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The equilibrium constant Kc for the following reaction is 4.59 × 10-7 at 730oC. 2HBr(g) ⇌...

The equilibrium constant Kc for the following reaction is 4.59 × 10-7 at 730oC. 2HBr(g) ⇌ H2(g) + Br2(g) Suppose 3.20 mol HBr and 1.50 mol H2 are added to a rigid 12.0-L flask at 730oC. What is the equilibrium concentration (in M) of Br2?

Solutions

Expert Solution

[HBr] = moles/L = 3.20 mol/12.0 L = 0.267 M

[H2] = 1.50 mol/12.0 L = 0.125 M

                              2HBr       <====>   H2       + Br2

initial (M)                 0.267                     0.125         0

change(M)                -2x                          +x          +x

equilibrium(M)      (0.267-2x)             (0.125 +x)     x

-----------------------------------------------------------------------

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

4.59 x 10^-7 = (x)(0.125 + x)/(0.267 - 2x)^2

4.59 x 10^-7 = 0.125x + x^2/(0.0713 - 1.068x + 4x^2)

1.84 x 10^-6x^2 - 4.902 x 10^-7x + 3.27 x 10^-8 = 0.125x + x^2

x^2 + 0.125x + 4.902 x 10^-7 = 0

x = 0.125 M

So the concentration of [Br2] = 0.125 M

queen_honey_blossom answered 2 years ago
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