For the reaction equilibrium- 2NOBrg-2NOg-Br2g-If PB1-P9 at equilibrium and P is total pressure- the ratio of KPP is equal to-

The correct option is A 181 K _ P = P _ NO ^ 2 . P _ Br P _ NOBr ^ 2 2NOBr⇌ 2NO+Br #160; #160; #160; #160; =( 2× P/9 ) #160 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

For the react...

Question

For the reaction equilibrium, $2 N O B r (g) ⇋ 2 N O (g) + B r_{2} (g)$ .
If $P_{B_{1}} = \frac{P}{9}$ at equilibrium and $P$ is total pressure, the ratio of $\frac{K_{P}}{P}$ is equal to:

\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 A $\frac{1}{81}$
$K_{P} = \frac{{P_{N O}}^{2} . P_{B r}}{{P_{N O B r}}^{2}} 2 N O B r ⇌ 2 N O + B r$

$= \frac{{(2 \times P / 9)}^{2} \times P / 9}{{(2 \times P / 3)}^{2}}$

$= P / 81$

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

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Similar questions

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 B r (g) + {B r}_{2} (g)

P_{{B r}_{2}} = \frac{P}{9}

at equilibrium and P is total pressure, the ratio of

\frac{K_{p}}{P}

is equal to:

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:

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)⇋2NO(g)+Br2(g).If PB1=P9 at equilibrium and P is total pressure, the ratio of KPP is equal to:

The correct option is A 181KP=PNO2.PBrPNOBr22NOBr⇌2NO+Br =(2×P/9)2×P/9(2×P/3)2 =P/81KPP=181

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:

The correct option is A $\frac{1}{81}$
$K_{P} = \frac{{P_{N O}}^{2} . P_{B r}}{{P_{N O B r}}^{2}} 2 N O B r ⇌ 2 N O + B r$

$= \frac{{(2 \times P / 9)}^{2} \times P / 9}{{(2 \times P / 3)}^{2}}$

$= P / 81$

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