Equilibrium/LeChatelier's question

[LetsGo352]

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Consider the following EXOTHERMIC rxn:

2CO(g) + O2 (g) --> 2CO2 (g)

If a container holds only CO and O2, what effect will raising the temperature have on the forward rxn?

a.) forward rate will increase
b.) forward rate will remain the same
c.) forward rate will decrease
d.) The reaction will not go forward.



I'm reading the BR books for review and doing the EK 1001 questions as a supplement. This is from the EK book.

The BR book would say that C would be the answer because since it's an exothermic reaction and the temperature has increased, we would want to take away heat from the system, meaning the reaction would tend to go in the reverse direction. However, since there is initially no products, the best answer would be a decrease in the forward rate, right?

EK, on the other hand, said the answer was A, the forward reaction will increase because "Rates always increase when temperature is increased, even when the reaction is exothermic."

Which explanation, in your opinion, is correct?

Thanks.

 

[MLT2MT2DO]

2CO(g) + O2 (g) --> 2CO2 (g) + HEAT

In an exothermic reaction... now apply Le Chatlier's and Berkeley is correct.

I think you misunderstood EK, as I don't ever remember taking that away from the lecture book

 

[LetsGo352]

2CO(g) + O2 (g) --> 2CO2 (g) + HEAT

In an exothermic reaction... now apply Le Chatlier's and Berkeley is correct.

I think you misunderstood EK, as I don't ever remember taking that away from the lecture book

Well this was from EK 1001, not the lecture book. Your explanation is what I would agree with, but I quoted the answer right out of EK 1001, which doesn't make sense to me.

 

[MLT2MT2DO]

Ah, I haven't delved into the 1001 books YET, but the lecture book agrees with Berkeley. Can't say I'm surprised by an EK mistake.

 

[LetsGo352]

Ah, I haven't delved into the 1001 books YET, but the lecture book agrees with Berkeley. Can't say I'm surprised by an EK mistake.

Yeah, I'll just assume it's a mistake. Thanks.

 

[Geekchick921]

2CO(g) + O2 (g) --> 2CO2 (g) + HEAT

In an exothermic reaction... now apply Le Chatlier's and Berkeley is correct.

I think you misunderstood EK, as I don't ever remember taking that away from the lecture book

Agreed. I think what the OP is saying though is that, since increasing the temperature increases the speed of the particles and, therefore, the amount of collisions, the rate will increase. I can see where the confusion comes in.

I think it's kind of a crappy question.

Well, wait... is this a reversible reaction? If not, well, then A makes sense as the answer, but if there's an equilibrium between the forward and reverse reactions then C would make more sense.

 

[Econ2MD]

Consider the following EXOTHERMIC rxn:

2CO(g) + O2 (g) --> 2CO2 (g)

If a container holds only CO and O2, what effect will raising the temperature have on the forward rxn?

a.) forward rate will increase
b.) forward rate will remain the same
c.) forward rate will decrease
d.) The reaction will not go forward.



I'm reading the BR books for review and doing the EK 1001 questions as a supplement. This is from the EK book.

The BR book would say that C would be the answer because since it's an exothermic reaction and the temperature has increased, we would want to take away heat from the system, meaning the reaction would tend to go in the reverse direction. However, since there is initially no products, the best answer would be a decrease in the forward rate, right?

EK, on the other hand, said the answer was A, the forward reaction will increase because "Rates always increase when temperature is increased, even when the reaction is exothermic."

Which explanation, in your opinion, is correct?

Thanks.

In your reaction, increasing the temperature increases the pressure of gas in the system. There are three moles of reactant and two moles of product. Therefore, the reaction will favor product to reduce the pressure change. At least initally.

At equilibrium, additional heat will push the reaction toward reactants, since heat is a product of the reaction. I think Le Chatlier's principle is at work in both situations, but I agree that the question is kind of contradictory.

 

[AnimusRelic]

The way that I am reading this, I think the EK explanation makes sense. The question stem is asking, in general terms, "what is going on to the foward reaction if we increase temperature?"

True, the reaction is exothermic and if the question was asking about which direction is favored more( forward or reverse) - the answer would be the reverse. BUT if we examine the Arrenhius equation, the rate constant k for the forward AND reverse increases with increasing temperature (and other things such as decreasing Eact). Since the question is only asking about the forward reaction, (A) seems to be the best answer. The reverse reaction WOULD increase more - but the forward is still increasing regardless.

 

[loveoforganic]

The way that I am reading this, I think the EK explanation makes sense. The question stem is asking, in general terms, "what is going on to the foward reaction if we increase temperature?"

True, the reaction is exothermic and if the question was asking about which direction is favored more( forward or reverse) - the answer would be the reverse. BUT if we examine the Arrenhius equation, the rate constant k for the forward AND reverse increases with increasing temperature (and other things such as decreasing Eact). Since the question is only asking about the forward reaction, (A) seems to be the best answer. The reverse reaction WOULD increase more - but the forward is still increasing regardless.

I agree with this.

 

[Quinquagenarian]

I concur with MLT2MT2DO explanations. Given your knowledge of two reaction types proceed to set up the equations as below:

2CO(g) + O2 (g) --> 2CO2 (g) + HEAT } Exothermic (+Heat Favors Rev.rxn)

HEAT + 2CO(g) + O2 (g) --> 2CO2 (g) } Endothermic (+Heat Favors Fwd.rxn)

You should be good to go from here.

All the best. I am in the game too with March mcat.👍