Given the following reactions and their respective ∆H values:1. 2H₂(g) + O₂(g) → 2H₂O(l); ∆H = -572.3 kJ2. N₂(g) + 3H₂(g) → 2NH₃ (g); ∆H = -91.8 kJWhat is the ∆H for the reaction: N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g)?Select one:a.-755.9 kJb.755.9 kJc.-664.1 kJd.664.1 kJ
Question
Given the following reactions and their respective ∆H values:1. 2H₂(g) + O₂(g) → 2H₂O(l); ∆H = -572.3 kJ2. N₂(g) + 3H₂(g) → 2NH₃ (g); ∆H = -91.8 kJWhat is the ∆H for the reaction: N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g)?Select one:a.-755.9 kJb.755.9 kJc.-664.1 kJd.664.1 kJ
Solution
The given reaction is: N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g)
This reaction can be obtained by reversing the first reaction and adding it to the second reaction.
The reverse of the first reaction is: 2H₂O(l) → 2H₂(g) + O₂(g); ∆H = 572.3 kJ
Adding this to the second reaction gives:
2H₂O(l) + N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g) + 2H₂(g); ∆H = 572.3 kJ - 91.8 kJ = 480.5 kJ
But we need to get rid of the 2H₂(g) on both sides, so we subtract the first reaction from this:
N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g); ∆H = 480.5 kJ - (-572.3 kJ) = 1052.8 kJ
So, none of the given options are correct. The ∆H for the reaction: N₂(g) + 3H₂(g) → 2NH₃ (g) + 2O₂(g) should be 1052.8 kJ.
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