Question

For the reaction

I2(g)+Br2(g)←−→2IBr(g),
Kc=280 at 150 ∘C. Suppose that 0.450 mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150 ∘C.

What is the equilibrium concentration of 2IBr, I2, Br2

Answer 1

Answer:The equilibrium concentration of
`[2IBr]=0.2010 mol/L, [I_2]=[Br_2]=0.01199 mol/L`

Step-by-step explanation:

`I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)`

Equilibrium constant =
`K_c=280`

Suppose that 0.450 mol IBr in a 2.00-L

`[IBr]=c=(0.450 mol)/(2 L)=0.225 mol/L`

`2IBr(g)\rightleftharpoons I_2(g)+Br_2(g)`

Initially c 0 0

At eq'm c-2x x x

`K_c'=(1)/(K_c)=([I_2][Br_2])/([Ibr]^2)`

`(1)/(280)=(x* x)/((c-2x)^2)=((x)/((c-2x)))^2`

`\sqrt{(1)/(280)}=(x)/(c-2x)`

`[I_2]=[Br_2]=x = 0.01199 mol/L`

`[IBr]=c-2x=0.450-2* 0.011887 mol/L=0.2010 mol/L`

The equilibrium concentration of
`[2IBr]=0.2010 mol/L, [I_2]=[Br_2]=0.01199 mol/L`