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

I₂ (g) + Br₂ (g) ↔ 2IBr (g)

Given is Kc = 280 at 150 degree C.

Kc = [IBr]² / [I₂][Br₂]

Initially [IBr] = 0.45 / 2 = 0.225 M

The actual reaction is:

2IBr ↔ I₂ + Br₂, Kc = 1/280 = 0.00357142

In equilibrium,

[IBr] = 0.225 - 2x

[I₂] = x

[Br₂] = x

By substituting the values we get,

K = [I₂] [Br₂] / [IBr]²

0.00357142 = x×x / (0.225 - 2x)²

√(0.00357142) = x / (0.225 - 2x)

0.0597613 (0.225 - 2x) = x

0.01344 - 2 × 0.0597613x = x

(1 + 2 × 0.0597613)x = 0.01344

x = 0.01344 / (1+2 × 0.0597613)

x = 0.01344 / 1.1195226

x = 0.012005

Substituting the values we get,

IBr = 0.225 - 2 × 0.012005 = 0.17698 M

I₂ = x = 0.012005 M

Br₂ = 0.012005 M

Answer 2

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

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

$[IBr]=c=\frac{0.450 mol}{2 L}=0.225 mol/L$

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

  Initially c       0    0

At eq'm   c-2x   x    x

$K_c'=\frac{1}{K_c}=\frac{[I_2][Br_2]}{[Ibr]^2}$

$\frac{1}{280}=\frac{x\times x}{(c-2x)^2}=(\frac{x}{(c-2x)})^2$

$\sqrt{\frac{1}{280}}=\frac{x}{c-2x}$

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

$[IBr]=c-2x=0.450-2\times 0.011887 mol/L=0.2010 mol/L$

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