If 0.500 mol each of phosphorus trichloride and chlorine are injected into a 1.00-L container at 60oC, find the equilibrium concentrations of all three species in the equilibrium mixture.

Question

If 0.500 mol each of phosphorus trichloride and chlorine are injected into a 1.00-L container at 60oC, find the equilibrium concentrations of all three species in the equilibrium mixture. PCl5(g) <===> PCl3(g) + Cl2(g) K - 12.5 at 60oC This is my ICE Table... ---PCl5(g) <===> PCL3(g) + Cl2(g) I--(0.00)-------(0.500)----(0.500) C--(x)----------(-x)-------(-x) E--(x)----------(0.500-x)--(0.500-x) K=([PCl3(g)][Cl2(g)])/[PCl5(g)] Subbing in the values I get 12.5 = (0.500-x)^2 / x 12.5x = 0.250 + x^2 -x^2 + 12.5x - 0.250 = 0 Then I sub this into the quadratic formula... x=(-12.5(+-)root((-12.5)^2-4(-1)(-0.250)))/2(-1) I get (using + root)x=0.02003210281 (using - root)x=12.4799679 My textbook says the correct answer should be 0.0185. Side Issue:I heard that using the negative root in the quadratic formula is useless in Chemistry, is that true?

Arora · Answer

Yes, this is because your 'x' value refers to a certain amount of concentration of a substance. And you cannot have a negative amount of substance.

Similarly, in mathematics and physics, when you solve quadratics related to something like length, you omit the negative root since you can't have negative lengths.

As for your answer, you aren't that inaccurate. I think your slight deviance from the answer is because you omitted the '2ab' term from '(a-b)^2' when forming your quadratic.

Anonymous · Answer

I solved another question similar to this and used the negative root in the quadratic equation and got the correct answer.

DrBob222 · Answer

I saw your new post above in which you talk about the negative root and I've addressed that there. I hope it is to your satisfaction. If you have further question about it you can follow up at that post. In this particular "rule", that business that you've heard, is right some of the time and wrong some of the time. It all depends upon what you let x stand for. IF x stands for a concentration, length, mass, etc then the rule is correct because you cn't have a negative mass, length, or concentration etc.. However, if x stands for something else and you must add or subtract something from the x to get the final answer, then that rule is wrong. Again, see more discussion at the other post you made.Hope this helps.