You wish to find the enthalpy for the reaction 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g) Given the following equations Equation 1: 2 Ge (s) + 3 H₂ (g) → Ge₂H₆ (l) ∆H = 137.3 kJ/mol Equation 2: Ge (s) + 4 HI (g) → GeI₄ (s) + 2 H₂(g) ∆H = -247.8 kJ/mol Equation 3: 2 NH₄I (s) → N₂ (g) + 2 HI (g) + 3 H₂(g) ∆H = 455.8 kJ/mol Having manipulated the three given equations to give you 6 Ge (s) + 9 H₂ (g) → 3 Ge₂H₆ (l) ∆H = 411.9 kJ/mol 6 GeI₄ (s) + 12 H₂(g) → 6 Ge (s) + 24 HI (g) ∆H = 1487 kJ/mol 14 NH₄I (s) → 7 N₂ (g) + 14 HI (g) + 21 H₂(g) ∆H = 3191 kJ/mol Calculate the enthalpy, in kJ/mol, for 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g).
You wish to find the enthalpy for the reaction 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g) Given the following equations Equation 1: 2 Ge (s) + 3 H₂ (g) → Ge₂H₆ (l) ∆H = 137.3 kJ/mol Equation 2: Ge (s) + 4 HI (g) → GeI₄ (s) + 2 H₂(g) ∆H = -247.8 kJ/mol Equation 3: 2 NH₄I (s) → N₂ (g) + 2 HI (g) + 3 H₂(g) ∆H = 455.8 kJ/mol Having manipulated the three given equations to give you 6 Ge (s) + 9 H₂ (g) → 3 Ge₂H₆ (l) ∆H = 411.9 kJ/mol 6 GeI₄ (s) + 12 H₂(g) → 6 Ge (s) + 24 HI (g) ∆H = 1487 kJ/mol 14 NH₄I (s) → 7 N₂ (g) + 14 HI (g) + 21 H₂(g) ∆H = 3191 kJ/mol Calculate the enthalpy, in kJ/mol, for 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g).
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter4: Energy And Chemical Reactions
Section4.7: Where Does The Energy Come From?
Problem 4.13CE
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You wish to find the enthalpy for the reaction 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g) Given the following equations Equation 1: 2 Ge (s) + 3 H₂ (g) → Ge₂H₆ (l) ∆H = 137.3 kJ/mol Equation 2: Ge (s) + 4 HI (g) → GeI₄ (s) + 2 H₂(g) ∆H = -247.8 kJ/mol Equation 3: 2 NH₄I (s) → N₂ (g) + 2 HI (g) + 3 H₂(g) ∆H = 455.8 kJ/mol Having manipulated the three given equations to give you 6 Ge (s) + 9 H₂ (g) → 3 Ge₂H₆ (l) ∆H = 411.9 kJ/mol 6 GeI₄ (s) + 12 H₂(g) → 6 Ge (s) + 24 HI (g) ∆H = 1487 kJ/mol 14 NH₄I (s) → 7 N₂ (g) + 14 HI (g) + 21 H₂(g) ∆H = 3191 kJ/mol Calculate the enthalpy, in kJ/mol, for 6 GeI₄ (s) + 14 NH₄I (s) → 3 Ge₂H₆ (l) + 7 N₂ (g) + 38 HI (g).
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