Asealed 1.0 l flask is charged with 0.500 mol of i2 and 0.500 mol of br2. an equilibrium reaction ensues: i2 (g) br2 (g) 2ibr (g) when the container contents achieve equilibrium, the flask contains 0.84 mol of ibr. the value of keq is

The value of Keq = 110

Let's know how to get it :

According to the equilibrium reaction equation :

I2(g) + Br2(g) → 2IBr(g)

now we need to get the value of Keq

when the Keq expression is:

Keq = [IBr]^2/ [I] [Br]

when we have [IBr] at equilibrium = 0.84 so we need to get [I] &[Br] at equilibrium .

when [Br2] = [I2] = 0.5 - 0.84 / 2

= 0.08 M

so, by substitution

∴ Keq = (0.84)^2/(0.08)^2

= 110

The equilibrium constant (Keq) for the reaction I₂(g) + Br₂(g) ⇄ 2IBr(g) when 0.500 mol of I₂ reacts with 0.500 mol of Br₂ in a 1.0 L flask is 110.2.  

The reaction is the following:

I₂(g) + Br₂(g) ⇄ 2IBr(g)

The initial concentrations of I₂ and Br₂ are:

At equilibrium we have:

I₂(g)  +  Br₂(g) ⇄ 2IBr(g)

(0.5-x)  (0.5-x)      2x

The equilibrium constant of the above reaction is:

We know that the flask contains 0.84 mol of IBr at equilibrium, so:

Now, knowing that x = 0.42 mol/L, we have:

Therefore, the value of Keq is 110.2.

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