The degree of dissociation α of the reaction N2O4(g)leftharpoons 2NO2(g) Can be related to Kp as [Given: Total pressure at equilibrium = P]

Question

The degree of dissociation α of the reaction N2O4(g)leftharpoons 2NO2(g) Can be related to Kp as [Given: Total pressure at equilibrium = P] (A) α =((KP/P)/4 + (KP)P) (B) α =(KP/4 + KP) (C) α =[(KP / P/4 + KP / P)]1 / 2 (D) α =[(KP/4 + KP)]1 / 2

Tardigrade Admin · Accepted Answer

N2O4(g)leftharpoons2NO2(g) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-q8rtojxvmuy1br9d.jpg" /> KP=((2 α )2/(1 - α ))× ((P/1 + α ))1 KP=(4 α 2/1 - α 2)P α =[(KP / P/4 + KP / P)]1 / 2

Q. The degree of dissociation $α$ of the reaction
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
Can be related to $K_{p}$ as [Given: Total pressure at equilibrium $= P$ ]

$α = \frac{\frac{K _{P}}{P}}{4 + \frac{K _{P}}{P}}$

$α = \frac{K _{P}}{4 + K _{P}}$

$α = [\frac{K _{P} / P}{4 + K _{P} / P}]^{1/2}$

$α = [\frac{K _{P}}{4 + K _{P}}]^{1/2}$

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$K_{P} = \frac{( 2 α ) ^{2}}{( 1 - α )} \times (\frac{P}{1 + α})^{1}$
$K_{P} = \frac{4 α ^{2}}{1 - α ^{2}} P$
$α = [\frac{K _{P} / P}{4 + K _{P} / P}]^{1/2}$

Q. The degree of dissociation α of the reaction N2​O4​(g)⇌2NO2​(g) Can be related to Kp​ as [Given: Total pressure at equilibrium =P]

α=4+PKP​​PKP​​​

α=4+KP​KP​​

α=[4+KP​/PKP​/P​]1/2

α=[4+KP​KP​​]1/2

N2​O4​(g)⇌2NO2​(g) KP​=(1−α)(2α)2​×(1+αP​)1 KP​=1−α24α2​P α=[4+KP​/PKP​/P​]1/2