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Derivation of Kp and Kc

Consider the ...

Question

Consider the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
The equilibrium constant of the above reaction is $K_{p} .$ If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by: (Assume that ) $P_{N H_{3}} << P_{t o t a l}$ at equilibrium)

\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{2^{2}}

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\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{2^{4}}

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\frac{K_{p}^{1 / 2} P^{2}}{2^{2}}

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\frac{K_{p}^{1 / 2} P^{2}}{2^{4}}

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Solution

The correct option is B $\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{2^{4}}$
$\begin{matrix} 2 N H_{3} & ⇌ & N_{2} & + & 3 H_{2} t = 0 & P_{1} & 0 & 0 t = e q & P_{N H_{3}} & \frac{P_{1}}{2} & \frac{3 P_{1}}{2} \end{matrix}$
$P = P_{t o t a l} = \frac{P_{1}}{2} + \frac{3 P_{1}}{2} = 2 P_{1}$
$∴ P_{1} = \frac{P}{2}$
$k_{e q} = \frac{(P_{N_{2}}) (P_{H_{2}})}{(P_{N H_{3}})^{2}} \Rightarrow \frac{1}{k_{p}} = \frac{(\frac{P_{1}}{2}) {(\frac{3 P_{1}}{2})}^{3}}{(P_{N H_{3}})^{2}}$
$∴ \frac{(P_{N H_{3}})^{2}}{K_{P}} = \frac{P^{4}}{4^{4}} \times 3^{3}$
$∴ P_{N H_{3}}^{2} = K_{p} \frac{P^{4}}{4^{4}} \times 3^{3}$
$P_{N H_{3}} = {(K_{p} \frac{P^{4}}{4^{4}} 3^{3})}_{p}^{\frac{1}{2}} = \frac{(K_{p})^{\frac{1}{2}} P^{2} 3^{\frac{3}{2}}}{2^{4}}$

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Similar questions

Q. Consider the reaction,

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

The equilibrium constant of the above reaction is

K_{P}

. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by:

(Assume that

P_{N H_{3}} << P_{t o t a l}

at equilibrium)

Q. A container containing ammonia is allowed to decompose and the following equilibrium is established.

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

.
Then, the equilibrium constant,

K_{p}

is expressed as:
(Where

p_{N_{2}}

= partial pressure of

N_{2}

and

p_{N H_{3}}

is the partial pressure of

N H_{3}

)

Q. If a mixture of

N_{2}

and

H_{2}

in the ratio 1 : 3 at 50 atmosphere and

650^{o} C

is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant (

K_{P}

) for the reaction.

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

Q. For the equilibrium,

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

the equilibrium constant,

K_{p}

is expressed as:

For the reaction 2 $N H_{3}$ (g) ⇌ $N_{2}$ (g) + 3 $H_{2}$ (g), if the total equilibrium pressure is P and degree of dissociation is α, the equilibrium constant $K_{p}$ is given by

$\frac{27 α^{4} p^{2}}{16 {(1 - α^{2})}^{2}}$

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Derivation of Kp and Kc

Standard XII Chemistry

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Consider the reaction N2(g)+3H2(g)⇌2NH3(g) The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by: (Assume that ) PNH3<<Ptotal at equilibrium)

The correct option is B 33/2K1/2pP2242NH3⇌N2+3H2t=0P100t=eqPNH3P123P12 P=Ptotal=P12+3P12=2P1 ∴P1=P2 keq=(PN2)(PH2)(PNH3)2⇒1kp=(P12)(3P12)3(PNH3)2 ∴(PNH3)2KP=P444×33 ∴P2NH3=KpP444×33 PNH3=(KpP44433)12=(Kp)12P233224

Consider the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
The equilibrium constant of the above reaction is $K_{p} .$ If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by: (Assume that ) $P_{N H_{3}} << P_{t o t a l}$ at equilibrium)