For the chemical reaction Ca (s) + ½ O₂ (g) → CaO (s) ∆H° = -635 kJ/mol. What is the molar enthalpy (in kJ/mol) for the reaction 2 CaO (s) → 2 Ca (s) + O₂ (g)?

Answer 1

Ca(s) + 1/2O2(g) ==> CaO(s) .. ∆Hº = -5635 kJ/mol

2CaO(s) ==> 2Ca(s) + O2(g) .. ∆Hº = ?

The equation in question is the reverse of the given equation, and it is also 2 x the given equation.

When you reverse a reaction, the new ∆H is the same value but opposite sign. So, here it would be +635

When you multiply a reaction by 2, the new ∆H is 2x the original. Show here it would be 2x635 = 1270 kJ

But they actually asked for the answer in kJ/mole. We now have 2 moles of CaO so the answer would be..

+1270 kJ / 2 moles = +635 kJ / mole