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Answer 1

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

Explanation :

Before, Solving the Question, let's understand the given reaction :

$\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)$

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,

$\sf\:p_{2NO}=2\times\dfrac{P}{9}$

Given : Total pressure of the Reaction is P

Let the partial pressure of 2NOBr be x, then

$\sf\:x=P-\dfrac{2P}{9}-\dfrac{P}{9}$

$\sf\implies\:x=\dfrac{9P-3P}{9}$

$\sf\implies\:p_{2NOBr}=\dfrac{6P}{9}$

Now,

Given Reaction :

$\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)$

Then,

$\rm\:K_p=\dfrac{p_{Br_2}\times\:(p_{2NO})^2}{(p_{2NOB}^2}$

$\sf\implies\:K_p=\dfrac{(\frac{P}{9})\times(\frac{2P}{9})^2}{(\frac{6P}{9})^2}$

$\sf\implies\:K_p=\dfrac{\frac{P}{9}\times\frac{4P^2}{81}}{\frac{36P^2}{81}}$

$\sf\implies\:K_p=\dfrac{4P^2\times\:81\times\:P}{81\times9\times\:P^2}$

$\sf\implies\:K_p=\dfrac{P}{81}$

We have to find the ratio of $\sf\dfrac{K_p}{P}$, then

$\sf\dfrac{K_p}{P}=\dfrac{\frac{P}{81}}{P}$

$\sf\implies\dfrac{K_p}{P}=\dfrac{1}{81}$

Therefore,

Correct option 2) 1/81

_____________

Theory :

Equilibrium constant in terms of pressure

It is denoted by $\sf\:K_p$

For a Genral Reaction :

$\sf\:aA+bB\:\rightleftharpoons\:cC+dD$

$\sf\:K_p=\dfrac{(p_C)^c\:\times\:(p_D)^d}{(p_A)^a\:\times\:(p_B)^b}$

Answer 2

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

Explanation :

Before, Solving the Question, let's understand the given reaction :

\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)2NOBr(g)⇌2NO(g)+Br2(g)

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,

\sf\:p_{2NO}=2\times\dfrac{P}{9}p2NO=2×9P

Given : Total pressure of the Reaction is P

Let the partial pressure of 2NOBr be x, then

\sf\:x=P-\dfrac{2P}{9}-\dfrac{P}{9}x=P−92P−9P

\sf\implies\:x=\dfrac{9P-3P}{9}⟹x=99P−3P

\sf\implies\:p_{2NOBr}=\dfrac{6P}{9}⟹p2NOBr=96P

Now,

Given Reaction :

\sf\:2NOBr(g)\rightleftharpoons\:2NO(g)+Br_2(g)2NOBr(g)⇌2NO(g)+Br2(g)

Then,

\rm\:K_p=\dfrac{p_{Br_2}\times\:(p_{2NO})^2}{(p_{2NOB}^2}Kp=(p2NOB2pBr2×(p2NO)2

\sf\implies\:K_p=\dfrac{(\frac{P}{9})\times(\frac{2P}{9})^2}{(\frac{6P}{9})^2}⟹Kp=(96P)2(9P)×(92P)2

\sf\implies\:K_p=\dfrac{\frac{P}{9}\times\frac{4P^2}{81}}{\frac{36P^2}{81}}⟹Kp=8136P29P×814P2

\sf\implies\:K_p=\dfrac{4P^2\times\:81\times\:P}{81\times9\times\:P^2}⟹Kp=81×9×P24P2×81×P

\sf\implies\:K_p=\dfrac{P}{81}⟹Kp=81P

We have to find the ratio of \sf\dfrac{K_p}{P}PKp , then

\sf\dfrac{K_p}{P}=\dfrac{\frac{P}{81}}{P}PKp=P81P

\sf\implies\dfrac{K_p}{P}=\dfrac{1}{81}⟹PKp=811

Therefore,

Correct option 2) 1/81

_____________

Theory :

Equilibrium constant in terms of pressure

It is denoted by \sf\:K_pKp

For a Genral Reaction :

\:\rightleftharpoons\:cC+dDaA+bB⇌cC+dD