Questions

[mohdateeq]

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Why does the K value change because of temperature only?

Why don't we use high pressures in the conversion of S02 to SO3 in the Contact Process?

When oxidising metals, such as Cu(II), does sulfuric acid get reduced to S02(g)?

Thanks.

[Undermyskin]

I'm wondering about the 1st one as well.
Last one: yes.

[The-Exiled]

Not sure about why concentration and pressure don't change the value of K...

But the reason we don't use high pressure during the conversion of SO2 to SO3 is because it is not enconomically viable. The equipment required costs too much. I think they use like 2 atmospheres.

[appletooth]

http://en.wikipedia.org/wiki/Equilib...ure_dependence

The reason why the temperature dependent nature of K was always simplified to "energy considerations." And why we were told not to question it.

[danz90]

Originally Posted by The-Exiled

Not sure about why concentration and pressure don't change the value of K...

But the reason we don't use high pressure during the conversion of SO2 to SO3 is because it is not enconomically viable. The equipment required costs too much. I think they use like 2 atmospheres.

Just always remember as a fact that ONLY temperature changes the value of K.

High pressure isn't economically viable, since the yield of SO3 is very high anyway (98%) with 1-2atm.

[tegzy26]

the value of k is not effected by changes in pressure or concentration because changes in pressure and/or concentration only change the position of the equilibrium.
an increase or decrease in temp effects the amount of heat produced

[Undermyskin]

Exactly what the text books say and exactly the way they avoid to explain further.

Can you actually give me more details? Like what they mean by 'involved concentrations of chemicals'.

I'm trying to satisfy myself with the explanation: this is the experimental aspect of science, we observe and here we have a law!

[minijumbuk]

This is outside the syllabus, so unless you really want to know why, just remember that K is only dependent on temperature.

There are actually lots of different types of equilibrium constants. There is one for concentration, one for pressure, one for temperature, pH.... LOTS
The "K" we are using right now is the K for temperature. When we use this K, it means the K value is only dependent on temperature. If we use the pressure K, then all the other factors, including temperature, would be independent, with K being only dependent on pressure.
It can a bit difficult to deal with, so for the sake of making the HSC course much simpler, they chose to only teach us the temperature dependent K.

[JasonNg1025]

Maybe it's because the equilibrium constant formula is dependant on the concentrations of the reactants and products. When you add or subtract to something's concentration, the system will shift to counteract that change. When you increase pressure, the concentration of both (gases, otherwise pressure doesn't have an effect) increases, which does not change the ratio. When you change temperature, however, you're not muddling with any concentrations, but the equilibrium does shift in a direction, which changes K.

Think of it like a set of scales (you know, like the two plates that balance each other out, then you add stuff to get equal weights). When you add the concentration of the reactants, it's like adding a weight to one side. Then Le Chatelier's principle comes along and tries to balance it out, adding another weight to the other side. The ratio of concentrations does not change. When you add pressure, it's like adding weight to both sides. The ratio still doesn't change. When you increase temperature, you're heating up the scales (which doesn't change any weight) but when Le Chatelier's principle comes in with the exo/endothermic things, a weight is randomly added to one side to balance out the heat (in the reaction). The balance is now different to the original, which means K has changed.

I'm probably wrong, it's just a guess :P
Cause this is not in the syllabus. As everyone else said, just remember it

EDIT: For your second question, the first reply is right - it's not economically viable, but also, higher pressures are not safe. A high pressure can cause leaks in the reaction chamber, causing release of gases into the atmosphere (especially the reactant SO₂.) And we all know that SO₂ is baddddd. If the pressure gets really really high, the thing just explodes.

[Pwnage101]

Originally Posted by JasonNg1025

Maybe it's because the equilibrium constant formula is dependant on the concentrations of the reactants and products. When you add or subtract to something's concentration, the system will shift to counteract that change. When you increase pressure, the concentration of both (gases, otherwise pressure doesn't have an effect) increases, which does not change the ratio. When you change temperature, however, you're not muddling with any concentrations, but the equilibrium does shift in a direction, which changes K.

Think of it like a set of scales (you know, like the two plates that balance each other out, then you add stuff to get equal weights). When you add the concentration of the reactants, it's like adding a weight to one side. Then Le Chatelier's principle comes along and tries to balance it out, adding another weight to the other side. The ratio of concentrations does not change. When you add pressure, it's like adding weight to both sides. The ratio still doesn't change. When you increase temperature, you're heating up the scales (which doesn't change any weight) but when Le Chatelier's principle comes in with the exo/endothermic things, a weight is randomly added to one side to balance out the heat (in the reaction). The balance is now different to the original, which means K has changed.

I'm probably wrong, it's just a guess :P
Cause this is not in the syllabus. As everyone else said, just remember it

EDIT: For your second question, the first reply is right - it's not economically viable, but also, higher pressures are not safe. A high pressure can cause leaks in the reaction chamber, causing release of gases into the atmosphere (especially the reactant SO₂.) And we all know that SO₂ is baddddd. If the pressure gets really really high, the thing just explodes.

Exactly right^^^^^^^^^^ - just think about it ffor a sec:

-what does the valuee of K tell you? TEH RATIO OF THE CONCENTRATION OF THE PRODUCTS COMPARED TO REACTANTS

So, when i increase teh conc of a reactant, the Eq shifts to make less of it, a DIRECT RESPONSE ACCORDING TO LE CHATELIERS PRINCIPLE - RIGHT? SO therefore the PRINCIPLE TELLS UIS THAT TEH RATIO WILL NOT CHANGE, IE K WILL NOT CHANGE

us ee the reason for it not changing - the Eq shifts to adjust so as the ratio of K is the same

When u deal with pressure and colume they affect conc

dunno how to put it clearly for u guys, but i hope u understand

concentration is defined as the amount of a substance in a given volume, right????

so when we change teh conc, volume, or pressure of a reactant, we change the conc of it, BUT le chatelier's restores teh ratio bak, so K stays the same

BUT when ur dealing with temp, notice how that has nothing to do with the definition of 'Concentration' - EG -if i increase or decresae temp of a glass of water from 20degrees to 30 degrees, it DOES NOT AFFECT the concentration of the water

So, by increasing teh temp in an exothermic reaction, for eg, since Le Chatelier will shift it to the left, K will get smaller - MORE REACTANTS ARE BEING PRODUCED INT HE SAME SPACE, WITH ELSS PRODUCTS

U SEE NOW??

When we change Concentration, make it higher for example, see how WE ARE CHANGING ITS CONC, tehrefore when LE CHATELIER PREDICTS IT WILL LOWER THE CONC OF THAT, IT IS JUST RE-ESTABLISHING THE RATIO THAT WAS THERE B4

When we change volume or pressure (inversely proportional), say increase pressure, we INCREASE THE CONC, as there is same amount in samller space, tehrefore WE HAVE CHANGED TEH CONC AND MADE IT HIGHER, thus by readjusting and shiting left, THE EQ IS REESTABOLISHING THE RATIO OF BEFORE - THE K THAT WAS TEHRE BEFORE

When we change temp, however, WE ARENT CHANGING THE CONC OF ANY SUBSTANCES, BUT ****AND THIS IS TEH CRUCIAL PART**** THE EQUILIBRIUM WILL SHIFT TO MINIMISE THIS CHANGE, AND THUS IF IT IS EXO AND WE INCREASE TEMP, IT WILL INCREASE THE CONC OF REACTANTS,A ND THUS K WILL DECREASE

SEEE???....WITH ALL THE OTHER ONES, WE INCREASED TEH CONC, THEN THE EQ DECREASED BY SHIFTING, THUS K STAYED THE SAME

BUT WITH TEMPERATURE, WE DIDNT INCREASE TEH CONC, BUT THE EQ SHIFTED ANYWAY, THUS K CHANGED

dunno fi that amde sense - so true the words of stoppard "we are tied down in a language that amkes up in obscurity wat it lacks in style"

/.

[Pwnage101]

lol i do get a tad excited dont i

[minijumbuk]

I disagree.
2NO + O₂ <==> 2NO₂

So K =
[NO₂]²
[O₂][NO]²

At equilibrium when the volume was 1.00 L, there were 0.02 mol of O₂, 0.05 mol of NO, and 0.2 mol of NO₂.

The K value would be 800.

When we increase the pressure (by decreasing volume to, say, 0.5L or something), then regardless of what happens mathematically, [NO₂] will increase, and reactants would decrease, since equilibrium shifts to the right. Correct?

Okay. Now, in the K expression,
[NO₂]²
[O₂][NO]²
If [NO₂]² increases, and [O₂][NO]² decreases, then if you simply substitute the new values of the equilibrium concentrations in, then the K value would mathematically be greater than 800. Is this not a "change in value of K"?

Unless you're trying to argue that adding pressure won't change the concentration of any substances, then of course it is a change in "K". But this would be a pressure dependent K, and as I said before, we don't learn this at HSC level.

[JasonNg1025]

I think that K probably changes both the top and the bottom...
Meaning the ratio doesn't change.
You're right in that [NO₂] takes a little leap due to Le Chatelier's principle, but you gotta remember that adding pressure inevitably increases concentration to all substances. Because it's much higher in the reactants (initially), Le Chatelier's ghost comes to make the product yield higher. Why did he do this? Because initially, the reactant concentration is much higher. Now, it's even. Or, as even as it was before.

[minijumbuk]

Yes, pressure increases the concentration of all substances.
It increases the products, in this example, more. There're less moles of reactants, and more moles of products. Why would this not change the ratio?

Let's use an example that's simpler to use:

N₂O_2(g)<==> 2NO_(g)
_______________________
Let's say there is 1 mole of reactant and 5 moles of product when it was at equilibrium. Then, pressure is increased by reducing the 1.00 L volume by half.
At 1.00 L:
Ratio of concentration of reactants to products = 1/1 : 5/1
= 1:5
_______________________
At 0.50 L:
What you're saying:
Ratio of moles of reactants to products = 1 : 5
= 1:5
Ratio of concentration of reactants to products = 1/0.5 : 1/0.5
= 1:5
So the ratio doesn't change.

So ratio didn't change.
_______________________
What I'm saying:
Ratio of moles of reactants to products = (1+x) : (5-2x)
= (1+x) : (5-2x)

Ratio of concentration of reactants to products = (1+x)/0.5 : (5-2x)/0.5
= (1+x) : (5-2x)
where x is the amount gained/lost after increasing pressure (by Le Chatelier's principle)
It would be ridiculous to say that the value of x is 0, as Le Chatelier's principle clearly indicates so.

So, for 0 < x < 1 (cannot have all of the 1 mole of N₂O₂ decomposing in an equilibrium reaction), the ratio becomes:
1.something : 4.something
= 1: something < 5
_______________________
Is that not a change in ratio? From 1:5 to 1<5) as a result of an increase in pressure?

If the ratio is changed, then K is changed. Correct?

[JasonNg1025]

What I'm saying is that before Le Chatelier's principle comes into play, the reactant concentration is already higher than normal. This is the reason for Le Chatelier's principle - to shift equilibrium to counteract some external change. It does not change it from the initial ratio, rather, it reverts it back to the initial ratio which was changed as a result of the pressure change.