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)
(I’m very confused and my teacher doesn’t help me).
I could help more if I knew what was confusing you. Some detail about what is causing the problem. As it is I'm flying in the blind. So I'll give the answer, reluctantly, but please follow up and tell me what is bugging you and how best I can help.
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 **
When delta H is - it means the reaction is "giving away" heat (that is, it has less heat than it had) and we could write the equation this way.
CO(g) + 2H2(g) <-> CH3OH(g) + heat
So if it is giving away heat and the reaction shows heat as a product you know it is heating up and that makes it exothermic. Think of it as exit heat; hence the word exothermic. Endothermic means "Inward heat"
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
Le Chatelier's Principle says "When we do something to a system in equilibrium, it will try to undo what we did to it." So the system gives off heat normally. If we heat it up, which upsets the equilibrium, the reaction will try to get rid of the heat we added. How can it do that? It can shift to the left. Going to the right produces heat; going to the left uses heat. If the rxn shifts to the left it will use up some of the CH3OH so less is the word that fits here.
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
Same principle. When increasing pressure, the system will shift to the side with fewer moles of gas.
4. If the reaction was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts
So you're adding H2. Guess what? You're adding H2 so the system will shift to get rid of H2. It will shift which way?
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'll leave this for you. Let me know if you're still confused and if so explain fully why?Good Luck.
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
Need helpppppp
Dont listen to Dr.BOB
Let's break down and answer each question:
1. The change in enthalpy for the forward reaction is -91KJ/mol. The enthalpy change gives us information about the heat flow of a reaction. If the change in enthalpy is negative, as in this case, it means that heat is released during the reaction. According to the reaction equation given, CO(g) + 2H2(g) <-> CH3OH(g), the forward reaction is the formation of CH3OH(g) from CO(g) and H2(g). Since heat is released during this forward reaction, we can conclude that it is an exothermic reaction. Therefore, the correct answer is B. Exothermic.
2. If the reaction is at equilibrium and then it was heated, we need to consider the Le Chatelier's principle. When we heat a system at equilibrium, the system will try to counteract the increase in temperature by favoring the endothermic reaction. In this case, the forward reaction is exothermic, so the reverse reaction is endothermic. The system will shift towards the endothermic side to consume some of the excess heat. As a result, the concentration of CH3OH would decrease, and the correct answer is B. less.
3. If the pressure in the container is increased, we need to apply Le Chatelier's principle again. For this reaction, we can see that there are three moles of gas on the left side (CO and H2) and one mole of gas on the right side (CH3OH). According to Le Chatelier's principle, if we increase the pressure, the system will shift towards the side with fewer moles of gas to reduce the pressure. Therefore, in this case, the concentration of CH3OH would decrease, and the correct answer is B. less.
4. If H2 is added to the reaction at equilibrium, we can apply Le Chatelier's principle. Adding more of a reactant will cause the system to shift towards the products to reduce the concentration of the added reactant. In this case, adding more H2 would shift the reaction towards the formation of CH3OH. Therefore, the concentration of CH3OH would increase, and the correct answer is A. more.
5. If H2 is added to the reaction at equilibrium, we can again apply Le Chatelier's principle. Adding more of a reactant will cause the system to shift towards the products to reduce the concentration of the added reactant. In this case, adding more H2 would shift the reaction towards the formation of CH3OH, resulting in the consumption of CO. Therefore, the concentration of CO would decrease, and the correct answer is B. less.
I hope this explanation helps you understand the concepts better!