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March 4, 2015

March 4, 2015

Posted by **Katherine** on Monday, September 30, 2013 at 4:20pm.

PCl5(g) -----> PCl3 (g)+ Cl2(g)

Kc= 1.80 at 250 degrees C

A 0.206 mol sample of PCl5(g) is injected into an empty 3.30 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

- Chemistry -
**Katherine**, Monday, September 30, 2013 at 4:22pmI don't know what to do after I set up the Kc equation. How do you isolate x from 1.80=(x^2)/(.062-x)

- Chemistry -
**DrBob222**, Monday, September 30, 2013 at 4:38pmFirst, I would check that 0.206/3.30. You shouldn't drop that last number; you are allowed three significant figures in 0.206 and 3.30 and that means you can have three s.f. in the answer. I have 0.0624. Second, your chemistry is right; you've done the hard part. Now it's just algebra. Here is what you start with.

(x^2)

--------- = 1.80

0.0624-x

Multiply both sides by 0.0624-x like this.

(0.0624-x)(x^2)

---------------- = 1.80*(0.0624-x)

(0.0624-x)

You see (0.0624-x) term on the left cancels since it is in the numerator and denominator and you are left with

x^2 = 1.80(0.0624-x)

Note: You could have done the same thing by cross multiplying but I don't know that you've seen that way.

Now multiply the right side to remove the parentheses, move all the terms to the left and you are left with a quadratic which you can solve for x. Let me know if you get stuck. Thanks for showing your work.

- Chemistry -
**Katherine**, Monday, September 30, 2013 at 4:57pmI was able to figure it out! Thank you very much for your help(:

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