5I.17

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Malia Shitabata 1F
Posts: 127
Joined: Sat Aug 17, 2019 12:17 am

5I.17

Postby Malia Shitabata 1F » Thu Jan 09, 2020 6:05 pm

The equilibrium constant Kc for the reaction N2(g) + O2 (g) 2NO (g) at 1200ºC is 1.00x10^-5. Calculate the equilibrium molar concentrations of NO, N2, and O2 in a reaction vessel of volume 1.00 L that initially held 0.114 mol N2 and 0.114 mol O2.

I set up the ICE table with all the right values and when I was solving for x, I disregarded its subtracted value for the equilibrium concentration of N2 and O2 so 0.114-x became 0.114 since Kc is less than 10^-4. This means my denominator when plugging in values to solve became (0.114)^2. I used the same values as the solution manual, except for that subtraction step, but my x value was 0.014 and the book says 1.8x10^-4. Can someone please tell me where I went wrong?

Connie Chen 1E
Posts: 51
Joined: Mon Jun 17, 2019 7:24 am

Re: 5I.17

Postby Connie Chen 1E » Thu Jan 09, 2020 6:38 pm

I'm not sure how you got x=0.0114, but if you plug in all the values into the Kc ratio so that it's 1.00*10-5=(2x)2/((0.114-x)(0.114-x)) and solve for x using the quadratic formula, you should get x=1.8*10-4.

Malia Shitabata 1F
Posts: 127
Joined: Sat Aug 17, 2019 12:17 am

Re: 5I.17

Postby Malia Shitabata 1F » Thu Jan 09, 2020 6:42 pm

Connie Chen 1E wrote:I'm not sure how you got x=0.0114, but if you plug in all the values into the Kc ratio so that it's 1.00*10-5=(2x)2/((0.114-x)(0.114-x)) and solve for x using the quadratic formula, you should get x=1.8*10-4.


I did 1.00x10^-5=(2x)^2/(0.114)^2 instead of (0.114-x)(0.114-x) for the denominator because I thought since K is less than 10^-4 x becomes insignificant but I got a different answer and I'm not sure why it didn't work out.

Luc Zelissen 1K
Posts: 57
Joined: Mon Jun 17, 2019 7:23 am

Re: 5I.17

Postby Luc Zelissen 1K » Thu Jan 09, 2020 6:52 pm

Malia Shitabata 1F wrote:
Connie Chen 1E wrote:I'm not sure how you got x=0.0114, but if you plug in all the values into the Kc ratio so that it's 1.00*10-5=(2x)2/((0.114-x)(0.114-x)) and solve for x using the quadratic formula, you should get x=1.8*10-4.


I did 1.00x10^-5=(2x)^2/(0.114)^2 instead of (0.114-x)(0.114-x) for the denominator because I thought since K is less than 10^-4 x becomes insignificant but I got a different answer and I'm not sure why it didn't work out.


I guess it was significant enough. You might aswell always include the (minus x) no matter how much the reaction favors the reactants, just to be sure you can get the right answer.


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