The question is incomplete, here is the complete question:
Consider the following reaction:
2NO(g) +Br₂(g) ⇔ 2NOBr(g) Kp= 28.4 at 298 K
In a reaction mixture at equilibrium, the partial pressure of NO is 107 torr and that of Br₂ is 166 torr. What is the partial pressure of NOBr in this mixture?
Answer: The equilibrium partial pressure of NOBr in the mixture is 6460.53 torr
Explanation:
We are given:
Equilibrium partial pressure of NO = 108 torr
Equilibrium partial pressure of bromine gas = 126 torr
The given chemical equation follows:
[tex]2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)[/tex]
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{(p_{NOBr})^2}{(p_{NO})^2\times p_{Br_2}}[/tex]
We are given:
[tex]K_p=28.4[/tex]
Putting values in above expression, we get:
[tex]28.4=\frac{(p_{NOBr})^2}{(108)^2\times 126}\\\\p_{NOBr}=6460.53,-6460.53[/tex]
Neglecting the negative value because partial pressure cannot be negative.
Hence, the equilibrium partial pressure of NOBr in the mixture is 6460.53 torr
