Question

The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 × 106 at 730°C. Starting with 5.20 moles of HBr in a 12.9−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

Answer 1

          H2(g) + Br2(g) <---> 2HBr(g)

initial     0 M        0 M       5.2/12.9 = 0.403 M

change      +x M        +x M       -2x

equil       x M         x M       0.403-2x M

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

(2.18*10^6) = (0.403-2x)^2/x^2

x = 2.72*10^-4

[HBr] at equilibrium = 0.403 - (2*2.72*10^-4) = 0.402 M

[H2] = X = 2.72*10^-4 M

[Br2] = x = 2.72*10^-4 M