Question

20. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇆ 2HBr(g) is 2.180 × 106 at 730°C. Starting with 4.20 moles of HBr in a 17.8−L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. [H2] = M [Br2] = M [HBr] = M

Answer 1

initial [HBr] = number of mol / volume
= 4.10 moles / 17.8 L
= 0.2303 M

H2(g) + Br2(g) <----> 2HBr(g)
0      0              0.2303 (initial)
x      x             0.2303-2x (at equilibrium)

Kc= [HBr]^2 / [H2][Br2]
2.180*10^6 = (0.2303-2x)^2 / (x*x)
2.180*10^6 = (0.2303-2x)^2 / (x)^2
sqrt (2.180*10^6) = (0.2303-2x) / (x)
1476.5 = (0.2303-2x) / (x)
1476.5*x = (0.2303-2x)
x = 1.558*10^-4 M

[H2] = x = 1.558*10^-4 M
[Br2] = x = 1.558*10^-4 M
[HBr] = 0.2303 - x = 0.2303 - 1.558*10^-4 = 0.2301 M