Question

At 25.0°C, the equilibrium constant K_c for Br_2(g) + I₂ <=> 2IBr_(g) is 282. If the initial concentration of bromine is 0.0100M, of Iodine is 0.0100M and of iodine monobromide is 0.200M, what is the equilibrium concentration of IBr?

Answer 1

i am assuming reaction move in forward direction
if we get x negative then , reaction will be moving in reverse direction

I2 + Br2 <--> 2IBr
0.01 0.01 0.2 intial
0.01-x 0.01-x 0.2+2x at equlibrium

Kc = [IBr]^2/([Br]*[I2])
282 = (0.2+2x)^2/(0.01-x)^2
282 = {(0.2+2x)/(0.01-x)}^2
16.8 = {(0.2+2x)/(0.01-x)}
16.8*(0.01-x) = (0.2+2x)
0.168 - 16.8x = 0.2 + 2x
18.8x = -0.032
x = -0.00171

we get x negative
so, reaction is moving in bacward direction
at equlibrium
[IBr] = 0.2+2x
= 0.2 + 2*(-0.0171)
= 0.2 - 0.0342
= 0.1658 M

Answer : 0.1658 M