For the equilibrium N2 - 3H2 - 2NH3- K6 at 1000 K is 2-37 - 10-3Equilibrium -N2- - 2-M -H2- - 3Mthe concentration of NH3 is-

The correct option is A 0.00358 Mwe have, K=[H_2]^3[N_2]^1[NH_3]^2 2.37× 10^ 3=[3M]^3[2M]^1[xM]^2 [xM]^2=[3M]^3[2M]^12.37× 10^ 3 solving the above ...

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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

For the equil...

Question

For the equilibrium $N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{6}$ at 1000 K is $2.37 \times 10^{- 3}$
Equilibrium $[N_{2}] = 2. M [H_{2}] = 3 M$ the concentration of $N H_{3}$ is:

0.00358 M

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0.0358 M

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0.358 M

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3.58 M

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Solution

The correct option is A 0.00358 M
we have,

$K = \frac{[H_{2}]^{3} [N_{2}]^{1}}{[N H_{3}]^{2}}$

$2.37 \times 10^{- 3} = \frac{[3 M]^{3} [2 M]^{1}}{[x M]^{2}}$

$[x M]^{2} = \frac{[3 M]^{3} [2 M]^{1}}{2.37 \times 10^{- 3}}$

solving the above equation, we get,

concentration of $N H_{3} = 0.00358 M$

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Similar questions

Q. For the equilibrium

N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{c}

at 1000

K

2.37 \times 10^{- 3}

. If at equilibrium

[N_{2}] = 2 M, [H_{2}] = 3 M

, the concentration of

N H_{3}

is:

For the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, the equilibrium expression may be written as

K \frac{{[N H_{3}]}_{3}^{2}}{[N_{2}] {[H_{2}]}_{2}^{3}}

For the reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, the equilibrium expression may be written as:

K = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}

Q. In a mixture of

N_{2}

and

H_{2}

in the ratio 1:3 at 30 atm and

300^{0} C

, the % of

{N H}_{3}

at equilibrium is 17.8.
Calculate

K_{P}

for

{N_{2} + {3 H}_{2} ⇌ 2 N H}_{3}

Q. A mixture of

N_{2}

and

H_{2}

in the volume ratio

1 : 3

is allowed to attain equilibrium :

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

670 K

and 1000 atm, 0.2491 mole fraction of

N H_{3}

is formed.

K_{p}

X \times 10^{- 6} {a t m}^{- 2}

100 X

is__________.

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For the equilibrium N2+3H2⇌2NH3,K6 at 1000 K is 2.37×10−3Equilibrium [N2]=2.M[H2]=3Mthe concentration of NH3 is:

The correct option is A 0.00358 Mwe have,K=[H2]3[N2]1[NH3]22.37×10−3=[3M]3[2M]1[xM]2[xM]2=[3M]3[2M]12.37×10−3solving the above equation, we get,concentration of NH3=0.00358M

For the equilibrium $N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{6}$ at 1000 K is $2.37 \times 10^{- 3}$
Equilibrium $[N_{2}] = 2. M [H_{2}] = 3 M$ the concentration of $N H_{3}$ is: