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

Question

Quizplus · Accepted Answer

We can use Hess's law and manipulate the given equations to obtain the desired reaction:

2 LiOH(s) → Li2B12S2B0O(s) + H2B12S2B0O(l)        ΔS° = 379.1 kJ/mol

2 H2B12S2B0O(g) + O2(g) → 2 H2B12S2B0O(l)            ΔS° = -285.9 kJ/mol

LiH(s) + H2B12S2B0O(l) → LiOH(s) + H2B12S2B0O(g)    ΔS° = -111.0 kJ/mol

By adding and canceling the intermediate species, we get:

2 LiH(s) + O2(g) → Li2B12S2B0O(s) + H2B12S2B0O(l)     ΔS° = -111.0 kJ/mol + 379.1 kJ/mol + (-285.9 kJ/mol) = -128.8 kJ/mol

Therefore, the answer is C, -128.8 kJ/mol.

2 LiOH(s)→ Li2O(s)+ H2O(l)ΔrH° = 379.1 KJ/mol

2 LiOH(s)→ Li₂O(s)+ H₂O(l)Δ_rH° = 379.1 KJ/mol