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From Summer 2007, Exam 2:

7. You have found the following value in a table of equilibrium constants:
2 C₂H₃F₃(g) + 3 Cl₂(g) 􀀧 2 C₂F₃Cl₃(g) + 3 H₂(g) K_c = 5.19x10¹⁸
What is the equilibrium constant for the reaction:
6 C₂F₃Cl₃(g) + 9 H₂(g) 􀀧 6 C₂H₃F₃(g) + 9 Cl₂(g)

We're manipulating an equilibrium constant here, let's start with the equilibrium constant expression for the original reaction:
K_c = { [C₂F₃Cl₃]² [H₂]³ } / { [C₂H₃F₃]² [Cl₂]³ } = 5.19x10¹⁸

To get the second reaction, we have to reverse the original and multiply it by 3. When we reverse the direction of the equilibrium, the roles of products and reactants change, so the equilibrium constant is inverted. When an equilibrium equation is multiplied by some constant, each of the concentrations is raised to that power, so the whole equilibrium constant expression is raised to that power. This means that the equilibrium constant for the "new" reaction in the problem is:
K_c' = { [C₂H₃F₃]⁶ [Cl₂]⁹ } / { [C₂F₃Cl₃]⁶ [H₂]⁹ } = (1 / K_c)³ = 7.15x10^-57

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