The equilibrium constant for the reaction 2IBr<->I2+Br2 is 8.5×10^-3 at 150°C. If 0.0300 mol of IBr is introduced to a 1L container, what is the concentration of the substance after equilibrium is established?

1 Answer
Dec 11, 2017

The is an equilibrium problem, about low-moderate difficulty. An #"ICE"# table is needed to efficiently solve this. Concentration calculations I will omit so my answer is more clear, however, be meticulous with these on exams, I can't count how many stupid mistakes I've made over these.

#2IBr(g) rightleftharpoons I_2(g) + Br_2(g)# where #K_c = 8.5*10^-3#, and

#K_c = ([I_2][Br_2])/([IBr]^2)#

Hence,

#2IBr(g) rightleftharpoons I_2(g) + Br_2(g)#
puu.sh

#K_c = 8.5*10^-3 = (x^2)/(0.03-x)^2#
#K_c = 9.2*10^-2 = x/(0.03 -x)#
#therefore x approx 0.0025#

The concentrations at equilibrium (which I've calculated per the #"ICE"# table) are,

#[IBr] = 0.0275M#
#[I_2] = [Br_2] = 0.0025M#

This is reasonable because #K_c < 1#, and the reaction won't be very active in the right direction, hence the small "product" concentrations.