Question

10. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 ×...

10. The equilibrium constant Kc for the reaction

H2(g) + Br2(g) ⇆ 2HBr(g)


is 2.180 × 106 at 730°C. Starting with 2.20 moles of HBr in a 13.7−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

[H2] =   
[Br2] =
[HBr] =
0 0
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Answer #1

The reaction is as shown below

The equilibrium constant

Calculate initial concentration of HBr

Let 2x M be the change in the concentration of HBr to reach equilibrium. Prepare an ICE table.

Initial concentration (M) 0.00 0.00 0.161
Change in concentration (M) x x -2x
Equilibrium concentration (M) x x

Substitute values in the equilibrium constant expression

Since the value of the equilibrium constant is very large, approximate 0.161-2x to 0.161

Take square root on both sides

Hence, the equilibrium concentrations are calculated as shown below

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