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First Order Reaction

N2O4 dissocia...

Question

$N_{2} O_{4}$ dissociates as $N_{2} O_{4} ⇌ 2 N O_{2}$ . At $55^{\circ} C$ and one atmosphere, $%$ decomposition of $N_{2} O_{4}$ is $50.3 %$ . At what $P$ and same temperature, the equilibrium mixture will have the ratio of $N_{2} O_{4} : N O_{2}$ as $1 : 8$ ?

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Solution

$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
At equilibrium $(1 - x)$ $2 x$
$p_{N_{2} O_{4}} = \frac{(1 - x)}{(1 + x)} \times P; p_{N O_{2}} = \frac{2 x}{(1 + x)} \times P$
$K_{p} = \frac{{(\frac{2 x}{(1 + x)} . P)}^{2}}{(\frac{(1 - x)}{(1 + x)} . P)} = \frac{4 x^{2} P}{1 - x^{2}}$
Given $x = 0.503$ and $P = 1$
$∴ K_{p} = 1.3548$
CASE II:
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
$(1 - x)$ $2 x$
given $\frac{1 - x}{2 x} = \frac{1}{8}$
$x = 0.8$
Let the new pressure be $P$ atm
$K_{p} = \frac{4 x^{2} P}{1 - x^{2}} = 1.3548$
$P = 0.19 a t m$

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Similar questions

N_{2} O_{4}

dissociates as,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

55^{o} C

1.2

%

decomposition of

N_{2} O_{4}

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At what pressure and same temperature, the equilibrium mixture has the ratio of

N_{2} O_{4} : N O_{2}

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?
(Report the answer upto two decimal)

N_{2} O_{4}

dissociates as,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

55^{o} C

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N_{2} O_{4}

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At what pressure and same temperature, the equilibrium mixture has the ratio of

N_{2} O_{4} : N O_{2}

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?
(Report the answer upto two decimal)

1

N_{2}, N O_{2}

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. On heating to a temperature at which all the

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