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The following equalibrium was set up in a 1.0 L flask at a temp of 735 K?
NOCL (g) <----> 2NO (g) + Cl2 (g)
a) You have been given a glass reaction vessel which has a 2.00 moles of NOCL and when equalibrium was established it was found that 33% of the NOCl had dissociated. From these data calculate the equalibrium constant for the Reaction.
b) After the reaction represented by the equalibrium reaction written above has reached the equalibrium condition at 298K, you decided to analyze the flask for the equalibrium concentrations of all species. You need to set the analyzer on the right sensitivity range for the analysis, so you decided to calculate what the equalibrium concentrations should be if the equilbrium constant for this process is 54.3 at 298 K. What are the concentrations of all species which exist in the reaction vessel at equalibrium.
Please show all work. (Thanks so much if you can please help; God Speed)
a)
The balanced equation is: 2NOCl <===> 2NO + Cl2
K = [NO]^2 [Cl2] / [NOCl]^2
2 mol NOCl produces 2 mol NO and 1 mol Cl2.
You start with 2.00 moles NOCl. At equilibrium, the concentrations would be:
[NOCl] = 0.667 x 2.00 = 1.33 M
[NO] = 0.667 M
[Cl2] = 0.333 M
K = (0.667x)^2 (0.333) / (1.33x)^2 = 0.0874 M
b)
Let 2x = amount of NOCl that dissociates. Then, at equilibrium:
[NOCl] = 2.00 - 2x
[NO] = 2x
[Cl2] = x
54.3 = (2x)^2(x) / (2.00 - 2x)^2
54.3 = 4x^3 / (2.00 - 2x)^2
4x^3 = 54.3 (4.00 - 8x + 4x^2)
x^3 = 54.3 (1 - 2x + x^2) = 54.3 - 108.6 x + 54.3 x^2
x^3 - 54.3 x^2 + 108.6 x - 54.3 = 0
I don't have a TI-84 or similar calculator with me. Solve this polynomial for x to arrive at the equilibrium concentrations.
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