Chemistry
Chemistry, 06.05.2021 18:40, YUKIT9l

From the following enthalpy changes, 2PCl3(l) β†’ 2P(s) + 3Cl2(g) βˆ†H = -640 kJ
2P(s) + 5Cl2(g) β†’ 2PCl5(s) βˆ†H = -886 kJ

calculate the value of βˆ†H for the reaction
PCl3(l) + Cl2(g) β†’ PCl5(s) βˆ†H = ??

answer
Answers: 1

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From the following enthalpy changes, 2PCl3(l) β†’ 2P(s) + 3Cl2(g) βˆ†H = -640 kJ
2P(s) + 5Cl2(g)...

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