Question
Calculate the enthalpy of the reaction of boron trioxide with steam: B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g)
Given:
H2O(l)
H2(g) + 1⁄2 O2(g)
2B(s) + 3H2(g)
2B(s) + 3/2 O2(g) → B2O3(s)
→ H2O(g) → H2O(l) → B2H6(g)
44 kJ/mol -286 kJ/mol 36 kJ/mol -1273 kJ/mol
Given:
H2O(l)
H2(g) + 1⁄2 O2(g)
2B(s) + 3H2(g)
2B(s) + 3/2 O2(g) → B2O3(s)
→ H2O(g) → H2O(l) → B2H6(g)
44 kJ/mol -286 kJ/mol 36 kJ/mol -1273 kJ/mol
Answers
dHrxn = (n*dHf products) - (n*dHf reactants)
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