TBR Chem Ex 3.9

[popchap]

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Q: At 92.2 C, the Kp for the following reaction is 0.2000 atm^-1. If you were to place exactly 0.200 atm of N2O4 (g) into a 1.00 L vessel, what would the partial pressure of NO2 (g) be once equilibrium was established?

2NO2 (g) --><-- N2O4 (g)


I'm having trouble understanding the part about partial pressure. The solution says that half of 0.200 atm of N2O4 shifts over then the partial pressure of NO2 is 0.200 atm.

I understand the calculation part but I would like to understand the process TBR used without calculating. I'm not understanding how the partial pressure of NO2 is 0.200 atm. Can someone please elaborate on this part? Thanks!

 

[Deadlifts]

Q: At 92.2 C, the Kp for the following reaction is 0.2000 atm^-1. If you were to place exactly 0.200 atm of N2O4 (g) into a 1.00 L vessel, what would the partial pressure of NO2 (g) be once equilibrium was established?

2NO2 (g) --><-- N2O4 (g)


I'm having trouble understanding the part about partial pressure. The solution says that half of 0.200 atm of N2O4 shifts over then the partial pressure of NO2 is 0.200 atm.

I understand the calculation part but I would like to understand the process TBR used without calculating. I'm not understanding how the partial pressure of NO2 is 0.200 atm. Can someone please elaborate on this part? Thanks!

Kp = Pn2o4 / Pno2^2

1/0.2 = 0.2 / Pno2^2

Pno2 = sqrt(0.2*0.2) = 0.2 atm

Where are you having trouble?

 

[popchap]

The explanation goes "Keq is less than 1.0, and the reaction starts with all products. The value of x is going to be significant (more than half shifts over). If half of the 0.200 atm of N2O4 shifts over, then the partial pressure of NO2 is 0.200 atm. Considering that more than half of the N2O4 (g) is going to shift, the value of NO2(g) is greater than 0.200 atm. However, not all of the N2O4 (g) can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm"

I might have missed something but I'm a bit confused on what this is saying.

 

[gettheleadout]

Kp = Pn2o4 / Pno2^2

1/0.2 = 0.2 / Pno2^2

Pno2 = sqrt(0.2*0.2) = 0.2 atm

Where are you having trouble?

Why did you set Kp equal to 1 over .2 when it's given as .2?

 

[popchap]

Why did you set Kp equal to 1 over .2 when it's given as .2?

It's given as 0.2000 atm^-1.

 

[gettheleadout]

It's given as 0.2000 atm^-1.

Right. 1/.2 is equal to 5, not .2.

The units are inverse atmospheres because the denominator contains a squared term, while the numerator contains only one atm term. The units of the equilibrium constant don't allow you to invert the value.

 

[popchap]

Right. 1/.2 is equal to 5, not .2.

The units are inverse atmospheres because the denominator contains a squared term, while the numerator contains only one atm term. The units of the equilibrium constant don't allow you to invert the value.

Well, Deadlifts never equaled 1/0.2 as 5. It's just inverted because it's 0.200 atm-1 which is 1/0.2.

Kp = Pn2o4 / Pno2^2

1/0.2 = 0.2 / Pno2^2

Pno2 = sqrt(0.2*0.2) = 0.2 atm

In the calculation that he did, he just cross multiplied to have Pno2 as the numerator.

I get that calculation part, I just don't understand what TBR is saying about this: "Keq is less than 1.0, and the reaction starts with all products. The value of x is going to be significant (more than half shifts over). If half of the 0.200 atm of N2O4 shifts over, then the partial pressure of NO2 is 0.200 atm. Considering that more than half of the N2O4 (g) is going to shift, the value of NO2(g) is greater than 0.200 atm. However, not all of the N2O4 (g) can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm."

Anyone explain that in simpler terms? I'm not so savvy with chem to begin with.

 

[gettheleadout]

Well, Deadlifts never equaled 1/0.2 as 5. It's just inverted because it's 0.200 atm-1 which is 1/0.2.

Kp = Pn2o4 / Pno2^2

1/0.2 = 0.2 / Pno2^2

Pno2 = sqrt(0.2*0.2) = 0.2 atm

In the calculation that he did, he just cross multiplied to have Pno2 as the numerator.

I get that calculation part, I just don't understand what TBR is saying about this: "Keq is less than 1.0, and the reaction starts with all products. The value of x is going to be significant (more than half shifts over). If half of the 0.200 atm of N2O4 shifts over, then the partial pressure of NO2 is 0.200 atm. Considering that more than half of the N2O4 (g) is going to shift, the value of NO2(g) is greater than 0.200 atm. However, not all of the N2O4 (g) can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm."

Anyone explain that in simpler terms? I'm not so savvy with chem to begin with.

I'm not sure how I can make it any more clear that deadlifts is committing (and you are accepting) a critical arithmetic error here, considering that 1) it's very basic math and I've stated it very clearly, and 2) you agree with him despite the fact that the solution contradicts the answer his method gives you.

The answer is .350 atm NO2, not .200. So whatever, as I stated in the thread I've elaborated on this question in before, TBR's reasoning here is quite complicated, and IMO inexplicable, but that is wholly beyond what's important if you can't manipulate the numbers correctly to begin with.

 

[popchap]

Ok, I get what you are saying about the calculation. My original thought was that the partial pressure that he got in the equation was if all the N2O4 shifts over then it would be 0.2 atm for each NO2. Since there are 2 moles of NO2, it would be 0.4 atm. However, since not all the N2O4 shifts over, the partial pressure for each NO2 would be less than 0.2. That's why I thought it would make sense for 0.2000 atm ^-1 to be 1/0.2 but later in the explanation, TBR says that Keq is less than 1.0 but 1/0.2 is 5 as you mentioned.

 

[Patassa]

Ok, I get what you are saying about the calculation. My original thought was that the partial pressure that he got in the equation was if all the N2O4 shifts over then it would be 0.2 atm for each NO2. Since there are 2 moles of NO2, it would be 0.4 atm. However, since not all the N2O4 shifts over, the partial pressure for each NO2 would be less than 0.2. That's why I thought it would make sense for 0.2000 atm ^-1 to be 1/0.2 but later in the explanation, TBR says that Keq is less than 1.0 but 1/0.2 is 5 as you mentioned.

I think you're missing the difference between .2 ^-1 atm vs .2 atm^-1 , those are not the same thing. Very critical that you get that, as LeadOut is point out.

 

[popchap]

Thanks for pointing that out guys. I'll be keeping that in mind when doing these problems.