For the dissociation reaction
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ . The degree dissociation (1) in terms of $K_{p}$ and total equilibrium pressure is:

α = \sqrt{\frac{4 P + K_{p}}{K_{p}}}

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α = \sqrt{\frac{K_{p}}{4 P}}

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α = \sqrt{\frac{K_{p}}{4 P + K_{p}}}

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none of these

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Solution

The correct option is B $α = \sqrt{\frac{K_{p}}{4 P}}$
$N_{2} O_{4} ⇌ 2 {N O}_{2}$
Initial $1$ $-$
at equilibrium $1 - α$ $2 α$
Total mass of gas $= 1 + α$
${P N}_{2} O_{4} = \frac{P (1 - α)}{(1 + α)} {P N O}_{2} = \frac{P (2 α)}{(1 + α)}$
$K_{P} = \frac{{({P N O}_{2})}^{2}}{({P N}_{2} O_{4})} = \frac{\frac{P (2 α)}{(1 + α)} . \frac{P (2 α)}{(1 + α)}}{\frac{P (1 + α)}{1 + α}}$
$K_{P} = \frac{P^{2} {(2 α)}^{2}}{P (1 - α)} α <<< 1$
$K_{P} = 4 {P α}^{2}$
$\frac{K_{P}}{4 P} = α^{2}$
$α = \sqrt{\frac{K_{P}}{4 P}}$

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Similar questions

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For the dissociation reactionN2O4(g)⇌2NO2(g). The degree dissociation (1) in terms of Kp and total equilibrium pressure is:

The correct option is B α=√Kp4P N2O4⇌2NO2Initial 1 −at equilibrium 1−α 2αTotal mass of gas =1+αPN2O4=P(1−α)(1+α)PNO2=P(2α)(1+α)KP=(PNO2)2(PN2O4)=P(2α)(1+α).P(2α)(1+α)P(1+α)1+αKP=P2(2α)2P(1−α)α<<<1KP=4Pα2KP4P=α2α=√KP4P

For the dissociation reaction
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ . The degree dissociation (1) in terms of $K_{p}$ and total equilibrium pressure is: