Answer : At equilibrium, the partial pressure of IBr, [tex]I_2[/tex] and [tex]Br_2[/tex] is 0.0211 atm, 0.00195 atm and 0.00195 atm respectively.
Solution : Given,
Initial partial pressure = 0.025 atm
[tex]K_p=8.5\times 10^{-3}[/tex]
The given equilibrium reaction is,
[tex]2IBr(g)\rightleftharpoons I_2(g)+Br_2(g)[/tex]
initial pressure 0.025 atm 0 0
At equilibrium (0.025 - 2x) x x
The expression of equilibrium constant for this reaction is,
[tex]K_p=\frac{[P_I_2][P_{Br}_2]}{[P_{IBr}]^2}[/tex]
Now put all the given values in this expression, we get value of 'x'.
[tex]8.5\times 10^{-3}=\frac{(x)(x)}{(0.025-2x)^2}[/tex]
By rearranging the terms, we get the value of 'x'.
x = 0.00195
Therefore,
The partial pressure of IBr = (0.025 - 2x) = (0.025 - 2(0.00195)) = 0.0211 atm
The partial pressure of [tex]I_2[/tex] = x = 0.00195 atm
The partial pressure of [tex]Br_2[/tex] = x = 0.00195 atm
