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Reaction Quotient

A 100 dm3 f...

Question

A $100 d m^{3}$ flask contains $10$ moles each of $N_{2}$ and $H_{2}$ at $777 K$ . After equilibrium was reached, the partial pressure of $H_{2}$ was $1 a t m$ . At this point, $5 l i t r e s$ of $H_{2} O_{(I)}$ was injected and the gas mixture was cooled to $298 K$ . Find out the pressure of the gaseous mixture left.

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Solution

$N_{2} + 3 H_{2} \to 2 N H_{3}$
At $t = 0;$ $1010$
At $t = t_{e q}$ $(10 - x) (10 - 3 x) 2 x$
Given,
$P_{H_{2}} = 1 a t m$

$P V = n R T$
$n_{H_{2}} = \frac{P V}{R T} = \frac{1 \times 100}{0.0821 \times 777} = 1.74 m o l e s$

$10 - 3 x = 1.74$

$x = 2.75 m o l e s$

$n_{N_{2}} = (10 - x) = 7.25 m o l e s$

$N H_{3} + H_{2} O \to N H_{4} O H$ (completely dissolved)

Volume remaining= $100 - 5 = 95 l i t r e s$

Total number of moles= $n_{H_{2}} + n_{N_{2}} = 1.74 + 7.25 = 8.99 m o l e s$
$P V = n R T$

Substituting the values we get $P = 2.26 a t m$
$P = 2 a t m$

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