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?
Answers
DrBob222
...........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).
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
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