44.44 The dissociation of molecular 10dine intó iðdine al- oms is represented as I(g) 21(g) At 1000 K, the equilibrium constant K. for the reaction is 3.80 x 10. Suppose you start with 0.0456 mole of I, in a 2.30-L flask at 1000 K. What are the concentrations of the gases at equilibrium?

Chemistry & Chemical Reactivity
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Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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Chapter15: Principles Of Chemical Reactivity: Equilibria
Section: Chapter Questions
Problem 15PS: The equilibrium constant for the dissociation of iodine molecules to iodine atoms I2(g) 2 I(g) is...
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14.44 The dissociation of molecular iodine into iodine at-
oms is represented as
L(8) = 21(g)
At 1000 K, the equilibrium constant K. for the
reaction is 3.80 x 10. Suppose you start with
0.0456 mole of I, in a 2.30-L flask at 1000 K. What
are the concentrations of the gases at equilibrium?
Transcribed Image Text:14.44 The dissociation of molecular iodine into iodine at- oms is represented as L(8) = 21(g) At 1000 K, the equilibrium constant K. for the reaction is 3.80 x 10. Suppose you start with 0.0456 mole of I, in a 2.30-L flask at 1000 K. What are the concentrations of the gases at equilibrium?
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