Question

In a 2.0 liters container are 3.2 moles of IBr. In the 40% of equilibrium the IBr is already disassociated. Determine for the reaction a. The moles de I2, Br2 and IBr on equilibrium? b. The value for the equilibrium constant?

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

Answer 1

a. At equilibrium, there are 0.64 mol `"Br"_2`, 0.64 mol `"I"_2`, and 1.9 mol `"IBr"`.
b. `K = 36`

Explanation:

The chemical equation is

`"I"_2 + "Br"_2 ⇌ "2IBr"`

Step 1. Calculate the initial concentration of `"IBr"`

`["IBr"]_0 = "3.2 mol"/"2.0 L" = "1.6 mol/L"`

Step 2. Calculate the equilibrium concentrations

We can set up an ICE table as we continue solving this problem.

`color(white)(mmmmmmm)"I"_2 + "Br"_2 ⇌ "2IBr"`
`"I/mol·L"^"-1": color(white)(mm)0 color(white)(mm)0color(white)(mmm)1.6`
`"C/mol·L"^"-1":color(white)(m)"+"x color(white)(m)"+"xcolor(white)(mmll)"-2"x`
`"E/mol·L"^"-1": color(white)(mll)xcolor(white)(mm)xcolor(white)(mml)"1.6-2"x`

At equilibrium, the `"IBr"` is 40 % dissociated, so 60 % remains.

The equilibrium concentration of `"IBr"` is

`["IBr"] = "0.60 × 1.6 mol/L = 0.96 mol/L"`

But

`["IBr"] = ("1.6-2"x)color(white)(l) "mol/L"`

∴ `0.96 = 1.6-2x`

`2x = 1.6-0.96 = 0.64`

`x = 0.64/2 = 0.32`

`["I"_2] = ["Br"_2] = xcolor(white)(l)"mol/L = 0.32 mol/L"`

`["IBr"] = "0.96 mol/L"`

Step 3. Calculate the moles of each compound at equilibrium

`n_text(I₂) = 2.0 color(red)(cancel(color(black)("L I"_2))) × "0.32 mol I"_2/(1 color(red)(cancel(color(black)("L I"_2)))) = "0.64 mol I"_2`

`n_text(Br₂) = 2.0 color(red)(cancel(color(black)("L Br"_2))) × "0.32 mol Br"_2/(1 color(red)(cancel(color(black)("L Br"_2)))) = "0.64 mol Br"_2`

`n_text(IBr) = 2.0 color(red)(cancel(color(black)("L IBr"))) × "0.96 mol IBr"/(1 color(red)(cancel(color(black)("L IBr")))) = "1.9 mol IBr"`

Step 5. Calculate the volume of the equilibrium constant

The equilibrium constant expression is

`K = ["IBr"]^2/(["I"_2]["Br"_2])`

∴ `K =1.9^2/(0.32 × 0.32) = 36`