[Ilragazzo]

Hi my name is Tony and i generally enjoy chemistry, lately i have been having troubles with my 1st year inorganic chem.

Back ground.
I'm an engineering major thinking of mechanical, materials, or electrical engineering or something in between. I have had intro courses to chem and a refresher before this one, but I feel as if i"m missing something.

My problems lie in equilibrium and possibly in thermodynamic chapters.

I just generally have a hard time starting EQ problems because I find it hard to identify what type of problem it is. this is where my biggest problem lies.

Thermodynamics problems like bomb calorimetery problems give me trouble because i seem to get the magnitudes correct but the sign is wrong. So I don't understand why and when I need to use the negative signs.  Also I am having a hard time keeping track of what units things should be expressed in. Jk,Kj/mol,Delta Heat, Delta Q  and things of that nature.

I have roughly a week and a half before I am tested on this material. Could someone help clear up some of my mistakes and possibly run me through a few problems?

Thanks,
Tony. 

[Dan]

Could someone help clear up some of my mistakes and possibly run me through a few problems?

Post a question you are struggling with, showing your attempt at the question, and we can see where you're going wrong and guide you through the problem.

[Ilragazzo]

Nitrogen oxide is a pollutant in the lower atmosphere that irritates the eyes and lungs and leads to the formation of acid rain. Nitrogen oxide forms naturally in atmosphere according to the endothermic reaction: N2(g) + O2(g) <-->2NO(g)


Kp=4.1*10^-31
@ T= 298K

Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 ATM  and a temperature of 298K . Assume that nitrogen composes 78% of air by volume and that oxygen composes 21% of air.

This one i understand i need an ice table and an assumption needs to be made. can one assume that we have a 1L sample?

   N2(g)    +     O2(g) <-->   2NO(g)
I= 0.78M        0.21M       0M

C= -X                 -X             +2X

E= 0.78-X         0.21-X         2X
 

to setup the equation I would use this.  ([NO]^2)/[N2][O2]= 4.1*10^-31

so ([2X]^2)/(0.78-X)(0.21-X)=4.1*10^-31   

A quick possibly not accurate positive only calc gives me 1.295*10^-16 

checking my work now....

ok so I wind up with 4X^2 = (6.7158*10^-32)  - (4.059*10^-31)X + (4.1*10^-31)X^2

and I'm stuck... when combining like terms my calc drops the tiny (4.1*10^-31)X^2 number . 

ok if I proceed further knowing that I have lost those numbers. then i will get...

+/- 1.2957*10^-16.

ok so now what does that number mean? it is the value of X in the ice table right?

so 0.78-1.2957*10^-16= 0.78M of N2
and 0.21 -1.2957*10^-16= 0.21M of O2

This number is so small it doesn't effect anything.

I have the right answers here in front of me ( obviously my attempts have been wrong) and I didn't get any credit for them from Mastering chem. I feel as if I am missing some thing. 




Edit: gas law needs to be used... it says it in the problem!
P= 1.0 ATM
v= assumed 1L
N=?
R=(0.08206)
T=298K

Solving for N gets me 0.04089 mol of gases
how do I use this to solve for N2 and O2

possibly like this?

P=1.0atm
V= .78L of N2
N=?
R=(0.08206)
T=298K
 solving for N gets me 0.03189 mols of N2
but now I have mols instead of concentration... divide that by a liter and I have morality again but I'm not sure that's the right thing to do.

[Sophia7X]

You should be plugging in volume into the ICE table, not molarity (the question says 78% nitrogen by VOLUME).

+/- 1.2957*10^-16.

ok so now what does that number mean? it is the value of X in the ice table right?

Yes, that is the value of x.

so 0.78-1.2957*10^-16= 0.78M of N2
and 0.21 -1.2957*10^-16= 0.21M of O2

This number is so small it doesn't effect anything.

Yes, it is true the number is very small. Don't forget to plug x in for NO, so 2(1.2957*10^-16) = 2.6*10^-16 liters.

The amount of NO is supposed to be very small because K is very small. If K < 1, then the reaction favors the left side (reverse). Since K is extremely extremely small, the reaction will lie to the far left. That makes sence that you will have almost no NO at all.

Then plug in to PV=nRT, except V should be 0.78 L N2, 0.21 L O2, and 2.6*10^-16 L NO, not 1 L.
Once you find moles, just divide that by total volume (0.78 L + 0.21 L + 2.6*10^-16 L = 0.99 L) to get the concentrations of each gas.

[Ilragazzo]

ok so if I were to re approach the problem from the top. my first step would still be ICE table, but using volume instead. correct?


next step solve for X.

next step?

Yes, it is true the number is very small. Don't forget to plug x in for NO, so 2(1.2957*10^-16) = 2.6*10-16 liters.

Do I use this volume of NO in my PV=NRT calc? If so why? What does that number represent? The volume on NO present in the are at 298k and 1 atm?

I'm still confused.
---------------------------------------------------------------------------------------------
First Step assuming 100L sample?

 N2(g)    +     O2(g) <-->   2NO(g)
I= 0.78L        0.21L              0L

C= -X                 -X             +2X

E= 0.78-X         0.21-X         2X


This nets me the same answer as previously attained.  +/- 1.2957*10^-16

next plug 1.2957*10^-16 in for x and solve for liters of each.

next use PV=NRT to find moles of each.

next add all initial volumes up to = 0.99L

divide the moles of each over total volume for concentrations correct? 

[Sophia7X]

Yes, you should get the same value for x, just in units of volume. Units matter because if you had used molarity, then there would be no point in using PV=nRT to convert since you already have the concentration. You plug the values for V, then solve for n. You must do 3 separate PV=nRT calculations, find the # of moles for each gas. And divide moles by 0.99 L (total volume) to get concentration (M) of each gas. Basically you're converting VOLUME -> MOLES -> CONCENTRATION.

Do I use this volume of NO in my PV=NRT calc? If so why? What does that number represent? The volume on NO present in the are at 298k and 1 atm?

Yup, plug that in for V. You need to figure out how many moles of NO are present. The 2.6*10^-16 liters of NO is the amount that is present after equilibrium has been established. Initially, you have 0 L of NO. So an equilibrium will be established to make some NO. Negligible amounts of N2 and O2 react (-x) to create a little bit of NO (+2x).

[Ilragazzo]

ok so today I finally got this problem right, and i ended up using Partial pressure /(RT)= M  for the ice table instead of liters. As using liters instead of some form of a concentration was slightly confusing.

ok step one: find Kp from Kc

step 2: find molarity from partial pressure using Pv=rnt. ( solving for P/(RT)=N/V )

step 3: make an ice table using your newly acquired molarity.

step 4:make a mass action expression.

step 5: set your mass action expression equal to the Kp, or kc. (which are the same for this one.) ( what is the difference between Kp and Kc?)

step 6: solve for x, interpret then convert to the units you need.   

[Sophia7X]

There's not a huge difference in Kp or Kc. Kp is the equilibrium constant when you are dealing with pressure, Kc is when you're dealing with concentration. You're probably aware of the equation to convert one to the other, Kp = Kc(RT)^delta n