EK GChem Help!!!!!!

[ScratchMan]

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Please view the attached question.

The book mentioned the answer is A. But, I dont agree with that.. but I feels I miss some important concept...

My reasoning is like followings.

@equilibrium, Keq = [CO2] since others are all solid. Thus, we can say Kp also should be same as partial pressure of CO2.

A. Kp is less than the partial pressure of CO2.
>> @eq. Kp = Pco2. If Kp < Pco2, the rxn shift to the left side. Even if we removed the beaker II, we still have the beaker I thus the beaker I will provide some CaO(s) and CO2(g). Therefore, the choice A should be possible true statement.

B. Kp is greater than the partial pressure of CO2.
>> @eq. Kp = Pco2. If Kp > Pco2, the rxn shift ot the right side. Even if we removed the beaker II, No problem since we have CaCO3(s) from the beaker I.
It is possible true statement.

C. Kp is equal to the partial pressure of CO2.
>> Obviously true statement.

D. Equilibrium could not be achieved under any conditions because solid CaO is required to achieve equilibrium.
>> Even if the beaker II is removed, there is still beaker I which provides CaCO3(s) <--> CaO(s) + CO2(g). Thus, equilibrium could be achieved.
It is false statement!!

Therefore, my answer is D. (the answer of the book is A.)

Please address me what I miss something critical concept or misunderstood if the book answer is right.

Thanks.

 

[ScratchMan]

Does anyone have an idea to resolve this question???

 

[chromosome21]

I agree with the answer. It should be choice A.

Understand the equilibrium.

here, Kp = Pco2 is our ideal equilibrium condition.

When Kp < Pco2, rxn will shift to left, means towards the reactants side.
when Kp = Pco2, rxn is in equilibrim
when Kp > Pco2, rxn will shift to the right, meaning towards the products side.

Now,
imp part of the que: IF beaker II is removed, in which condition eui. will NOT be achieved.
choice A: Kp is less than pp of of CO2

IF Kp is LESS thaan Pco2, means the reaction MUST shift to the left towards the reactants, but here, we don't have the material to make this shift happen since we REMOVED the beaker II for this to be true! Since we don't have CaO (beaker II), we can't make more CaCO3. There is no forward reaction is happening here. YOur thinking of CaCO3 to make CaO+CO2 is not right, b/c the reaction can't go forward since Kp is LESS than Pco2. It means we don't have CaO at all! And therefore, the equilibrium cannot be reached.

makes sense?

 

[addis4ever]

I would go with C as well because since CaO and CaCO3 are both solids we don't count them in the equilibrium.

 

[ScratchMan]

I agree with the answer. It should be choice A.

Understand the equilibrium.

here, Kp = Pco2 is our ideal equilibrium condition.

When Kp < Pco2, rxn will shift to left, means towards the reactants side.
when Kp = Pco2, rxn is in equilibrim
when Kp > Pco2, rxn will shift to the right, meaning towards the products side.

Now,
imp part of the que: IF beaker II is removed, in which condition eui. will NOT be achieved.
choice A: Kp is less than pp of of CO2

IF Kp is LESS thaan Pco2, means the reaction MUST shift to the left towards the reactants, but here, we don't have the material to make this shift happen since we REMOVED the beaker II for this to be true! Since we don't have CaO (beaker II), we can't make more CaCO3.👍 There is no forward reaction is happening here. YOur thinking of CaCO3 to make CaO+CO2 is not right, b/c the reaction can't go forward since Kp is LESS than Pco2. It means we don't have CaO at all! And therefore, the equilibrium cannot be reached.

makes sense?

Thanks for reply. 🙂

The key point is that we do not have CaO(s) thus the reaction cannot do reverse reaction. Therefore, the choice D should be true statement. Make sense!!!

Choice B: Ksp > Pco2 means that the reaction shift to right side. It could be happened since we have the beaker I. As long as we have CaCO3, it can have forward rxn. thus, it is possible.
Choice C: Ksp = Pco2 means that the reaction reaches to equilibrium status. It could be happened since we do not consider any SOLID thus removing Beaker II does not matter.

Choice A: Ksp < Pco2. Under this condition, the reaction is willing to shift to left side, but it cannot be happened since we do not have CaO(s). It means that it cant reach to equilibrium state.

Thanks!!!

 

[chromosome21]

Thanks for reply. 🙂

The key point is that we do not have CaO(s) thus the reaction cannot do reverse reaction. Therefore, the choice D should be true statement. Make sense!!!

Choice B: Ksp > Pco2 means that the reaction shift to right side. It could be happened since we have the beaker I. As long as we have CaCO3, it can have forward rxn. thus, it is possible.
Choice C: Ksp = Pco2 means that the reaction reaches to equilibrium status. It could be happened since we do not consider any SOLID thus removing Beaker II does not matter.

Choice A: Ksp < Pco2. Under this condition, the reaction is willing to shift to left side, but it cannot be happened since we do not have CaO(s). It means that it cant reach to equilibrium state.

Thanks!!!

Actually, choice D is a wrong statement b/c with choices B and C conditions, we could still achieve equilibrium. Choice A is the only condition where we can't able to achieve equilibrium b/c Kp < Pco2, so the shift will be towards the reactants and it can't happen b/c the material that is required has been removed.. hope this helps