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

Answer 1

The reaction in question 2CaO(s) ==> 2Ca(s) + O₂(g) is the reverse of the given equation, and it is also twice the number of moles of the original equation.

The fact that is is the reverse, means we must change the sign of ∆H, thus making it positive. So, at this point it would be +635 kJ/mole. Now, since we doubled the number of moles, we have to multiply the ∆H by 2 to account for this change.

Thus, the ∆H for the new reaction is 2 x +635 = 1270 kJ = ∆H for 2CaO(s) ==> 2Ca(s) + O₂(g)