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a) K = [NO]^2 [Cl2] / [NOCl]^2 2 mol NOCl produces 2 mol NO and 1 mol Cl2. b) I don't have a TI-84 or similar calculator with me. Solve this polynomial for x to arrive at the equilibrium concentrations. EXFO Announces Acquisition of Global Wireless Test Specialist NetHawk Oyj Thanks for visiting!
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Chemistry Analyzer
The balanced equation is: 2NOCl <===> 2NO + 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
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
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