8. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇌ 2 HBr(g) is 2.18×106...

8. The equilibrium constant Kc for the reaction H₂(g) + Br₂(g) ⇌ 2 HBr(g) is 2.18×106 at 730°C. Starting 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H₂, Br₂, and HBr at equilibrium.

the answer is

[H2] = [Br2] = 1.81×10^-4 M

[HBr] = 0.267 M

but how and why?

Solutions

Expert Solution

Equilibrium expression:

Find concentration of Hbr from the moles and volume:

= 0.2667 M Hbr(initial)

Set up the equilibrium table:

H2 + Br2 = 2Hbr

Initial M : 0, 0, 0.2667

Change in M: -x, -x, +2x

Equilibrium M: x, x, 0.2667+2x

Plug these into the equilibrium equation:

solving this quadratic equation x = 1.81 x 10^-4

Now plug this into all the x-values to find the equilibrium concentrations:

[h2]=x= 1.81x10^-4 M

[Br2=x]= 1.81x10^-4 M

[br2]=0.2667+2x= 0.267 M

queen_honey_blossom answered 1 year ago

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