TBR Section III Passage XII question 80

[orangetea]

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If reaction 1 is carried out in a closed piston system with an external pressure of 1 atm, what occurs when o.1 atm of NO gas is added to the system?

reaction 1: 2NO(g) + O2(g) -->(forward and reverse) 2NO(g)


a. the volume would decrease by more than ten percent
b. the volume would decrease by less than ten percent
c. the volume would increase by more than ten percent
d. the volume would increase by less than ten percent


My reasoning was that since it's increasing pressure the volume would decrease by less than ten percent since it's shifting.. but why is the answer C =/

 

[DrknoSDN]

reaction 1: 2NO(g) + O2(g) -->(forward and reverse) 2NO(g)/

I believe you want the second 2NO to be 2(NO2).
2 NO + O2 <--> 2 NO2

If you start adding NO on the left it will cause the reaction to shift right due to Le Chatelier's principle.
When this happens, every O2 molecule on the left that reacts forms 2 NO2 molecules on the right.
If you keep adding NO to the left until you get 0.1 more atms, you will have 10% more volume from the NO(g), but the conversion of O2 to 2x NO2 would cause an even greater increase in volume because 2 molecules of NO2 take up more space than 1 molecule of O2.

My reasoning was that since it's increasing pressure the volume would decrease by less than ten percent since it's shifting.. but why is the answer C =/

The pressure is not increasing. Possibly poor word choice but they are implying adding a volume equal to 0.1 atm causing volume to increase and pressure returns to equilibrium with the 1atm surroundings.

 

[orangetea]

I believe you want the second 2NO to be 2(NO2).
2 NO + O2 <--> 2 NO2

If you start adding NO on the left it will cause the reaction to shift right due to Le Chatelier's principle.
When this happens, every O2 molecule on the left that reacts forms 2 NO2 molecules on the right.
If you keep adding NO to the left until you get 0.1 more atms, you will have 10% more volume from the NO(g), but the conversion of O2 to 2x NO2 would cause an even greater increase in volume because 2 molecules of NO2 take up more space than 1 molecule of O2.


The pressure is not increasing. Possibly poor word choice but they are implying adding a volume equal to 0.1 atm causing volume to increase and pressure returns to equilibrium with the 1atm surroundings.

Thank you for your answer but if were adding more gas pressure is increasing which is why the piston has to increase the volume but that that's where I am confused because I thought pv=nrt .. or maybe I am not understanding that equation correctly.. i forgot all my chem

 

[DrknoSDN]

You are right that this follows ideal gas behavior but in this example the piston would either not move (constant volume and increased pressure), or it would equilibrate with the external pressure (increase volume) and return to 1 atm. I believe the latter is what is occurring based on the answer choices.

Again they are not increasing the pressure by 0.1 atm. They are adding enough gas molecules where if volume was held constant, pressure would increase by 0.1 atm, but here volume is not constant.

 

[MedHopeful20134]

If reaction 1 is carried out in a closed piston system with an external pressure of 1 atm, what occurs when o.1 atm of NO gas is added to the system?

reaction 1: 2NO(g) + O2(g) -->(forward and reverse) 2NO(g)


a. the volume would decrease by more than ten percent
b. the volume would decrease by less than ten percent
c. the volume would increase by more than ten percent
d. the volume would increase by less than ten percent


My reasoning was that since it's increasing pressure the volume would decrease by less than ten percent since it's shifting.. but why is the answer C =/

The answer is D here.
The volume would increase by less than 10 %.

 

[Czarcasm]

The answer is D here.
The volume would increase by less than 10 %.

Yeah, I'm interested to know why the answer wasn't D as well. There are two possible ways this system can re-establish equilibrium: increasing volume to alleviate the initial pressure increase or consuming reactants to produce products - the side with fewer moles (LeChatlier's principle). I believe in this instance, both could happen. To alleviate the pressure increase, the piston will rise (increasing volume). However, because some reactants will be consumed to products and because ultimately it's a 3 mole (reactants) to 2 mole (products) ratio, there is a net decrease in moles; this would imply a decrease in pressure which is directly proportional to moles gas. Had only volume changed proportionally to the pressure increase, volume would have to increase by 10%. In this instance, because some moles can react as well to alleviate the pressure increase, I would assume the net effect is an increase in volume but by less than 10% (choice D). In other words, converting some reactant to product alleviates the need to increase the volume all the way to 10%.

 

[DrknoSDN]

The question was copied down wrong. It actually reads:

80. If Reaction 1 is carried out in a closed piston system with an external pressure of 1 atm, what occurs when 0.1 atm of NO gas is added to the system?
2 NO(g) + 02(g) <--> 2 N02(g)
A. The volume decreases by more than ten percent.
B. The volume decreases by less than ten percent.
C. The volume increases by less than ten percent.
D. The volume increases by more than ten percent.

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Strangely the answer could be D if the system were in a dynamic equilibrium. The original post said volume would increase by more than 10 percent. A dynamic system would allow that to be true.

As you are adding NO gas to to increase the partial pressure by 0.1 atm, some of the NO would be converted in the forward reaction consuming O2 and forming 2 NO2. Thus, you would need to add more volume of NO gas than expected because some of the added NO would be lost to the forward reaction while you are adding it, until equilibrium is reached with the partial pressure of NO being 0.1 atm higher than it was originally.

Goes to show if you are given an answer it changes how to interpret a question. 😕