The change in enthalpy for the forward reaction is -91KJ/mol
CO(g) + 2H2(g) <-> CH3OH(g)
1. The forward reaction is?
A. Endothermic
B. Exothermic **
You're right. delta H is negative means exothermic; delta H is positive means endothermic. Let me rewrite the equation to show that.
CO(g) + 2H2(g) <-> CH3OH(g) + heat
EXOTHERMIC so heat is given off so I've added + heat to show that.
2. If the reaction is at equilibrium and then was heated _____ CH3OH would be present after the reaction adjusted to the new temperature
A.more
B.less
C.same amount
Here is a succinct statement about Le Chatelier's Principle (which is what this question is all about). It won't win any Pulitizer Prizes but students understand it better than the formal textbook statement. "When a system in equilibrium is subjected to a stress it will shift in such a way as to undo what we did to it." So here is the equilibrium. CO(g) + 2H2(g) <-> CH3OH(g) + heat
So if we ADD HEAT the system will shift to use up the added heat. How can it do that? It can cause CH3OH to be heated and produce CO and H2. That is it will reverse the direction and go backwards (shift to the left) in order to use up the heat that was added. So you have CH3OH being used up (making it less) and it produces MORE CO and MORE H2 than was there when it was first equilibrated. Notice that the AMOUNT of CO is increased, the AMOUNT of H2 is increased and the AMOUNT of CH3OH is decreased.
3. If the reaction was at equilibrium and then the pressure in its container was increased_____CH3OH would be present after the reaction adjusted to the new pressure
See below
A.more
B.less
C.same amount
4. If the reaction was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts
A.more
B.less
C.same amount
5. If reaction was at equilibrium and then H2 was added, ____CO would be present after the reaction adjusted
A.more
B.less
C.same amount
(I’m very confused and my teacher doesn’t help me)
______________
For question 1. I got the same answer. I think you meant question 2. Question 1 was endothermic or exothermic.
I am confused because it says “after the reaction adjusts” so I thought when the reaction goes back to equilibrium it would all be the same amount again? I’m not sure I’m still confused. You're partly right. Yes it shifts to the left and produces less CH3OH and leaves more CO and H2 but OF COURSE it changes the amout as I discussed above.
Let me try question 3.
If the pressure increases then it would shift towards the side where less molecules are. Therefore it would shift to the right. I just don’t know how to tell if there will be more, less or same amount.
Since the pressure is upsetting the equilibrium and shifting to the right. There would be “more” CH3OH(g).?
Sure you do. You have the shift correct. If it shifts to the right you KNOW more CH3OH is produced and you KNOW CO and H2 are decreased. Yes, the amounts change. They MUST change. Why? Because Kc (or Kp) is a constant. A constant is a constant. In this case Kp = pCH3OH/pCO*p^2 H2. If you upset the equilibrium by changing the pressure and it shifts to the right, you KNOW the values for CO, H2 and CH3OH must change so that Kp, a constant, remains Kp, a constant. You can change pressure, volume, concentration and temperature. Changing T PROBABLY will change Kp/Kc but in all of the others Kp/Kd remains the original Kp and does not change. P and V will not cause the reaction to shift if moles on left = moles on the right.
Be sure and follow up if any of this is unclear. By the way, thank you so much for showing your confusion. It helped me answer because it immediately told me the problem you were having. I think you had Le Chatelier's Principle well in hand; however, you though the values for pressure or the values for cocentration or moles did not change. They do. Rmember, Kp or Kc is a constant and it reamins the same EXCEPT when changing the temperature. That will change Kp or Kc MOST of the time.
Question 4. If H2 was added how much CH3OH?
If we add H2 it would shift to the right. Therefore they would be the same amount because it is taking H2 to make more CH3OH?
Yes it will shift to the right because it wants use up the added H2 gas so now you will have more CH3OH and less H2.
Question 5. If H2 was added.
If we add H2 it would shift to the right to try and make up that. So therefore, there would be less CO because it is getting taken to make up for CH3OH
You're right. You add H2, the reaction shifts to the right, the amout of CH3OH increases and the amont of CO decreases.
The change in enthalpy for the forward reaction is -91KJ/mol
CO(g) + 2H2(g) <-> CH3OH(g)
1. The forward reaction is?
A. Endothermic
B. Exothermic **
2. If the reaction is at equilibrium and then was heated _____ CH3OH would be present after the reaction adjusted to the new temperature
A.more
B.less
C.same amount
3. If the reaction was at equilibrium and then the pressure in its container was increased_____CH3OH would be present after the reaction adjusted to the new pressure
A.more
B.less
C.same amount
4. If the reaction was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts
A.more
B.less
C.same amount
5. If reaction was at equilibrium and then H2 was added, ____CO would be present after the reaction adjusted
A.more
B.less
C.same amount
(I’m very confused and my teacher doesn’t help me)
______________
For question 1. I got the same answer.
I am confused because it says “after the reaction adjusts” so I thought when the reaction goes back to equilibrium it would all be the same amount again? I’m not sure I’m still confused.
Let me try question 3.
If the pressure increases then it would shift towards the side where less molecules are. Therefore it would shift to the right. I just don’t know how to tell if there will be more, less or same amount.
Since the pressure is upsetting the equilibrium and shifting to the right. There would be “more” CH3OH(g).?
Question 4. If H2 was added how much CH3OH?
If we add H2 it would shift to the right. Therefore they would be the same amount because it is taking H2 to make more CH3OH?
Question 5. If H2 was added.
If we add H2 it would shift to the right to try and make up that. So therefore, there would be less CO because it is getting taken to make up for CH3OH
1 answer
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