For the reaction equilibrium- 2NOBr g - 2NOBr g - Br 2 g If P Br 2 - P 9 at equilibrium and P is total pressure- the ratio of K p P is equal to-

The correct option is D 1 81 2NO Br _ 2( g ) #160; ⇌ 2NO+ Br _ 2 It is given that Br_2 has a partial pressure of P/9 , therefore partia ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

For the react...

Question

For the reaction equilibrium, $2 N O B r (g) ⇌ 2 N O B r (g) + {B r}_{2} (g)$
If $P_{{B r}_{2}} = \frac{P}{9}$ at equilibrium and P is total pressure, the ratio of $\frac{K_{p}}{P}$ is equal to:

\frac{1}{9}

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\frac{1}{81}

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\frac{1}{27}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{81}$
$2 N O {B r}_{2 (g)} ⇌ 2 N O + {B r}_{2}$

It is given that $B r_{2}$ has a partial pressure of $P / 9$ , therefore partial pressure for $N O B r$ must be $2 P / 9$
Given total pressure $P = \frac{2 P}{9} + \frac{P}{9} + P_{N O B r_{2}}$

$P_{N O B r_{2}} = \frac{2 P}{3}$

$K_{P} = \frac{{P_{N O}}^{2} . P_{{B r}_{2}}}{{P_{{N O B r}_{2}}}^{2}} = \frac{{(2 \times P / 9)}^{2} \times P / 9}{{(2 \times P / 3)}^{2}}$

$\frac{K_{P}}{P} = \frac{1}{81}$

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Similar questions

For the reaction equilibrium; $2 N O B r_{(g)} ⇌ 2 N O_{g} + B r_{2 (g)}$ , if $P_{B r_{2}} = \frac{P}{9}$ at equilibrium and 'P' is total pressure, the ratio is equal to:

Q. For the reaction equilibrium

2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)}

, if

P_{B r_{2}} = \frac{P}{9}

at equilibrium and P is the initial total pressure, then the ratio

\frac{K_{P}}{P}

is equal to :

Q. For the reaction equilibrium,

2 N O B r (g) ⇋ 2 N O (g) + B r_{2} (g)

.
If

P_{B_{1}} = \frac{P}{9}

at equilibrium and

P

is total pressure, the ratio of

\frac{K_{P}}{P}

is equal to:

2 N O B r (g) ⇌ 2 N O (g) + {B r}_{2} (g)

. At equilibrium if nitrosyl bromide (

N O B r

) is

40

% dissociated at certain temp. and a total pressure of

0.30 a t m

K_{p}

for the reaction

2 N O (g) + {B r}_{2} (g) ⇌ 2 N O B r (g)

is:

2 N O B r (g) ⇌ 2 N O (g) + B r_{2} (g)

. If nitrosyl bromide (NOBr) is 40% dissociated at a certain and a total pressure of 0.30 atm.

K_{p}

for the reaction is:

2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)

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For the reaction equilibrium, 2NOBr(g)⇌2NOBr(g)+Br2(g)If PBr2=P9 at equilibrium and P is total pressure, the ratio of KpP is equal to:

The correct option is D 1812NOBr2(g)⇌2NO+Br2It is given that Br2 has a partial pressure of P/9, therefore partial pressure for NOBr must be 2P/9Given total pressure P=2P9+P9+PNOBr2 PNOBr2=2P3KP=PNO2.PBr2PNOBr22=(2×P/9)2×P/9(2×P/3)2KPP=181

For the reaction equilibrium, $2 N O B r (g) ⇌ 2 N O B r (g) + {B r}_{2} (g)$
If $P_{{B r}_{2}} = \frac{P}{9}$ at equilibrium and P is total pressure, the ratio of $\frac{K_{p}}{P}$ is equal to: