Confused about which values Keq refers to

[deleted647690]

There is a question about equilibrium in TBR in a passage that has a table showing increasing temperature with the increasing Kp value for a reaction. The question asks, according to the table, the reaction is which?
a.Exothermic as written
b.Endothermic as written
c.adiabatic
d.Isothermal


The reaction is written as 2CO + O2 --->2CO2

I chose exothermic as written. My reasoning was that, as temperature increases, you are increasing heat. The Kp value is products / reactants, and if that is increasing, that must mean you are increasing the products. Therefore, since heat is increasing, that means that heat is on the products side, so it is an exothermic reaction..


However, the answer was endothermic. "Because the value of Keq increases as the temperature increases, products increase upon the addition of heat to the reaction."

I'm just getting really confused about what Keq is referring to. I thought temperature would appear on the products side, and thus that is why the temperature is increasing along with increasing Keq




So is my error in thinking that heat is not included in the Keq expression? Rather, the heat is affecting Keq but not written in the actual expression?




Also, I guess a misunderstanding of what Keq is exactly caused me to also miss this other question. "Addition of Co2 to the equilibrium mixture results in which of the following?" Refers to the same reaction as above.
A. Decrease in moles of O2
B. Decrease in moles of CO
C. Increase in Kp value
D.No change in value of Kp



I said C. I know that the only thing that changes Keq is temperature. But isn't Keq = [products] / [reactants]? Thus changing the equilibrium concentration of the products would cause the keq value to increase?

The correct answer here was no change. I don't understand how changing the values in the expression would have no effect on the Keq.

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[aldol16]

For the first question, it's endothermic because of the following. First, raising the temperature increases Kp, which means that as the temperature goes up, you have more products forming. Put another way, putting heat into the system drives the system to the right. So which side of the equation does heat have to be on? Think about Le Chatelier's principle here.

For the second question, the equilibrium constant only varies with temperature, not with how many moles you have in the system. If you add stress to the system in terms of more reactants or products isothermally, the system will rearrange itself according to Le Chatelier's principle until the Keq (constant) is reached again. So in this case, putting in more CO2 would drive the reaction to the left until the Keq constant is reached again. So CO and O2 would increase, not decrease. So A and B are out. C is out because you know that equilibrium constant depends only on temperature.

 

[deleted647690]

For the first question, it's endothermic because of the following. First, raising the temperature increases Kp, which means that as the temperature goes up, you have more products forming. Put another way, putting heat into the system drives the system to the right. So which side of the equation does heat have to be on? Think about Le Chatelier's principle here.

For the second question, the equilibrium constant only varies with temperature, not with how many moles you have in the system. If you add stress to the system in terms of more reactants or products isothermally, the system will rearrange itself according to Le Chatelier's principle until the Keq (constant) is reached again. So in this case, putting in more CO2 would drive the reaction to the left until the Keq constant is reached again. So CO and O2 would increase, not decrease. So A and B are out. C is out because you know that equilibrium constant depends only on temperature.

For the second question, when they say "addition of CO2 to the equilibrium mixture" I take that as meaning they are changing the equilibrium concentration. So that would mean the values that go into the Keq expression are different. So what they mean here is that equilibrium has already been set and they are adding an additional stress to the system?

 

[sazerac]

For the second question, when they say "addition of CO2 to the equilibrium mixture" I take that as meaning they are changing the equilibrium concentration. So that would mean the values that go into the Keq expression are different. So what they mean here is that equilibrium has already been set and they are adding an additional stress to the system?

Be careful with your reasoning here. Equilibrium implies a steady state after sufficient time has passed. So the system was at equilibrium, then extra CO2 was added. At this instant, the system wasn't in equilibrium anymore. Then the system migrated to equilibrium again, and when it was finally there the concentration (pressure) of products over reactants was the same constant it was before the stress was added.

 

[aldol16]

For the second question, when they say "addition of CO2 to the equilibrium mixture" I take that as meaning they are changing the equilibrium concentration. So that would mean the values that go into the Keq expression are different. So what they mean here is that equilibrium has already been set and they are adding an additional stress to the system?

Your misunderstanding stems from the fact that the "equilibrium mixture" is determined solely by temperature and there is nothing they can do about it - they can't change it arbitrarily because it's set by temperature.