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

Question 59

Multiple Choice

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)


A) +125.2 kJ
B) -17.7 kJ
C) -128.8 kJ
D) -303.6 kJ
E) + 128.8 kJ

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