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Determine the enthalpy of reaction for P4010(s) + 6PCIs(g) - 10POCI3(g) ->> P4(s) + 6Cl₂(g) ->> 4PC3(g) AH = -1225.6 kJ/mol P4(s) + 502(g) →→ P4O10(S) AH = -2967.3 kJ/mol PCI3(g) + Cl₂(g) → PCIs(g) AH° = -84.2 kJ/mol PC 3(g) + 1/2O2(g) → POCI3(g) AH° = -285.72 kJ/mol

Determine the enthalpy of reaction for P4010(s) + 6PCIs(g) - 10POCI3(g) ->> P4(s) + 6Cl₂(g) ->> 4PC3(g) AH = -1225.6 kJ/mol P4(s) + 502(g) →→ P4O10(S) AH = -2967.3 kJ/mol PCI3(g) + Cl₂(g) → PCIs(g) AH° = -84.2 kJ/mol PC 3(g) + 1/2O2(g) → POCI3(g) AH° = -285.72 kJ/mol

Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter15: Energy And Chemical Change
Section: Chapter Questions
Problem 92A
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Determine the enthalpy of reaction for P4O10(s) + 6PCIs(g) → 10POCI3(g) P4(s) + 6Cl₂(g) ->>>> 4PC|³(g) AH° = -1225.6 kJ/mol P4(s) + 50₂(g) → P4O10(S) AH° = -2967.3 kJ/mol PCl3(g) + Cl₂(g) → PCIs(g) AH° = -84.2 kJ/mol PC3(g) + 1/2O2(g) → POCI3(g) AH° = -285.72 kJ/mol
Transcribed Image Text:Determine the enthalpy of reaction for P4O10(s) + 6PCIs(g) → 10POCI3(g) P4(s) + 6Cl₂(g) ->>>> 4PC|³(g) AH° = -1225.6 kJ/mol P4(s) + 50₂(g) → P4O10(S) AH° = -2967.3 kJ/mol PCl3(g) + Cl₂(g) → PCIs(g) AH° = -84.2 kJ/mol PC3(g) + 1/2O2(g) → POCI3(g) AH° = -285.72 kJ/mol
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