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A mixture consisting of .10 mol H2 and .10 mol Br2 is placedi n a container...

A mixture consisting of .10 mol H2 and .10 mol Br2 is placedi n a container of volume 2.0 L. The reaction H2 + Br --> 2 HBr is allowed to come to equilbrium. Then .20 mol HBr is placed into a second sealed container of volume 2.0 L at the same temperature and is allowed to reach equilibrium with H2 and Br2. Which of the following will be different in the two containers at equilbrium? Which will be the same? (a) AMount of Br2 (b) partial pressure of H2 (c) the ration PHBR^2 / (PH2)(PBr2) (d) the ratio PHBr / PBr2 (e) the ration (PHBr)^2 / (PH2) (PBr2) (f) total pressure in the container.

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Answer #1

      H2 + Br2 --> 2 HBr

1 mol H2= 1 mol Br2 =2 mol HBr

Kc = [HBr]/[H2][Br2]

(a) AMount of Br2   = different in the two containers (Due difference in quantity of reactant,products)

(b) partial pressure of H2 = different in the two containers (Due difference in quantity of reactant,products)


(c) the ratio PHBR^2 / (PH2)(PBr2) = same in the two containers (at constant T , the ratio will be constant)


(d) the ratio PHBr / PBr2 = same in the two containers (in the equation, mol ratio of H2,Br2 IS SAME)


(e) the ratio (PHBr)^2 / (PH2) (PBr2) = same in the two containers


(f) total pressure in the container = different in the two containers (Due difference in quantity of reactant,products)

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