Question

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?

Answer 1

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)`

`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.