Posted by **j** on Saturday, December 3, 2011 at 9:38pm.

If the initial concentration of NH3(g) is 4.643 mol/L, calculate the % of NH3(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kc at 773.0 K is 16.70. The initial concentration of the reaction products is 0 mol/L.

2NH3(g) = N2(g)+3H2(g)

I know that we're supposed to do an ICE chart. I got (x^2)/ 6.745-2x . But I don't know how to solve for the x and the rest I would appreciate if you could write out the steps to this :)

- Chemistry -
**DrBob222**, Saturday, December 3, 2011 at 11:13pm
You need work on your ICE charts.

...............2NH3 ==> N2 + 3H2

initial.......4.643.....0......0

change........-2x.......x......3x

equil.......4.643-2x.....x......3x

Kc = 16.70 = (N2)(H2)^3/(NH3)2

16.70 = (x)(3x)^3/(4.643-2x)^2

16.70 = 27x^4/(etc)

I didn't check my work (so I could have made a math error) but I expanded the denominator, multiplied by 16.70, and ended up with

27x^4 -66.8x^2 +310.15x -360 = 0

First I ignored the 27x^4 and solved the quadratic and obtained an answer of 2.31, substituted 2.31 back into the equation to see if I obtained zero. I didn't; the value was quite positive. Several more trials and I think the answer is >1.279 but < 1.280. You can try your hand at the trial and error method or find a computer program at school that will solve a quartic or look on the web for a calculator that will do it.

- Chemistry -
**DrBob222**, Saturday, December 3, 2011 at 11:23pm
I looked on the web and found, and used, two calculators but I wasn't satisfied with either answer. One answer was 1.03 something but that didn't give me a zero when I substituted. The other one gave a value of 1.276 and that didn't give a zero either. Both showed negative numbers which means the answer is higher than 1.276.

- Chemistry -
**j**, Sunday, December 4, 2011 at 1:39am
Thank you so much :)

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