[ghostanime2001]

Given:

1. Na(s)  Na(g) ΔH = 109 kJ
2. ΔH^°_diss = 251 kJ for O₂
3. ΔH^°_diss = 435 kJ for H₂
4. ΔH^°_diss = 465 kJ for O--H
5. ΔH^°_diss = 255 kJ for Na--O
6. ΔH_soln = -46 kJ for NaOH_(s)
7. ΔH^°_f = -427 kJ for NaOH_(s)

Predict ΔH for NaOH_(s) NaOH_(g)

Ans: 159 kJ

I understand the standard states of all substances for #1-3 but I am unsure of #6,7. I don't understand what the standard states are for #5,6 (both reactants and products). I don't understand what the physical state of sodium is for #7 (is it solid or gas ?). Also how will the aqueous ions produced from dissociation of sodium hydroxide in #6 cancel out ? 

Thanks!! 

[ghostanime2001]

sorry I meant to say "for #1-3 but I am unsure of #4,5,7. I don't understand what the standard states are for #4,5 (both reactants and products)."

[ghostanime2001]

Any ideas ?

[ghostanime2001]

Any ideas about this question anyone ?

[ghostanime2001]

it's been a long time now since any replies, should I make a new topic?

[Borek]

For bond dissociation I would assume state doesn't matter, for formation it is always from the substances in the standard state (so solid Na).

[ghostanime2001]

what do I do with enthalpy of solution, does that act as linking dissociation from solid to gaseous state? particularly for hydroxide?

[Borek]

No, that should be for NaOH(s) NaOH(aq).

[ghostanime2001]

what about  NaOH(aq) to NaOH(g) ?

[ghostanime2001]

In my worksheet, it says enthalpy of solution is the change in enthalpy which occurs when one mole of a substance is dissolved in water. The reaction in the worksheet also indicates that a substance like KOH_(s) would dissociate into K⁺_(aq) and OH^-_(aq) instead of KOH_(aq)

I am actually to the point of wondering if I actually need to make use of all equations I listed in my first post. It seems that Na--O can't be cancelled so do I really need to incorporate it into my calculations ?

[ghostanime2001]

I have obtained the correct answer using all but one equation, the enthalpy of solution, which I don't know how to use at all. This is how I arranged my equations to give the correct answer:

1. 1/2 O_2(g)  O_(g) ΔH = 125.5 kJ
2. 1/2 H_2(g)  H_(g)  ΔH = 217.5 kJ
3. Na_(s)  Na_(g)  ΔH = 109 kJ
4. O_(g) + H_(g)  O--H  ΔH = -465 kJ
5. Na_(g) + O_(g)  Na--O  ΔH = -255 kJ
6. NaOH_(s)  Na_(s) + 1/2 O_2(g) + 1/2 H_2(g)  ΔH = 427 kJ

I don't know how to show the slant symbol for cancelling terms but if you add all equations together on paper, the final equation should be NaOH(s) + O(g)  O--H + Na--O. This is where I get confused, how do you remove the O(g) term on the reactant side so that the net reaction is NaOH(s)  O--H + Na--O OR NaOH(s)  NaOH(g)