When sodium burns in air it produces a bright white light and heat energy

2Na + O2 2Na2O ?H = -414kJ/mol

What would you expect to be the heat energy associated with the reverse process

2Na2O 2Na + O2

Can you please show the entire calculation with units please?

This is what I have but it is wrong:
2 moles Na2O x +414 kJ/mol /2moles Na2O = +828 kJ

If it is -414 kJ/mol going in the forward direction it must be +414 kJ/mol going in the opposite direction.

the enthalpy -414kJ/mol is the enthalpy of combustion. i.e. how much energy released for one mole of Na metal to burn in air. so the actual equation is;

Na + 1/2O2 --> NaO ?H = -414kJ/mol.

However, the equation given in your question is written in whole number coefficients just to get rid of the 1/2 but keep in mind that the ?H still remain the same in this case.

so simply reverse the sign

To determine the heat energy associated with the reverse process, you can use the concept of Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same regardless of the path taken.

Given the forward reaction:
2Na + O2 -> 2Na2O ΔH = -414 kJ/mol

We want to find the reverse reaction:
2Na2O -> 2Na + O2 ΔH = ?

To find the enthalpy change for the reverse reaction, we need to reverse the sign of the forward reaction's ΔH value. Hence:

ΔH_reverse = -(-414 kJ/mol) = +414 kJ/mol

So the heat energy associated with the reverse process is +414 kJ/mol. The sign change indicates that heat is absorbed in the reverse reaction, opposed to being released in the forward reaction.

Your calculation was incorrect because it seems like you multiplied the forward reaction's ΔH value by 2 instead of reversing the sign. Here's the correct calculation:

2 moles Na2O x (+414 kJ/mol) / 2 moles Na2O = +414 kJ/mol

Therefore, the heat energy associated with the reverse process is +414 kJ/mol.