Question

The equilibrium constant Kc for the following reaction is 2.18 x10^6 at 730°C.?

H2(g)+Br2(g) <--->2HBr(g)  
Starting with 1.20 moles of HBr in a 19.5 L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

M?

Answer 1

Initial concentration of HBr = number of moles / volume in L

= 1.20 mol / 19.5 L

= 0.0615 M

Given Equilibrium constant for H2(g)+Br2(g) <--->2HBr(g) is , Kc = 2.18x10⁶

So Equilibrium constant for reverse reaction is Kc' = 1/Kc = 1/2.18x10⁶ = 4.58x10^-7

   2HBr (g) <----> H2(g) + Br2(g)

initial conc 0.0615 0 0

change -2c +c +c

Equb conc 0.0615-2c c c

Equilibrium constant , Kc' = ( c x c ) / ( 0.0615 - 2c ) = 4.58x10^-7

On solving we get c = 1.675x10^-4 M

Equilibrium concentration of H2 = c = 1.675x10^-4 M

Equilibrium concentration of Br2= c = 1.675x10^-4 M

Equilibrium concentration of HBr =   0.0615-2c = 0.0612 M