TBR Gen Chem Example 3.9

[gkdubey]

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Alright guys so I have seen a couple other threads on this exact example, but unfortunately, none of them had any decisive answers or explanations on there and each one left me more confused than the last. So here goes again:

Okay this example makes no sense. TBR Example 3.9 Gen Chm Chapter 3 pg. 174

At 92.2C, the Kp for the following rxn is .2000 atm. If you were to place exactly 0.200atm of N2O4 into a 1.00 liter vessel, what would the partial pressure of NO2 be once equilibrium was established?

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

They solved this just by what was given by saying "if more than half of the 0.200atm of N2O4 shifts over, then the partial pressure of nitrogen dioxide is 0.200atm (How? is it because half would mean 0.100 would shift over and then you multiply it by two bc of the coefficient?). Considering that more than half of the N204(g) is going to shift (how did they know that?), the value of NO2(g) is greater than 0.200. However, not all of the N2O4 can shift over so the answer must be less than 0.400"


Can someone please tell me how they figured this out? I know you can compare Q values to K to figure out how a reaction will shift but how do you get the Q value? There is nothing given for the NO2 at the beginning so if you try to do reactants over products you just get an unknown in denominator.

I ALSO don't understand when x should be considered and when it should be ignored. The rules say things like if K and Q are similar and x is SMALL ignore but if K and Q are different and x is large then consider x. But Q often times if not all times is NOT given. So how do I determine this? I am confused.

Please help!

 

[sps27]

Alright guys so I have seen a couple other threads on this exact example, but unfortunately, none of them had any decisive answers or explanations on there and each one left me more confused than the last. So here goes again:

Okay this example makes no sense. TBR Example 3.9 Gen Chm Chapter 3 pg. 174

At 92.2C, the Kp for the following rxn is .2000 atm. If you were to place exactly 0.200atm of N2O4 into a 1.00 liter vessel, what would the partial pressure of NO2 be once equilibrium was established?

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

They solved this just by what was given by saying "if more than half of the 0.200atm of N2O4 shifts over, then the partial pressure of nitrogen dioxide is 0.200atm (How? is it because half would mean 0.100 would shift over and then you multiply it by two bc of the coefficient?). Considering that more than half of the N204(g) is going to shift (how did they know that?), the value of NO2(g) is greater than 0.200. However, not all of the N2O4 can shift over so the answer must be less than 0.400"


Can someone please tell me how they figured this out? I know you can compare Q values to K to figure out how a reaction will shift but how do you get the Q value? There is nothing given for the NO2 at the beginning so if you try to do reactants over products you just get an unknown in denominator.

I ALSO don't understand when x should be considered and when it should be ignored. The rules say things like if K and Q are similar and x is SMALL ignore but if K and Q are different and x is large then consider x. But Q often times if not all times is NOT given. So how do I determine this? I am confused.

Please help!

1) I read this example in TBR so it was a good revision for me. In the book at the bottom of the page, they have given an eqn like Keq = (0.2-x)/4x^^2 and then they plugged in the number 0.1 for 'x' and I think u understand why it is like that because the ans choices are actually 2x. So if you evaluate Keq = (0.2-0.1) / 4x(0.1)x(0.1) = 2.5. Now this value of Keq is larger than given in the problem which is 0.2. So the question is, how do you bring this value down. Obviously by increasing the denominator or increasing x. We assumed x to be 0.1 which means concentration of NO2 is 2(0.1) = 0.2. So in reality if we want to reach Keq = 0.2, we must have NO2 more than 0.2. The mathematical treatment of this problem helped me understand this better as opposed to stating that more than half of N2O4 should be converted which like you, I found a bit confusing as well.

2) I couldn't understand the relevance of K and Q to this problem. Regarding the 'x' issue as they have mentioned in the table, if the Keq >> 10e3 or K eq << 10e-3, then ignore x. So I use the Keq as a guide line rather than K and Q.

Not sure if any of this helps but I hope it does....

 

[gkdubey]

Thank you SO much. Your clarification along with a couple others on a different thread I posted this in has really helped me a lot. It makes much more sense now looking at it mathematically. Thanks again! I'm so glad we have a resource like this forum! I don't know what I would do without it!