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:
[tex] [I_{2}] = \frac{n}{V} = \frac{0.500 \:mol}{1.0 L} = 0.500 M [/tex]
[tex] [Br_{2}] = \frac{n}{V} = \frac{0.500 \:mol}{1.0 L} = 0.500 M [/tex]
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:
[tex] Keq = \frac{[IBr]^{2}}{[I_{2}][Br_{2}]} [/tex]
We know that the flask contains 0.84 mol of IBr at equilibrium, so:
[tex] 2x = 0.84 \:mol/L [/tex]
[tex] x = 0.42 \:mol/L [/tex]
Now, knowing that x = 0.42 mol/L, we have:
[tex] Keq = \frac{(2x)^{2}}{(0.5 -x)(0.5 -x)} [/tex]
[tex] Keq = \frac{(0.84)^{2}}{(0.5 - 0.42)^{2}} = 110.2 [/tex]
Therefore, the value of Keq is 110.2.
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