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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ghostanime2001 on March 05, 2013, 03:08:59 AM

Title: Reaction Heats
Post by: ghostanime2001 on March 05, 2013, 03:08:59 AM
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!!  :)
Title: Re: Reaction Heats
Post by: ghostanime2001 on March 05, 2013, 02:37:10 PM
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)."
Title: Re: Reaction Heats
Post by: ghostanime2001 on March 08, 2013, 12:08:24 AM
Any ideas ?
Title: Re: Reaction Heats
Post by: ghostanime2001 on March 10, 2013, 06:23:53 PM
Any ideas about this question anyone ?
Title: Re: Reaction Heats
Post by: ghostanime2001 on May 26, 2013, 01:22:35 AM
it's been a long time now since any replies, should I make a new topic?
Title: Re: Reaction Heats
Post by: Borek on May 26, 2013, 03:38:21 AM
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).
Title: Re: Reaction Heats
Post by: ghostanime2001 on May 26, 2013, 04:33:04 PM
what do I do with enthalpy of solution, does that act as linking dissociation from solid to gaseous state? particularly for hydroxide?
Title: Re: Reaction Heats
Post by: Borek on May 26, 2013, 04:46:13 PM
No, that should be for NaOH(s) :rarrow: NaOH(aq).
Title: Re: Reaction Heats
Post by: ghostanime2001 on May 26, 2013, 05:04:12 PM
what about  NaOH(aq) to NaOH(g) ?
Title: Re: Reaction Heats
Post by: ghostanime2001 on June 27, 2013, 02:30:29 AM
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 ?
Title: Re: Reaction Heats
Post by: ghostanime2001 on June 27, 2013, 07:18:54 PM
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)