
66. Calculate AH®for the process Co304(s)3Co(s)+202(g) from the following information: Co(s) + % O2 (g) → CoO(s) 3C0O(s) + ½ O2 (g) → Co3O4 AH 298 =-237.9 kJ AH298 =-177.5 kJ Solution Reverse the directions of the two equations so that Co3O4(s) will be on the left-hand side and multiply the first equation by 3: addition of the two resulting equation gives: 68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: (a) N2(g)+O2(g)–2NO(g) (b) Si(s)+2C12(g)→SİC14(g) (c) Fe203(s)+3H2(g)→2FE(s)+3H2O(1) (d) 2LİOH(s)+CO2(g)→Lİ2CO3(s)+H2O(g) Solution (a) the reaction of formation of NO(g). the reaction is twice (b) This reaction is the reaction of formation of SICL(g); thus, its enthalpy is the enthalpy of formation of SIC14(g), which is -662.75 (from Appendix G), MacBook Air
Transcribed Image Text:66. Calculate AH®for the process Co304(s)3Co(s)+202(g) from the following information: Co(s) + % O2 (g) → CoO(s) 3C0O(s) + ½ O2 (g) → Co3O4 AH 298 =-237.9 kJ AH298 =-177.5 kJ Solution Reverse the directions of the two equations so that Co3O4(s) will be on the left-hand side and multiply the first equation by 3: addition of the two resulting equation gives: 68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: (a) N2(g)+O2(g)–2NO(g) (b) Si(s)+2C12(g)→SİC14(g) (c) Fe203(s)+3H2(g)→2FE(s)+3H2O(1) (d) 2LİOH(s)+CO2(g)→Lİ2CO3(s)+H2O(g) Solution (a) the reaction of formation of NO(g). the reaction is twice (b) This reaction is the reaction of formation of SICL(g); thus, its enthalpy is the enthalpy of formation of SIC14(g), which is -662.75 (from Appendix G), MacBook Air
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