Le Chatelier's Principle will answer all of the "shifting" questions. For example,
A, increasing heat means the rxn must shift so as to relieve the heat. That means the rxn will shift to the left (too much heat on the right makes it go to the left to use the heat).
B, increased heat means too much heat on the left so it will shift to the right to use the heat.
For pressure increases, the reaction will shift to the side with the smaller number of moles. For example, increasing pressure for reaction A will move the reaction to the left (2 moles on the left and 3 on the right). Etc.
I have the 1st part of the answer to my question I guess what I don't understand is the 2nd part of the question. I am supposed to 1. predict how ke increases in temperature, 2. predict how equilibrium will shift when pressure is decreased and 3. predict how equilibrium will shift when the concentration of the substance with the asterisk is in creased, 4. and predict how equlibrium will shift when the temperature is decreased. How do you figure this out because doing the math calculation was easier.
Here are the 3 problems;
A) 2✴O3(g) ↔ 3O2(g) + heat
[6.0 x 10-1] [0.21]
B) 2CO2(g) + heat ↔ 2✴CO(g) + O2(g)
[0.103] [0.024] [1.18 X 10-2]
C) NO2(g) + O2(g) + heat ↔ ✴NO(g) + O3(g)
[0.072] [0.083] [6.73 X 10-2] [6.73X10-2]
4 answers
Thanx Dr. Bob I get it a little guess I will have to call my professor.
Le Chatelier's Principle says, "When a reaction in equilibrium is subjected to a stress, the reaction will shift in such a way as to relieve the stress."
So for 2O3 ==> 3O2 + heat.
Just remember the reaction tries to undo what we did.
Adding O3 will shift the rxn to the right (to use up the extra O3 that was added). Adding O2 will shift the reaction to the left (to use up the extra O2 that was added).
Adding heat will shift it to the left (to use the heat that was added).
Increasing pressure will shift the reaction to the left (because when pressure is increased the reaction wants to occupy the smaller volume and there are fewer moles on the left).
Just think of it this way. Let's assume I am relatively normal and "at equilibrium." Suddenly I get a phone call saying I have won a million dollars. That represents a stress to my system and I will react so as to relieve the stress. How you say? If I suddenly have too much money I will try to get rid of it so as to return to equilibrium. So I go out and spend spend spend. When all of the money is spent (at least most of it), I will again be at equilibrium. So adding a stress (extra money) and my system reacts to get rid of the extra money. I hope this helps.
So for 2O3 ==> 3O2 + heat.
Just remember the reaction tries to undo what we did.
Adding O3 will shift the rxn to the right (to use up the extra O3 that was added). Adding O2 will shift the reaction to the left (to use up the extra O2 that was added).
Adding heat will shift it to the left (to use the heat that was added).
Increasing pressure will shift the reaction to the left (because when pressure is increased the reaction wants to occupy the smaller volume and there are fewer moles on the left).
Just think of it this way. Let's assume I am relatively normal and "at equilibrium." Suddenly I get a phone call saying I have won a million dollars. That represents a stress to my system and I will react so as to relieve the stress. How you say? If I suddenly have too much money I will try to get rid of it so as to return to equilibrium. So I go out and spend spend spend. When all of the money is spent (at least most of it), I will again be at equilibrium. So adding a stress (extra money) and my system reacts to get rid of the extra money. I hope this helps.
A great deal! Thanks
1 answer