2. Calculate the value of AH° for the following reaction: P4010(s) + 6PC15(g) 10C13PO(g) using the following four equations: a) P4(s) + 6Cl₂(g) ---> 4PC13(g) b) P4(s) + 50₂(g) ---> P4O10(s) c) PC13(g) + Cl₂(g) ---> PC15(g) d) PC13(g) + 1/2O2(g) ---> C13PO(g) AH = -285.7 kJ AH° = -1225.6 kJ AH° = -2967.3 kJ AH° -84.2 kJ =

2. Calculate the value of AH° for the following reaction: P4010(s) + 6PC15(g) 10C13PO(g) using the following four equations: a) P4(s) + 6Cl₂(g) ---> 4PC13(g) b) P4(s) + 50₂(g) ---> P4O10(s) c) PC13(g) + Cl₂(g) ---> PC15(g) d) PC13(g) + 1/2O2(g) ---> C13PO(g) AH = -285.7 kJ AH° = -1225.6 kJ AH° = -2967.3 kJ AH° -84.2 kJ =

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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