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:The equilibrium concentration of [tex][2IBr]=0.2010 mol/L, [I_2]=[Br_2]=0.01199 mol/L[/tex]
Explanation:
[tex]I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)[/tex]
Equilibrium constant = [tex]K_c=280 [/tex]
Suppose that 0.450 mol IBr in a 2.00-L
[tex][IBr]=c=\frac{0.450 mol}{2 L}=0.225 mol/L[/tex]
[tex]2IBr(g)\rightleftharpoons I_2(g)+Br_2(g)[/tex]
Initially c 0 0
At eq'm c-2x x x
[tex]K_c'=\frac{1}{K_c}=\frac{[I_2][Br_2]}{[Ibr]^2}[/tex]
[tex]\frac{1}{280}=\frac{x\times x}{(c-2x)^2}=(\frac{x}{(c-2x)})^2[/tex]
[tex]\sqrt{\frac{1}{280}}=\frac{x}{c-2x}[/tex]
[tex][I_2]=[Br_2]=x = 0.01199 mol/L[/tex]
[tex][IBr]=c-2x=0.450-2\times 0.011887 mol/L=0.2010 mol/L[/tex]
The equilibrium concentration of [tex][2IBr]=0.2010 mol/L, [I_2]=[Br_2]=0.01199 mol/L[/tex]
