2 LiOH(s) → Li₂O(s) + H₂O(l) ΔH° = 379.1 kJ LiH(s) + H₂O(l) → LiOH(s) + H₂(g) ΔH° = -111.0 kJ
2 H₂(g) + O₂(g) → 2 H₂O(l) ΔH° = -285.9 kJ
Compute ΔH° in kJ for 2 LiH(s) + O₂(g) → Li₂O(s) + H₂O(l)

Question

Quizplus · Accepted Answer

First we need to determine the overall reaction by adding the given reactions: 2 LiOH(s) → Li₂O(s) + H₂O(l) ΔH° = 379.1 kJ LiH(s) + H₂O(l) → LiOH(s) + H₂(g) ΔH° = -111.0 kJ 2 LiH(s) + H₂(g) + O₂(g) → Li₂O(s) + H₂O(l) + O₂(g) We need to cancel the LiOH and H₂O(l) so we multiply the second equation by 2 and reverse it, then we obtain: 2 LiH(s) + O₂(g) → Li₂O(s) + 2H₂(g) ΔH° = -222.0 kJ Finally, we add the given reaction: 2H₂(g) + O₂(g) → 2H₂O(l) ΔH° = -285.9 kJ We can see that the two sulfuric acid molecules (H₂(g)) are cancelled, so we're left with: 2 LiH(s) + O₂(g) → Li₂O(s) + 2H₂(g) ΔH° = -128.8 kJ Therefore, the answer is C) -128.8 kJ.

2 LiOH(s) → Li2O(s) + H2O(l) ΔH° = 379

2 LiOH(s) → Li₂O(s) + H₂O(l) ΔH° = 379