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5. Given the following data: + %O2(g) B2H6(g) + 302(g) + ½O2(g) H2O(1) 2B(s) B2O3(s) AH = -1273 kJ/mole B2O3(s) + 3H20(g) H2O(1) H2O(g) AH = -2035 kJ/mole H2(g) AH = -286 kJ/mole %3D AH = 44 kJ/mole Calculate AHrxn for the following reaction: 2B(s) + 3H2(g) B2H6(g) -> + 3/202(g) Reverse (b) B2O3(s) + 3H2O(g) 3x (c) 3H2(g) + 3/202(g) 3H20(1) (a) 2B(s) B2O3(s) B2H6(g) + 302(g) 3H2O(1) 3H20(g) AH = -1273 kJ/mole AH = +2035 kJ/mole %3D AH = -858 kJ/mole %3D Зx (d) AH = 132 kJ/mole 2B(s) + 3H2(g) - B2H6(g) AH = +36 kJ/mole %3D The reaction is ENDOTHERMIC because AH is positive

5. Given the following data: + %O2(g) B2H6(g) + 302(g) + ½O2(g) H2O(1) 2B(s) B2O3(s) AH = -1273 kJ/mole B2O3(s) + 3H20(g) H2O(1) H2O(g) AH = -2035 kJ/mole H2(g) AH = -286 kJ/mole %3D AH = 44 kJ/mole Calculate AHrxn for the following reaction: 2B(s) + 3H2(g) B2H6(g) -> + 3/202(g) Reverse (b) B2O3(s) + 3H2O(g) 3x (c) 3H2(g) + 3/202(g) 3H20(1) (a) 2B(s) B2O3(s) B2H6(g) + 302(g) 3H2O(1) 3H20(g) AH = -1273 kJ/mole AH = +2035 kJ/mole %3D AH = -858 kJ/mole %3D Зx (d) AH = 132 kJ/mole 2B(s) + 3H2(g) - B2H6(g) AH = +36 kJ/mole %3D The reaction is ENDOTHERMIC because AH is positive

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
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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5. Given the following data: + ¾O2(g) B2H6(g) + 302(g) + ½O2(g) H2O(I) 2B(s) B2O3(s) AH = -1273 kJ/mole %3D B2O3(s) + 3H2O(g) H2O(1) H2O(g) AH = -2035 kJ/mole %3D H2(g) AH = -286 kJ/mole %3D AH = 44 kJ/mole Calculate AHrxn for the following reaction: 2B(s) + 3H2(g) B2H6(g) + 3/202(g) Reverse (b) B2O3(s) + 3H20(g) 3x (c) 3H2(g) + 3/202(g) 3H2O(I) (a) 2B(s) B2O3(s) → B2H6(g) + 3O2(g) 3H2O(1) → 3H2O(g) AH =-1273 kJ/mole %3D AH = +2035 kJ/mole %3D AH = -858 kJ/mole %3D Зx (d) AH = 132 kJ/mole %3D 2B(s) + 3H2(g) → B2H6(g) AH = +36 kJ/mole %3D The reaction is ENDOTHERMIC because AH is positive
Transcribed Image Text:5. Given the following data: + ¾O2(g) B2H6(g) + 302(g) + ½O2(g) H2O(I) 2B(s) B2O3(s) AH = -1273 kJ/mole %3D B2O3(s) + 3H2O(g) H2O(1) H2O(g) AH = -2035 kJ/mole %3D H2(g) AH = -286 kJ/mole %3D AH = 44 kJ/mole Calculate AHrxn for the following reaction: 2B(s) + 3H2(g) B2H6(g) + 3/202(g) Reverse (b) B2O3(s) + 3H20(g) 3x (c) 3H2(g) + 3/202(g) 3H2O(I) (a) 2B(s) B2O3(s) → B2H6(g) + 3O2(g) 3H2O(1) → 3H2O(g) AH =-1273 kJ/mole %3D AH = +2035 kJ/mole %3D AH = -858 kJ/mole %3D Зx (d) AH = 132 kJ/mole %3D 2B(s) + 3H2(g) → B2H6(g) AH = +36 kJ/mole %3D The reaction is ENDOTHERMIC because AH is positive
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