For the reaction equilibrium- 2 NOBr g - 2 NO g - Br 2 g - if P Br 2- P -9 at equilibrium and - P - is total pressure- the ratio is equal to-A- 1 - 9B- 1 - 27C- 1 - 81D- 1 - 3

The correct option is C 1/81 For the reaction 2NOB_r(g) ⇌ 2NO_g + Br_2(g) If P_Br_2 = P/9 then P_NO = 2P/9 (∴ NO. of moles NO = twice the no. of moles ...

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Standard XII

Chemistry

Effective Equilibrium Constant

For the react...

Question

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

1/9

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1/81

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1/27

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1/3

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Solution

The correct option is B
1/81

For the reaction $2 N O B_{r} (g) ⇌ 2 N O_{g} + B r_{2 (g)}$

If $P_{B r_{2}} = \frac{P}{9} t h e n P_{N O} = \frac{2 P}{9}$

$(∴ N O . o f m o l e s N O = t w i c e t h e n o . o f m o l e s o f B r_{2})$

$P_{N O B r} = P - (P_{N O} + P_{B r_{2}}) = P - (\frac{2 p}{9} + \frac{P}{9}) = \frac{2 P}{3}$

$K_{P} = \frac{P_{N O}^{2} \times P_{B r_{2}}}{P_{N O B r}^{2}} = \frac{(\frac{2 P}{9})^{2} \times \frac{P}{9}}{(\frac{2 P}{3})^{2}}; \frac{K_{P}}{P} = \frac{1}{81}$

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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)

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For the reaction equilibrium; 2NOBr(g)⇌2NOg+Br2(g), if PBr2=P9 at equilibrium and 'P' is total pressure, the ratio is equal to:

The correct option is B 1/81 For the reaction 2NOBr(g)⇌2NOg+Br2(g) If PBr2=P9 then PNO=2P9 (∴NO. of moles NO=twice the no. of moles of Br2) PNOBr=P−(PNO+PBr2)=P−(2p9+P9)=2P3 KP=P2NO×PBr2P2NOBr=(2P9)2×P9(2P3)2;KPP=181

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: