Achieve HW #9

[Shannon Clark 1F]

I'm stuck halfway through this problem:

"At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.300 M and [NO]=0.600 M.
N2(g)+O2(g)↽−−⇀2NO(g)
If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re‑established?"

I have that Kc=(0.6-2x)^2/(0.3+x)^2, but what do I do from here, I know I have to find the Kc in order to then use the quadratic formula, but can someone help me with the steps for the rest of the problem? How do I find the Kc value if I still have the x in my equation?

[Ramya_Paravastu_1H]

So you found your Kc in terms of x from an ICE chart, right?

You can find the Kc value (numeric) using the equilibrium concentrations given: "At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.300 M and [NO]=0.600 M".

After that, just equate the value you get above to the Kc equation you have in terms of x. Then, you can find x and plug it into the equation for NO in the E row of your ICE chart to get the answer.

Good luck :)

[RJ Lopez 2l]

Hi
First you would calculate Kc using the initial molarities that are given in the question, so (0.600)^2/0.300^2. That would be your Kc value, then you would set the expression you created equal to Kc and solve for x. Once you find x, do (0.6-2x) to find the new concentration of NO.

Hope this helps

[Claire_Sabol_2G]

Hi there! Remember that the kc value is the ratio of your products and reactants at equilibrium. At the beginning of the question, you are already given the concentrations of both product and reactants at equilibrium, meaning you can calculate the k value from the start. Once you find the k value, you will set it equal to the equation you found through the ICE box. Set these both equal to each other and use the quadratic formula to find x. Once you find x, substitute that value into your equilibrium equation for NO :)

[Sunny Hou 2I]

Hi,
So for this problem, you would first have to determine the equilibrium constant with the information that is given to you in the question first. So calculate Kc with the 0.300M of N2 and O2 and 0.600M of NO. Once you have your Kc value in your equilibrium, notice that because the only thing that can change Kc is either temperature or pressure, it means that no matter how you change the reactants or products the equilibrium constant will not change. And because of that, you can then equate your Kc equation with the 'x' to the Kc value you already found. Then solve for x using the quadratic formula. Once you have the x value because you have established with your ICE table that the value at the new equilibrium for NO will be 0.6-2x. All you would have to do is to plug in the value for x.
Hope this helps.

[Shannon Clark 1F]

Hi
First you would calculate Kc using the initial molarities that are given in the question, so (0.600)^2/0.300^2. That would be your Kc value, then you would set the expression you created equal to Kc and solve for x. Once you find x, do (0.6-2x) to find the new concentration of NO.

Hope this helps

Hi, this helped me a little! I got that my Kc=4, but do I set up the equation to solve for x as 4=(0.9 -2x)^2/(0.3+x)^2 OR 4=(0.6-2x)^2/(0.3+x)^2?? I'm very stuck

[Alaura Dis 1H]

Wouldn't the equation be dealing with .3 and .9 rather than .3 and .6? We know that equilibrium is achieved at .3 and .6 because it is given in the problem and this is how we can calculate the Kc. Kc will be the same as long as temperature and pressure remain constant. However, since NO is now .9M, this would be the new initial molarity. So I believe the equation for NO at equilibrium would be .9-2x? I think I had a different question than this one, but this is the logic I applied to solving other questions like this.

[Shannon Clark 1F]

Hi
First you would calculate Kc using the initial molarities that are given in the question, so (0.600)^2/0.300^2. That would be your Kc value, then you would set the expression you created equal to Kc and solve for x. Once you find x, do (0.6-2x) to find the new concentration of NO.

Hope this helps

Hi, this helped me a little! I got that my Kc=4, but do I set up the equation to solve for x as 4=(0.9 -2x)^2/(0.3+x)^2 OR 4=(0.6-2x)^2/(0.3+x)^2?? I'm very stuck

Nevermind! I figured it out! Thank you everyone for the replies!