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### Homework 5I.25

Posted: Tue Jan 14, 2020 10:18 am
I've set up an ice table so that the equilibrium concentrations are as follows:
[SO2] = 0.02M - x
[NO2] = 0.04M - x
[NO] = 0.02 + x
[SO3] = 0.03 + x

And I've set up the equilibrium concentration as Kc = [NO][SO3]/[SO2][NO2], and letting these equal 85. My expanded equation for Kc is (x^2 + 0.05x
+ 0.0006)/(x^2-0.06+0.0008). I feel like I am setting it up right but I keep getting a different answer than what the answer key shows. Can someone explain to me how to do this problem correctly or help me identify any mistakes that I made? Thanks

### Re: Homework 5I.25

Posted: Tue Jan 14, 2020 10:31 am
Ok, I realized that I was dividing by the wrong number and I got the same answers as the book. However, I came up with 2 different values of x: 0.0423, and 0.0189. These numbers are both positive, so how do I determine which one to use? The book uses 0.0189

EDIT: I think I figured it out, it's because if you use 0.0423, you will get a negative concentration for the products when you subtract x. Can someone confirm?

### Re: Homework 5I.25

Posted: Tue Jan 14, 2020 2:02 pm
Also having trouble with this problem. I follow your work until you plug into the Kc formula. Could you explain how you solved for x? Could you also explain why you omitted the x variable in the denominator with 0.06?

### Re: Homework 5I.25

Posted: Tue Jan 14, 2020 11:45 pm
You use 0.0189 instead of 0.0424 since this value is larger than the concentration of NO2.

### Re: Homework 5I.25

Posted: Wed Jan 15, 2020 9:58 am
405318478 wrote:Also having trouble with this problem. I follow your work until you plug into the Kc formula. Could you explain how you solved for x? Could you also explain why you omitted the x variable in the denominator with 0.06?

I didn't mean to omit it; there should be an x variable there. I solved for x by getting everything to one side and having one side equal 0, then used the quadratic equation to solve for x.