For the dissociation reaction N 2 O 4 g - 2 NO 2 g - the degree of dissociation - in terms of K p and total equilibrium pressure P is-A- -4 P - K P - K p B- - K P -4 P - K p C- -KP-4 PD- None of these

The correct option is B α = √(K_P4P + K_P)[ amp; N_2O_4(g) amp; ⇌ amp; 2NO_2(g); ...

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Byju's Answer

Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

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 $α = \sqrt{\frac{K_{P}}{4 P + K_{P}}}$
$\begin{matrix} N_{2} O_{4} (g) & ⇌ & 2 N O_{2} (g) t = 0 & 1 m o l & 0 a t e q . & 1 - α & 2 α & n_{f} = 1 + α \end{matrix}$
$mole fraction 1−α1+α 2α1+α$
$P a r t i a l p r e s s u r e (\frac{1 - α}{1 + α}) P (\frac{2 α}{1 + α}) P$
$K_{P} = \frac{(P_{N O_{2}})^{2}}{P_{N_{2} O_{4}}} = \frac{{(\frac{2 α}{1 + α})}^{2} P^{2}}{(\frac{1 - α}{1 + α}) P}$
$= \frac{4 α^{2}}{(1 + α) (1 - α)} P$
on solving we will get,
$α^{2} = \frac{K_{p}}{4 P + K_{p}}$
$α = \sqrt{\frac{K_{p}}{4 P + K_{p}}}$

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Similar questions

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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:

The correct option is B α=√KP4P+KP N2O4(g)⇌2NO2(g)t=0 1 mol0at eq. 1−α 2αnf=1+α mole fraction 1−α1+α 2α1+α Partial pressure(1−α1+α)P (2α1+α)P KP=(PNO2)2PN2O4=(2α1+α)2P2(1−α1+α)P =4α2(1+α)(1−α)P on solving we will get, α2=Kp4P+Kp α=√Kp4P+Kp

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: