At 800 K the equilibrium constant for I2(g) 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.56 10-2 g of I(g), how many grams of I2 are in the mixture?

Question

At 800 K the equilibrium constant for I2(g)  2 I(g) is Kc = 3.1  10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.56  10-2 g of I(g), how many grams of I2 are in the mixture?

DrBob222 · Answer

...........I2 ==> 2I equil mols I = 0.0356g/about 127 = about 2.8E-4. M = mols/L = about 2.8E-5. You need to confirm these numbers. Kc = (I)^2/(I2) You know equil (I) and Kc, calculate (I2).

lee · Answer

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