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Author Topic: Reaction Heats  (Read 1027 times)

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ghostanime2001

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Reaction Heats
« on: March 04, 2013, 09:08:59 PM »

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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ghostanime2001

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Re: Reaction Heats
« Reply #1 on: March 05, 2013, 08:37:10 AM »

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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ghostanime2001

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Re: Reaction Heats
« Reply #2 on: March 07, 2013, 06:08:24 PM »

Any ideas ?
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ghostanime2001

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Re: Reaction Heats
« Reply #3 on: March 10, 2013, 12:23:53 PM »

Any ideas about this question anyone ?
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ghostanime2001

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Re: Reaction Heats
« Reply #4 on: May 25, 2013, 07:22:35 PM »

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

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Re: Reaction Heats
« Reply #5 on: May 25, 2013, 09:38:21 PM »

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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ghostanime2001

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Re: Reaction Heats
« Reply #6 on: May 26, 2013, 10:33:04 AM »

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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Borek

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Re: Reaction Heats
« Reply #7 on: May 26, 2013, 10:46:13 AM »

No, that should be for NaOH(s) :rarrow: NaOH(aq).
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ghostanime2001

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Re: Reaction Heats
« Reply #8 on: May 26, 2013, 11:04:12 AM »

what about  NaOH(aq) to NaOH(g) ?
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ghostanime2001

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Re: Reaction Heats
« Reply #9 on: June 26, 2013, 08:30:29 PM »

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 ?
« Last Edit: June 26, 2013, 08:55:57 PM by ghostanime2001 »
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ghostanime2001

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Re: Reaction Heats
« Reply #10 on: June 27, 2013, 01:18:54 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)
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