Using the following thermochemical data: 2Er(s) + 6HF(g) → 2ErF3(s) + 3H2(g) ΔH° = -1795.4 kJ/mol 2Er(s) + 6HCl(g) → 2ErCl3(s) + 3H2(g) ΔH° = -1443.6 kJ/molcalculate ΔH° for the following reaction: ErF3(s) + 3HCl(g) → ErCl3(s) + 3HF(g)
Using the following thermochemical data: 2Er(s) + 6HF(g) → 2ErF3(s) + 3H2(g) ΔH° = -1795.4 kJ/mol 2Er(s) + 6HCl(g) → 2ErCl3(s) + 3H2(g) ΔH° = -1443.6 kJ/molcalculate ΔH° for the following reaction: ErF3(s) + 3HCl(g) → ErCl3(s) + 3HF(g)
General Chemistry - Standalone book (MindTap Course List)
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Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.51QP: The process of dissolving ammonium nitrate, NH4NO3, in water is an endothermic process. What is the...
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Using the following thermochemical data:
calculate ΔH° for the following reaction:
ErF3(s) + 3HCl(g) → ErCl3(s) + 3HF(g)
2Er(s) + 6HF(g) → 2ErF3(s) + 3H2(g) | ΔH° = -1795.4 kJ/mol |
2Er(s) + 6HCl(g) → 2ErCl3(s) + 3H2(g) | ΔH° = -1443.6 kJ/mol |
ErF3(s) + 3HCl(g) → ErCl3(s) + 3HF(g)
351.8 kJ/mol
-3239.0 kJ/mol
-1619.5 kJ/mol
703.6 kJ/mol
175.9 kJ/mol
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