Equilibrium 2AB3(g) <-------- A2(g)+3B(g) if 0.87 moles of AB3 are injected into

Question

Equilibrium 2AB3(g) <-------- A2(g)+3B(g) if 0.87 moles of AB3 are injected into a 5.0 L container at 25 celsius, at equilibrium the final [A2] is found to be 0.070. a) calculate the equilibrium concentration of AB3. b) Calculate the equilibrium [A2] c) Calculate the equilibrium [B2] Different Q. At a high temperature, 0.50 mole of HBr was placed in a 1.0 L container and allowed to decomposed according to the reaction: 2HBr(g)----- H2(g) +Br2(g) At equilibrium the [Br2] was measured to be 0.13M. What is Keq for this reaction at this temperature?

Explanation / Answer

Part A) 2AB3(g) <==> A2(g) + 3B(g)

Initial [AB3] = 0.87 mol/5 L = 0.174 M

Equilibrium [A2] = 0.070 mol/5 L = 0.014 M

a) Equilibrium concentration of [AB3] = 0.174 - 2 x 0.014 = 0.146 M

b) Equilibrium [B] = 3 x 0.014 = 0.042 M

Part B) 2HBr <==> H2 + Br2

initial [HBr] = 0.50 mol/1 L = 0.5 M

Equilibrium [Br2] = 0.13 M

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

        = (0.13)^2/(0.5 - 2 x 0.13)^2

        = 0.293

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