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Author Topic: Calculating Equilibrium Constants - States; and Spontaneous Reactions  (Read 1911 times)  Share 

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tcg93

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Calculating Equilibrium Constants - States; and Spontaneous Reactions
« on: November 05, 2010, 12:24:42 pm »
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Hi,

We were told at school, when calculating equilibrium constants (not Ka) that you only include species with the main state of the equation. However, after looking through an exam this MCQ answers state otherwise.

Could someone please clear this up for me

(The attached question from an exam says that the answer is D - I picked B, but IF the answer is D then wouldn't C be right too (units weren't asked for))


Also,
why doesn't a +ve EMF always mean a reaction is spontaneous (Q17 - answer A)?
« Last Edit: November 05, 2010, 05:22:02 pm by tcg93 »
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Mao

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Re: Calculating Equilibrium Constants - states and Spontaneous Reactions
« Reply #1 on: November 05, 2010, 03:54:27 pm »
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For the first q, what is the difference between c and d? I don't see any..
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toshibaj

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Re: Calculating Equilibrium Constants - states and Spontaneous Reactions
« Reply #2 on: November 05, 2010, 04:00:29 pm »
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For the first, even if units are asked for, wouldn't C be right? The units cancel out don't they? 
 
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98.40_for_sure

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Re: Calculating Equilibrium Constants - states and Spontaneous Reactions
« Reply #3 on: November 05, 2010, 04:41:57 pm »
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The chemistry teacher at scotch said that putting in units for a K equation would be technically wrong? So based on what he said... it would be C? Can someone clarify this?
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Souljette_93

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Re: Calculating Equilibrium Constants - states and Spontaneous Reactions
« Reply #4 on: November 05, 2010, 04:51:19 pm »
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Quote from: toshibaj on November 05, 2010, 04:00:29 pm
For the first, even if units are asked for, wouldn't C be right? The units cancel out don't they? 
 

Yeah i thought C as well. As far as i know, you have to put units.That's what my teacher said, as well as TSFX i believe.
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tcg93

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Re: Calculating Equilibrium Constants - states and Spontaneous Reactions
« Reply #5 on: November 05, 2010, 05:11:18 pm »
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Quote from: Mao on November 05, 2010, 03:54:27 pm
For the first q, what is the difference between c and d? I don't see any..

hey Mao,

why don't we exclude the water in the equilibrium constant because it is in a different state? - i was told that by my teacher
(there is no diff between C and D)

and also can u explain why Q17 is A?

Cheers
« Last Edit: November 05, 2010, 05:13:03 pm by tcg93 »
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98.40_for_sure

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #6 on: November 05, 2010, 06:01:00 pm »
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Rate of reaction too slow?
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tcg93

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #7 on: November 05, 2010, 06:05:32 pm »
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Quote from: 99.95_for_sure on November 05, 2010, 06:01:00 pm
Rate of reaction too slow?

Doesn't that mean the reaction is still spontaneous, only it is really slow?
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fady_22

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #8 on: November 05, 2010, 06:11:42 pm »
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Quote from: tcg93 on November 05, 2010, 05:11:18 pm
Quote from: Mao on November 05, 2010, 03:54:27 pm
For the first q, what is the difference between c and d? I don't see any..

hey Mao,

why don't we exclude the water in the equilibrium constant because it is in a different state? - i was told that by my teacher
(there is no diff between C and D)

and also can u explain why Q17 is A?

Cheers

I think they are going with the definition, which is wrong. It should be which will always be positive in a spontaneous reaction.
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physics

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #9 on: November 05, 2010, 06:14:28 pm »
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spontaneous just means reactions from chemical-->electircal energy
non-spontaneoue just means that electricity needs to be supplied to start the reaction.
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tcg93

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #10 on: November 05, 2010, 06:24:07 pm »
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Quote from: physics on November 05, 2010, 06:14:28 pm
spontaneous just means reactions from chemical-->electircal energy
non-spontaneoue just means that electricity needs to be supplied to start the reaction.

so since you're producing electricity for spontaneous, why is the EMF not always +Ve?
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Mao

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #11 on: November 05, 2010, 07:16:12 pm »
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Q4:

Difficult to deal with. Very difficult to deal with. Strictly speaking K doesn't have a unit (if you are working with 'activity'), but since we're not working with activity, units should be included, always.
Which brings the question, what 'units' does a liquid have?
I think C as the correct answer. liquids have a molar concentration (water has a concentration of ~55M), the square brackets still denote 'concentration', so the unit for this K is unitless.
D is definitely incorrect, since the numerator has the unit M2, the denominator must have the unit M3 to give the M-1 they claim, which is absurd.


Q17

This question is a bit stupid. When EMF is positive, the reaction is
- exothermic
- generates electricity
The EMF and rate will be different if we change the conditions (different conc, etc). Thus all four options are correct. Ahh I see what they did there.

The reaction may not spontaneously proceed. Exothermic does not always mean spontaneity, this reaction may have a ridiculously large activation energy to start, thus it may not start spontaneously.
« Last Edit: November 05, 2010, 07:18:27 pm by Mao »
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toshibaj

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #12 on: November 05, 2010, 08:12:37 pm »
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Sorry for sounding dumb, but how do we know the reaction is exothermic?
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Mao

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #13 on: November 06, 2010, 01:08:00 pm »
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Quote from: toshibaj on November 05, 2010, 08:12:37 pm
Sorry for sounding dumb, but how do we know the reaction is exothermic?

From the EMF. A positive EMF means it is thermodynamically favoured, i.e. exothermic.
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Martoman

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Re: Calculating Equilibrium Constants - States; and Spontaneous Reactions
« Reply #14 on: November 06, 2010, 02:35:26 pm »
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Quote from: Mao on November 06, 2010, 01:08:00 pm
Quote from: toshibaj on November 05, 2010, 08:12:37 pm
Sorry for sounding dumb, but how do we know the reaction is exothermic?

From the EMF. A positive EMF means it is thermodynamically favoured, i.e. exothermic.

lol wut? I was thinking its giving off electricity == heat.
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