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Question :
For the reaction equilibrium, 2NOBr (g) ⇌ 2NO (g) + Br2 (g), If P(Br₂) = P/9 at equilibrium and P is total pressure. The ratio Kp/P is equal to:
Solution :
Ratio of Kp /P is 1 : 81
Step by step Explanatìon :
Before , Solving the Question , let's understand the given reaction :
[tex]\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)[/tex]
Here , In the product we have two moles of NO and One mole of Br₂ .
It is given that the partial pressure of Br₂ is P/9 ,Then according to the given reaction , the partial pressure of 2NO will be two times more than that of Br₂ .
Thus ,
[tex]\sf\:p_{2NO}=2\times\dfrac{P}{9}[/tex]
(Given ) : Total pressure of the Reaction is P
Let the partial pressure of 2NOBr be x , then
[tex]\sf\:x=P-\dfrac{2P}{9}-\dfrac{P}{9}[/tex]
[tex]\sf\implies\:x=\dfrac{9P-3P}{9}[/tex]
[tex]\sf\implies\:p_{2NOBr}=\dfrac{6P}{9}[/tex]
Now ,
Given Reaction :
[tex]\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)[/tex]
Then ,
[tex]\rm\:K_p=\dfrac{p_{Br_2}\times\:(p_{2NO})^2}{(p_{2NOB}^2}[/tex]
[tex]\sf\implies\:K_p=\dfrac{(\frac{P}{9})\times(\frac{2P}{9})^2}{(\frac{6P}{9})^2}[/tex]
[tex]\sf\implies\:K_p=\dfrac{\frac{P}{9}\times\frac{4P^2}{81}}{\frac{36P^2}{81}}[/tex]
[tex]\sf\implies\:K_p=\dfrac{4P^2\times\:81\times\:P}{81\times9\times\:P^2}[/tex]
[tex]\sf\implies\:K_p=\dfrac{P}{81}[/tex]
We have to find the ratio of [tex]\sf\dfrac{K_p}{P}[/tex] , then
[tex]\sf\dfrac{K_p}{P}=\dfrac{\frac{P}{81}}{P}[/tex]
[tex]\sf\implies\dfrac{K_p}{P}=\dfrac{1}{81}[/tex]
Therefore,
Correct option 2) 1/81
_____________
Theory :
Equilibrium constant in terms of pressure
It is denoted by [tex]\sf\:K_p[/tex]
For a Genral Reaction :
[tex]\sf\:aA+bB\:\rightleftharpoons\:cC+dD[/tex]
[tex]\sf\:K_p=\dfrac{(p_C)^c\:\times\:(p_D)^d}{(p_A)^a\:\times\:(p_B)^b}[/tex]