Question

The enthalpy change under standard conditions for which of the reactions below would be equal to the ∆H°f of NaOH(s)? (A) Na(s) + H2O(l)  NaOH(s) + 1/2 H2(g) (B) Na(s) + 1/2 O2(g) + 1/2 H2(g)  NaOH(s) (C) Na(s) + 1/2 H2O2(l)  NaOH(s) (D) Na+(aq) + OH–(aq)  NaOH(s) please explain why you chose that answer and why the other answers are wrong. If it has to do with them being in their standard states, how do I know what the standard states of the elements are.

Answer 1

Start from the basic. Standard state of the substance is defined at 25 deg C and 1atm pressure. So at these conditions if the substance exists in say Solid state then its standard state is Solid.

So we know all of the metals except Mercury are solid at these condistions hence for all the metals except Mercury standard state is Solid. So Standard state of Na is Solid.

Now delta H formation at standard state means the heat required to form that substance at above standard conditions from its constituent molecules at standard state. Which simply means that if you want to form Water then Water is H2O that means you must react H2 and O2 gases Hydrogen and Oxygen exists in this forms at standard condition. The heat required to form liquid water (since standard state of water liquid ) at these conditions from H2 and O2 gases is its Standard heat of formation.

Now lets look at our problem. We are required to find reaction for the formation of NaOH.

From the discussion above we can say to form NaOH we must react Na at its standard state which is Solid. Hence we found one reactant Na (s).

Now we are left with OH. O must come from O2 and H from H2 as discussed above. and these will be in the gas phase. Hence our reqction must be.

Na(s) + 1/2 H2(g) + 1/2 O2(g) ===> NaOH (s)