$N H_{3}$ is heated at 15 atm from $27^{0} C$ to $347^{0} C$ assuming volume constant. The new pressure becomes to atm at equilibrium of the reaction $2 N H_{3} ⇌ N_{2} + 3 H_{2}$ % of mole of $N H_{3}$ actually decomposed in the reaction is:

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Solution

$2 N H_{3}$ $⇌$ $N_{2}$ $+$ $3 H_{2}$
Initial mole $a$ $0$ $0$
Moles at eqm. $a - 2 x$ $x$
$3 x, T o t a l = a + 2 x$
Pressure of $a$ moles of $N H_{3}$ at $27^{o} C = 15 a t m$
Pressure of $a$ moles of $N H_{3}$ at $347^{o} C$
$= P$ atm (say)
As volume remains constant, $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\frac{15}{300} = \frac{P}{620}$ or $P = 31 a t m$
Now at $327^{o} C$ and constant volume,
Pressure $\propto$ No. of moles
$∴ 31 \propto a$
$50 α a + 2 x$
$∴ \frac{a + 2 x}{a} = \frac{50}{31}$ or $x = \frac{19}{62} a$
$%$ of $N H_{3}$ decompose
$= \frac{2 x}{a} \times 100 = 2 \times \frac{19 a}{62} \times \frac{1}{a} \times 100 = 61.3 %$

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Ammonia kept under pressure of 15 atm at 27⁰C is heated to 347⁰C in a closed vessel in the presence of catalyst. Under these conditions, NH₃ is partially decomposed according to the equation, $2 N H_{3} ⇋ N_{2} + 3 H_{2}$ . The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the percentage of NH₃ actually decomposed.

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15 a t m

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347^{o} C

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50 a t m

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Ammonia under a pressure of 15 atm at 27⁰C is heated to 347⁰C in a closed vessel in the presence of catalyst. Under the conditions, NH₃ is partially decomposed according to the

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atm at

27^{\circ} C

is heated to

347^{\circ} C

in a closed vessel in the presence of a catalyst. Under the conditions,

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	$2 N H_{3}$	$⇌$	$N_{2}$	$+$	$3 H_{2}$
Initial mole	$a$		$0$		$0$
Moles at eqm.	$a - 2 x$		$x$		$3 x, T o t a l = a + 2 x$

NH3 is heated at 15 atm from 270C to 3470C assuming volume constant. The new pressure becomes to atm at equilibrium of the reaction 2NH3⇌N2+3H2 % of mole of NH3 actually decomposed in the reaction is:

$N H_{3}$ is heated at 15 atm from $27^{0} C$ to $347^{0} C$ assuming volume constant. The new pressure becomes to atm at equilibrium of the reaction $2 N H_{3} ⇌ N_{2} + 3 H_{2}$ % of mole of $N H_{3}$ actually decomposed in the reaction is: