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Concentration Terms - An Introduction (w/w etc)

1 mole of N...

Question

$1$ mole of $N_{2}$ and $3$ moles of $H_{2}$ are placed in a closed container at a pressure of $4$ atm. The pressure falls to $3$ atm at the same temperature when the following equilibrium is attained: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ .
The equilibrium constant $K_{p}$ for dissociation of $N H_{3}$ is:

\frac{1}{0.5} \times (1.5)^{3}

atm

^{- 2}

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0.5 \times (1.5)^{3}

atm

^{2}

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\frac{(0.5) \times (1.5)^{3}}{3 \times 3}

atm

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\frac{3 \times 3}{0.5 \times (1.5)^{3}}

atm

^{- 2}

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Solution

The correct option is A $0.5 \times (1.5)^{3}$ atm $^{2}$
Initially, total $4$ moles are present at $4$ atm pressure.
At equilibrium, the pressure is $3$ atm.
Total number of moles at equilibrium are $\frac{4 a t m}{4 m o l e s} \times 3 a t m = 3 m o l e s$

$N_{2}$
$H_{2}$
$N H_{3}$
Initial moles
$1$
$3$
$0$
Moles at equilibrium $1 - x$
$3 - 3 x$
$2 x$

Because, total number of moles at equilibrium $= 3 = 1 - x + 3 - 3 x + 2 x$
or, $x = 0.5$

Moles of $N_{2}$ = 0.5, moles of $H_{2} = 1.5,$ moles of $N H_{3} = 1$

Hence, partial pressure of $N_{2} = 0.5, H_{2} = 1.5, N H_{3} = 1$

The expression for the equilibrium constant is $K_{P} = \frac{P_{N_{2}} \times P_{H_{2}}^{3}}{P_{N H_{3}}^{2}} = \frac{0.5 \times (1.5)^{3}}{1^{2}} = 0.5 \times (1.5)^{3}$ atm $^{2}$

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Similar questions

Q. 1 mole

N_{2}

and 3 mol

H_{2}

are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

. The equilibrium constant

K_{p}

for dissociation of

N H_{3}

is:

1

mole

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and

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are placed in a closed container at a pressure of

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

. The equilibrium constant

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Q. For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, one mole of

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and three moles of

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Q. In a reaction mixture containing

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Q. If

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	$N_{2}$	$H_{2}$	$N H_{3}$
Initial moles	$1$	$3$	$0$
Moles at equilibrium	$1 - x$	$3 - 3 x$	$2 x$

1 mole of N2 and 3 moles of H2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained: N2(g)+3H2(g)⇌2NH3(g). The equilibrium constant Kp for dissociation of NH3 is: