Equilibrium between $SO_3$(product) and $SO_2,O_2$ (reactants)

[Dhamnekar Win,odd]

Are these above answers correct? specifically last one (c)?

[exchemist]

  On 3/30/2022 at 12:55 PM, Dhamnekar Win,odd said:

Are these above answers correct? specifically last one (c)?

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Haven't you forgotten to square some things in ( b )? 

P.S. Oh I see, it looks as if you have squared them but not shown you are doing so in the fraction you wrote down.

Edited March 30, 2022 by exchemist

[Dhamnekar Win,odd]

 Yes, I forgot to square the concentration of SO₃ in the numerator and the concentration of SO₂  in the denominator of the fraction in (b). But R.H.S. 125 is correctly computed.

 

 But what is your opinion about my answer to (c)?   

Edited March 30, 2022 by Dhamnekar Win,odd
new corrected answer was added

[exchemist]

  On 3/30/2022 at 2:42 PM, Dhamnekar Win,odd said:

 Yes, I forgot to square the concentration of SO₃ in the numerator and the concentration of SO₂  in the denominator of the fraction in (b). But R.H.S. 125 is correctly computed.

 

 But what is your opinion about my answer to (c)?   

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Your logic looks right to me, certainly (well done for allowing for the decrease in free oxygen), though I don't pretend to do this sort of thing every day. You probably have a lot more practice at doing problems like this than I do.  

I seem to recall you have asked for confirmation of your answers before and they were OK then.  I have a feeling you may be quite good at this - unless I'm mixing you up with someone else.  

[Dhamnekar Win,odd]

My corrected answer to (c) is $\frac{[1.6]^2}{[ 0.3 - 0.1 + x]^2 [0.20 - 0.05]}=125 \Rightarrow  x = 0.17$ mol SO₂. 

 

 ∴(0.17 mol SO₂/L) (10 L)= 1.7 mol of SO₂.

Edited March 31, 2022 by Dhamnekar Win,odd

[joigus]

Flawless application of Le Chatelier's principle.

I've no objection to there being good practical reasons to remove oxygen, and I totally trust @exchemist with this. But, why do you assume oxygen reducing its concentration --as per the exercise's statement? Are you re-calculating concentrations due to total volumes changing? Can you specify the whole chemical reaction equation with the phases?

I'm just curious.

[exchemist]

  On 3/31/2022 at 3:56 PM, joigus said:

Flawless application of Le Chatelier's principle.

I've no objection to there being good practical reasons to remove oxygen, and I totally trust @exchemist with this. But, why do you assume oxygen reducing its concentration --as per the exercise's statement? Are you re-calculating concentrations due to total volumes changing? Can you specify the whole chemical reaction equation with the phases?

I'm just curious.

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I by no means exclude the possibility I may get problems like wrong, but my understanding of the scenario from the description is you have a number of moles of O2, SO2 and SO3 at equilibrium in this fixed volume vessel, and then you shove in some more SO2, thereby causing a bit more SO3 to form as equilibrium is re-established.....which will inevitably absorb a bit of O2 as it does so. 

The volume is fixed so it is the pressures that will alter as more gas is added, but it is all expressed in moles, so you can work with concentration in moles/litre, rather than pressures - though it comes to the same thing as long as you have close to ideal gas behaviour.  

[joigus]

Oh, I see; all of them are necessarily gases.

The bit I don't understand is, what assumption from the exercise's statement is at the basis of O₂ going down from 0.2 mol/L to 0.2-0.05 = 0.15 mol/L when the extra SO₂ is put there?

I seem to be missing something here...

[exchemist]

  On 3/31/2022 at 7:42 PM, joigus said:

Oh, I see; all of them are necessarily gases.

The bit I don't understand is, what assumption from the exercise's statement is at the basis of O₂ going down from 0.2 mol/L to 0.2-0.05 = 0.15 mol/L when the extra SO₂ is put there?

I seem to be missing something here...

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Yes all three are gases. If the SO3 in this 10l vessel increases from 15moles to 16 moles, one mole of O, that is, half a mole of O2, must be consumed. So on a mol/l basis, 0.05mol/l is consumed, isn't it?  

[joigus]

  On 3/31/2022 at 8:22 PM, exchemist said:

Yes all three are gases. If the SO3 in this 10l vessel increases from 15moles to 16 moles, one mole of O, that is, half a mole of O2, must be consumed. So on a mol/l basis, 0.05mol/l is consumed, isn't it?  

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Silly me. You're right. So the total mass is constant? I didn't see that in the initial statement. By,

"How many moles of sulfur dioxide must be forced into the reaction vessel", I understood new moles of sulfur dioxide are added to the equilibrium. From what you say, the new SO₃ must come from the pre-existing equilibrium, right?

I had difficulties implying that from the statement. Sorry if I sound obtuse.

[exchemist]

  On 3/31/2022 at 8:35 PM, joigus said:

Silly me. You're right. So the total mass is constant? I didn't see that in the initial statement. By,

"How many moles of sulfur dioxide must be forced into the reaction vessel", I understood new moles of sulfur dioxide are added to the equilibrium. From what you say, the new SO₃ must come from the pre-existing equilibrium, right?

I had difficulties implying that from the statement. Sorry if I sound obtuse.

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No, there is 1 mole of new SO2 added. But that will put the system out of equilibrium, so more SO3 will form, consuming some of the oxygen as it does so.  As you say, it's Le Chatelier's principle (the chemist's version of Lenz's Law 😀 ).

Edited March 31, 2022 by exchemist

[studiot]

Dhamnekar Win's fraction with the quadratic is correct for (c)

I have not checked the arithmetic of his solution.

[joigus]

  On 3/31/2022 at 8:52 PM, exchemist said:

No, there is 1 mole of new SO2 added. But that will put the system out of equilibrium, so more SO3 will form, consuming some of the oxygen as it does so.  As you say, it's Le Chatelier's principle (the chemist's version of Lenz's Law 😀 ).

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Ah, OK. Got it!

So to me what's going on, schematically is,

                                                               2SO₂         +            O₂       <>      2SO₃

------------------------------------------------------------------------------

First equilibrium:                                    0.3                          0.2                     1.5

Added SO₂ (out of equilibrium):        0.3+0.1x                     0.2                     1.5

Second equilibrium:                        0.3+0.1x-0.1              0.2-0.05             1.5+0.1                

So I think the calculation should be,

$$125=\frac{1.6^{2}}{\left(0.2+0.1x\right)^{2}\left(0.2-0.05\right)}$$

Which is, I think, what @Dhamnekar Win,odd meant when they wrote,

  On 3/31/2022 at 2:54 PM, Dhamnekar Win,odd said:

My corrected answer to (c) is [1.6]2[0.3−0.1+x]2[0.20−0.05]=125⇒x=0.17 mol SO₂. 

 

 ∴(0.17 mol SO₂/L) (10 L)= 1.7 mol of SO₂.

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That is, 0.3-0.1+0.1x instead of 0.3-0.1 +x

Am I right?

[studiot]

  On 3/31/2022 at 10:37 PM, joigus said:

Am I right?

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Yes, your meticulousness does you credit. +1

[Dhamnekar Win,odd]

  On 3/31/2022 at 10:37 PM, joigus said:

Ah, OK. Got it!

So to me what's going on, schematically is,

                                                               2SO₂         +            O₂       <>      2SO₃

------------------------------------------------------------------------------

First equilibrium:                                    0.3                          0.2                     1.5

Added SO₂ (out of equilibrium):        0.3+0.1x                     0.2                     1.5

Second equilibrium:                        0.3+0.1x-0.1              0.2-0.05             1.5+0.1                

So I think the calculation should be,

 

125=1.62(0.2+0.1x)2(0.2−0.05)

 

Which is, I think, what @Dhamnekar Win,odd meant when they wrote,

That is, 0.3-0.1+0.1x instead of 0.3-0.1 +x

Am I right?

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  Whichever and whatever computations of K_c   you use, eventually, final answer would be 0.17 mol /L SO₂ should be added to the existing 0.2 mol/L SO_2.