Question

At certain temperature Kc is 4.13x10^-2 for the equilibrium 2IBr(g)<—> I2(g) + Br2(g). Assuming that equilibrium is established at the above temperature by adding only IBr(g) to the reaction flask what are the concentrations of I2(g) and Br(g) in equilibrium with 0.0124M of IBr(g)? Explain and show steps

Answer 1

Kc= [I2][Br2]/[IBr]²;

[I2]=[Br2]; [I2][Br2]=[Br2]²;

[Br]²=Kc*[IBr]²=4.13*10^(-2)*0.0124^2=6.35*10^-6;

[I2]=[Br2]=2.52*10^-3;

Answer: 2.52*10^-3