[redpepper007]

Hello, geniuses!

I am having a problem in chemistry...



What is this systems equilibrium constant expression? And under what conditions (temperature, pressure, concentrations) you can most fully get rid of SO2?

I really have no idea with what to start... :/

[igloo5080]

The expression for K_p is the standard one (remember to include only gases)

You can either use the expression for K_p or, more straightforwardly, Le Chatelier's Principle to deduce the effect of pressure and changes in concentration of each of the components on the position of equilibrium, which in your case needs to be as far over to the right as possible.

As for temperature it is probably most straightforward using Le Chatelier's Principle, realizing that this reaction has a negative enthalpy change of reaction.

[redpepper007]

Thanks for your reply

But I didnt understand anything... maybe you can just simply solve this? (Im really bad at chemistry)

[skbuncks]

Thanks for your reply

But I didnt understand anything... maybe you can just simply solve this? (Im really bad at chemistry)

You don't need to be good at chemistry to understand Le Chatlier's principle. The wiki page on this is really quite good, in particular for the problem at hand the section on temperature

skb

[redpepper007]

You don't need to be good at chemistry to understand Le Chatlier's principle. The wiki page on this is really quite good, in particular for the problem at hand the section on temperature

skb

Well, I already figured out how to write systems equilibrium constant expression. But Le Chatlier's principle doesnt seem to be able to calculate exact temperature, pressure and concentration, to remove SO2 from equatation...

No hope for me, right?

[skbuncks]

Is the question asking for exact conditions?

What effect do you think decreasing the reaction temperature would have over increasing the reaction temperature?

What effect do you think decreasing the concentration of H2S would have over increasing it?

What effect do you think increasing the reaction volume (and thus decreasing the pressure) would have over decreasing the reaction volume (and thus increasing the pressure)?

skb

[redpepper007]

Is the question asking for exact conditions?

What effect do you think decreasing the reaction temperature would have over increasing the reaction temperature?

What effect do you think decreasing the concentration of H2S would have over increasing it?

What effect do you think increasing the reaction volume (and thus decreasing the pressure) would have over decreasing the reaction volume (and thus increasing the pressure)?

skb

Actually I dont understand what should I calculate... I typed task text in my first post. I need to get that equilibrium to the right side in order to get rid of SO2, is that correct? Then, refering to Le Chatlier, answer will be just a text, and no exact numbers?

[skbuncks]

If the information you gave in the OP is all that you have been given then no, you will not have to calculate anything. Your answer will be just text and based on Le Chatelier's principles.
Answering the questions I posed in my last post should set you on your way.

skb

[redpepper007]

If the information you gave in the OP is all that you have been given then no, you will not have to calculate anything. Your answer will be just text and based on Le Chatelier's principles.
Answering the questions I posed in my last post should set you on your way.

skb

Yes, thats all i've been given. But it still gives me absolutely no clarification of what to do, how do I know to which side I need to move the balance in order to get rid of SO2?

[skbuncks]

The wiki page I linked to will tell you all you need to know.

Firstly take temperature.

1) add temp to the equation on the side of the products

SO2 + H2S = 3S + H2O + heat

If you remove heat from the system then it will compensate by generating more heat, thus shifting the equilibrium to the right. Therefore should you run the reaction hot or cold?

skb

[redpepper007]

The wiki page I linked to will tell you all you need to know.

Firstly take temperature.

1) add temp to the equation on the side of the products

SO2 + H2S = 3S + H2O + heat

If you remove heat from the system then it will compensate by generating more heat, thus shifting the equilibrium to the right. Therefore should you run the reaction hot or cold?

skb

Ohh.... I think ill get mad trying to figure out chemistry. Deem, why didnt I learn it while still at school...

Well, to remove heat I should run the reaction cold, decrease temperature. Then balance will shift to right side, but does this mean that the amount of SO2 will be removed?

And I studied that wiki page, I have the same thing in my notes in my native language, but I still dont get it...

[redpepper007]

skbuncks, please... 

[skbuncks]

The wiki page I linked to will tell you all you need to know.

Firstly take temperature.

1) add temp to the equation on the side of the products

SO2 + H2S = 3S + H2O + heat

If you remove heat from the system then it will compensate by generating more heat, thus shifting the equilibrium to the right. Therefore should you run the reaction hot or cold?

skb

Ohh.... I think ill get mad trying to figure out chemistry. Deem, why didnt I learn it while still at school...

Well, to remove heat I should run the reaction cold, decrease temperature. Then balance will shift to right side, but does this mean that the amount of SO2 will be removed?

And I studied that wiki page, I have the same thing in my notes in my native language, but I still dont get it...

That's right yes. By taking heat out of the system (ie running it cold) it will shift the equilibrium to the right so as to generate more heat. By shifting the equilibrium to the right  the amount of SO2 remaining will be reduced.

If you now move onto concentration of H2S, what effect do you think increasing this will have? If you consider that there will be more H2S available to react with the SO2, ie more collisions between molecules, will more or less SO2 react? Will it shift the equilibrium to the right or the left?

skb

[redpepper007]

Ohh.... I think ill get mad trying to figure out chemistry. Deem, why didnt I learn it while still at school...

Well, to remove heat I should run the reaction cold, decrease temperature. Then balance will shift to right side, but does this mean that the amount of SO2 will be removed?

And I studied that wiki page, I have the same thing in my notes in my native language, but I still dont get it...

That's right yes. By taking heat out of the system (ie running it cold) it will shift the equilibrium to the right so as to generate more heat. By shifting the equilibrium to the right  the amount of SO2 remaining will be reduced.

If you now move onto concentration of H2S, what effect do you think increasing this will have? If you consider that there will be more H2S available to react with the SO2, ie more collisions between molecules, will more or less SO2 react? Will it shift the equilibrium to the right or the left?

skb

Well, if I add more H2S, it will use all available SO2, and there wont even be enough of SO2... So balance will shift to the left side, right?

BTW thanks that you are so patient

[skbuncks]

Well, if I add more H2S, it will use all available SO2, and there wont even be enough of SO2... So balance will shift to the left side, right? Smiley

BTW thanks that you are so patient Smiley

Almost. By adding more H2S you will have a large excess of H2S, the equilibrium will shift to the RIGHT so as to reduce the excess. A consequence of this is that more of the SO2 will be consumed thus increasing the conversion and yield.
wiki puts it quite succinctly

If we are to add a species to the overall reaction, the reaction will favour the side opposing the addition of the species

So, add H2S to the left and the reaction will shift to right, conversely if you were to add SO2 to the system the equilibrium would shift to the left.

Now, on to pressure/volume...

skb