The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.95 x...

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.95 x 103 at a certain temperature. Find the equilibrium pressure of HBr if 10.70 atm of HBr is introduced into a sealed container at this temperature.

Expert Solution

H2(g) + Br2(g) ↔ 2HBr (g)

I 0 0 10.7

C +x +x -2x

E +x +x 10.7-2x

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

1.95*10^3 = (10.7-2x)^2/x*x

1.95*10^3 *x^2 = (10.7-2x)^2

1.95*10^3 *x^2-4x^2-114.19+42.8x=0

1946x^2+42.8x-114.19=0

x = 0.23

[HBr] = 10.7-2x = 10.7-2*0.23 = 10.24atm

[H2] = x = 0.23atm

[Br2] = x = 0.23atm

queen_honey_blossom answered 2 years ago

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