Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ghostanime2001 on March 05, 2013, 03:08:59 AM
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Given:
1. Na(s) :rarrow: 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. ΔHsoln = -46 kJ for NaOH(s)
7. ΔH°f = -427 kJ for NaOH(s)
Predict ΔH for NaOH(s) :rarrow: 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!! :)
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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)."
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Any ideas ?
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Any ideas about this question anyone ?
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it's been a long time now since any replies, should I make a new topic?
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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).
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what do I do with enthalpy of solution, does that act as linking dissociation from solid to gaseous state? particularly for hydroxide?
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No, that should be for NaOH(s) :rarrow: NaOH(aq).
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what about NaOH(aq) to NaOH(g) ?
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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 ?
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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 O2(g) :rarrow: O(g) ΔH = 125.5 kJ
2. 1/2 H2(g) :rarrow: H(g) ΔH = 217.5 kJ
3. Na(s) :rarrow: Na(g) ΔH = 109 kJ
4. O(g) + H(g) :rarrow: O--H ΔH = -465 kJ
5. Na(g) + O(g) :rarrow: Na--O ΔH = -255 kJ
6. NaOH(s) :rarrow: Na(s) + 1/2 O2(g) + 1/2 H2(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) :rarrow: 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) :rarrow: O--H + Na--O OR NaOH(s) :rarrow: NaOH(g)