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Standard XII

Chemistry

Law of Mass Action

In a three li...

Question

In a three litre vessel, 2 moles of $N_{2}$ and 3 moles of length $H_{2}$ are taken. If at equilibrium x moles of $N_{2}$ were obtained then the concentration of $N H_{3}$ at equilibrium will be:

2 - x

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4 - 2 x

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\frac{2 x}{3}

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\frac{4 - 2 x}{3}

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Solution

The correct option is B $\frac{2 x}{3}$
$N_{2} + 3 H_{2} \to 2 N H_{3}$

given data volume $= 3 l i t$

Initial 2mole 3(3mole) 0mole
equilibrium $2 - x$ $9 - x$ $2 x$

concentration of $N_{2} = \frac{2 - x}{3}$

concentration of $N H_{3} = \frac{2 x}{3}$

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Similar questions

Q. At a certain temperature (T), the equilibrium constant

(K_{c})

is 1 for the reaction

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

If 2 moles of

N_{2},

4 moles of

H_{2},

6 moles of

N H_{3}

and 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equilibrium concentration of

H_{2}

and

N H_{3}

are equal then what is the equilibrium concentration of

N_{2}

(in M)?

Q. 1 mole of

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and 3 moles of

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are placed in 1L vessel. Find the concentration of

{N H}_{3}

at equilibrium if the equilibrium constant

(K_{c})

at 400K is

\frac{4}{27}

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

Q. A mixture of

2

moles of

N_{2}

and

8

moles of

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are heated in a

2

lit vessel, till equilibrium is established. At equilibrium

0.4

moles of

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was present. What will be the equilibrium concentration of

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Q. 1 mole of

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, and 3 moles of

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are placed in 1L vessel. Find the concentration of

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at equilibrium, it 4 equilibhum pressure is 1 atm and the equilibrium constant at 400 K is

\frac{4}{27}

Q. A mixture of two moles of

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In a three litre vessel, 2 moles of N2 and 3 moles of length H2 are taken. If at equilibrium x moles of N2 were obtained then the concentration of NH3 at equilibrium will be:

The correct option is B 2x3N2+3H2→2NH3 given data volume =3litInitial 2mole 3(3mole) 0mole equilibrium 2−x 9−x 2x concentration of N2=2−x3 concentration of NH3=2x3

In a three litre vessel, 2 moles of $N_{2}$ and 3 moles of length $H_{2}$ are taken. If at equilibrium x moles of $N_{2}$ were obtained then the concentration of $N H_{3}$ at equilibrium will be:

The correct option is B $\frac{2 x}{3}$
$N_{2} + 3 H_{2} \to 2 N H_{3}$

given data volume $= 3 l i t$

Initial 2mole 3(3mole) 0mole
equilibrium $2 - x$ $9 - x$ $2 x$

concentration of $N_{2} = \frac{2 - x}{3}$

concentration of $N H_{3} = \frac{2 x}{3}$