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Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

1 mole of N...

Question

1 mole of $N_{2}$ , and 3 moles of $H_{2}$ are placed in 1L vessel. Find the concentration of $N H_{3}$ at equilibrium, it 4 equilibhum pressure is 1 atm and the equilibrium constant at 400 K is $\frac{4}{27}$ .

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Solution

Given, $v o l u m e = 1 l$ $T = 400 K$ $K_{c} = \frac{4}{27}$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
Identitely: $1$ $3$ -
Eq $1 - x$ $3 (1 - x)$ $2 x$
$K_{c} = \frac{[N H_{3}]^{2}}{[H_{2}]^{3} [N_{2}]} \Rightarrow \frac{4}{27} = \frac{(2 x)^{2}}{(1 - x) (3 - 3 x)^{3}}$
On solving we get $x = 0.381 M$
Concentration of $N H_{3} = 2 x = 2 (0.381) M$
$= 0.762 M$

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1 mole of N2, and 3 moles of H2 are placed in 1L vessel. Find the concentration of NH3 at equilibrium, it 4 equilibhum pressure is 1 atm and the equilibrium constant at 400 K is 427.

Given, volume=1l T=400K Kc=427 N2+3H2⇌2NH3Identitely: 1 3 -Eq 1−x 3(1−x) 2xKc=[NH3]2[H2]3[N2]⇒427=(2x)2(1−x)(3−3x)3On solving we get x=0.381MConcentration of NH3=2x=2(0.381)M=0.762M

1 mole of $N_{2}$ , and 3 moles of $H_{2}$ are placed in 1L vessel. Find the concentration of $N H_{3}$ at equilibrium, it 4 equilibhum pressure is 1 atm and the equilibrium constant at 400 K is $\frac{4}{27}$ .