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Consider the following reaction: Consider the 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 119 torr and that of Br2 is 151 torr . What is the partial pressure of NOBr in this mixture?

Respuesta :

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

[tex]2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)[/tex]

Equilibrium constant is given as:

[tex]K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}[/tex]

[tex]28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}[/tex]

[tex][p_{NOBr}]=7792[/tex] atm

Partial pressure of NOBr is 7792 atm

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