Bromine is one of the few elements that is a liquid at room temperature. It vapo

Question

Bromine is one of the few elements that is a liquid at room temperature. It vaporizes easily and has a beautiful red color in both the liquid and gaseous states. Bromine reacts readily with hydrogen gas to make hydrogen bromide. One way to represent this equilibrium is: 2 HBr(g) doubleheadarrow H_2(g) + Br_2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) HBr(g) doubleheadarrow 1/2 H_2(g) + 1/2 Br_2(g) K_1= 2) H_2(g) + Br_2(g) doubleheadarrow 2HBr(g) K_2 = 1/2 H_2(g) + 1/2 Br_2(g) doubleheadarrow HBr(g) K_3 =

Explanation / Answer

2HBr ( g ) <-------> H2(g) + Br2(g)

K = [ H2 ] [ Br2 ]/[ HBr]^2

Other reactions

1) HBr(g ) <--------> 1/2H2(g) + 1/2Br2(g)

K1 = ([H2]^1/2× [Br]^1/2)/[HBr]

K1^2 = K

K1 = K

2) H2(g) + Br2(g) <------> 2HBr

K2 = [ HBr ]^2/([H2] × [Br2])

K2 = 1/K

3) 1/2 H2(g) + 1/2Br2(g) <--------> HBr(g)

K3 = [ HBr ]/([ H2 ]^1/2 × [ Br2 ]^1/2)

K3^2 = [HBr]^2/([H2] × [ Br2 ])

= 1/K

K3 = 1/K

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