For the dissociation reaction N2O4g - 2NO2g- the degree of dissociation - in terms of Kp and total equilibrium pressure P is -

The correct option is B α = √(K_P4P) C N2O4(g)NX2OX4(g) #160;say initially is #160;P #160;atmP #160;atm.PV=nRTPV=nRTP(1)=(36.8/92)(0.082 ...

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Standard XII

Chemistry

Effective Equilibrium Constant

For the disso...

Question

For the dissociation reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the degree of dissociation $(α)$ in terms of $K_{p}$ and total equilibrium pressure P is :

α = \sqrt{\frac{4 P + K_{P}}{K_{P}}}

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α = \sqrt{\frac{K_{P}}{4 P + K_{P}}}

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α = \sqrt{\frac{K_{P}}{4 P}}

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none of these

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Solution

The correct option is B $α = \sqrt{\frac{K_{P}}{4 P}}$
$C$
$
N2O4(g) say initially is P atm.
$P V = n R T$
$P (1) = (36.8 / 92) (0.082) (300 K)$
$P = 9.84 a t m$
Now we have:
$N 2 O 4 (g) ↽ - - ⇀ 2 N O 2 (g)$
Considering initial pressure of N2O4 as p, at equilibrium, assuming 'x' as the degree of dissociation, we will have pressure of N2O4 to be p(1-x) and pressure of NO2 as 2px.
So,
$K p = [ N O 2 ] 2 [ N 2 O 4 ]$
$K p = ( 2 p x ) 2 p ( 1 - x )$
$K p = 4 p x 2 ( 1 - x )$
assuming 1−x approximately as 1,
$K p = 4 p x 2$ $$$

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For the dissociation reaction N2O4(g)⇌2NO2(g), the degree of dissociation (α) in terms of Kp and total equilibrium pressure P is :

For the dissociation reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the degree of dissociation $(α)$ in terms of $K_{p}$ and total equilibrium pressure P is :