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Derivation of Kp and Kc

2NH3g ⇌ N2g +...

Question

$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ If 'a' moles of $N H_{3}$ are given initially. Let x moles of $N H_{3}$ are decomposed at equilbrium and $V l i t r e$ be the capacity of vessel in which reaction is being studied. Find out the value of $K_{c}$ .

\frac{27 x^{2}}{16 V^{2} (a - x)}

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\frac{27 x^{4}}{16 V^{2} (a - x)^{2}}

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\frac{27 x^{4}}{16 V^{4} (a - x)^{2}}

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\frac{27 x^{2}}{16 V^{2} (a - x)^{2}}

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Solution

The correct option is B $\frac{27 x^{4}}{16 V^{2} (a - x)^{2}}$
$\begin{matrix} 2 N H_{3} & ⇌ & N_{2} & + & 3 H_{2} Initial moles & a & 0 & 0 Moles at equilbrium & a - x & \frac{x}{2} & \frac{3 x}{2} Concentration at equilbrium & \frac{a - x}{V} & \frac{x}{2 V} & \frac{3 x}{2 V} \end{matrix}$
$K_{c} = \frac{[N_{2}] [H_{2}]^{3}}{[N H_{3}]^{2}} = \frac{(\frac{x}{2 V}) {(\frac{3}{2} \frac{x}{V})}^{3}}{{(\frac{a - x}{V})}^{2}} = \frac{27 x^{4}}{16 V^{2} (a - x)^{2}}$

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Similar questions

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

If 'a' moles of

N H_{3}

are given initially. Let x moles of

N H_{3}

are decomposed at equilbrium and

V l i t r e

be the capacity of vessel in which reaction is being studied. Find out the value of

K_{c}

Q. In the decomposition reaction of ammonia:

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

2

moles of

{N H}_{3}

are introduced in the vessel of

1

litre. At equilibrium,

1

mole

{N H}_{3}

was left, the value of

K_{c}

will be:

2

moles of

N_{2}

are mixed with

6

moles of

H_{2}

in closed vessel of one litre capacity. If

50

N_{2}

is converted into

{N H}_{3}

at equilibrium, the value of

K_{c}

for the reaction,

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

Q. In the reversible reaction

2 N H_{3} (g) \Leftrightarrow N_{2} (g) + 3 H_{2} (g)

when 1 mol of

N H_{3}

is taken initially in a litre flask and 0.2 moles of

N_{2}

are formed at equilibrium. Its equilibrium constant

K_{c}

is:

Q. Two moles of

{N H}_{3}

gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature as

{2 N H}_{3} (g) ⇋ . N_{2} (g) + {3 H}_{2} (g)

. At equilibrium, one mole of

{N H}_{3} (g)

remain. The value of

K_{C}

is:

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Derivation of Kp and Kc

Standard XII Chemistry

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2NH3(g)⇌N2(g)+3H2(g) If 'a' moles of NH3 are given initially. Let x moles of NH3 are decomposed at equilbrium and V litre be the capacity of vessel in which reaction is being studied. Find out the value of Kc.

The correct option is B 27x416V2(a−x)22NH3⇌N2+3H2Initial molesa00Moles at equilbriuma−xx23x2Concentration at equilbriuma−xVx2V3x2V Kc=[N2][H2]3[NH3]2=(x2V)(32xV)3(a−xV)2=27x416V2(a−x)2

$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ If 'a' moles of $N H_{3}$ are given initially. Let x moles of $N H_{3}$ are decomposed at equilbrium and $V l i t r e$ be the capacity of vessel in which reaction is being studied. Find out the value of $K_{c}$ .