For the equilibrium- N 2 g -3 H 2 g - 2 N H 3 g - the equilibrium constant- K p is expressed as-K p -33- p N 22-3 P N H 3B- Kp-33- pN2 P H 23- P NH 32C- Kp-33- pN24-pN2 H3Kp-33 - 2- pN24-p NH 32

The correct option is C K_p = 3^3× p_N_2^4p_NH_3^2N_2 + 3H_2 ⇌ 2NH_3 K_p = p_NH_3^2p_N_2.p_H_2^3 ⇒ K_p = p_NH_3^2p_N_2× (3p_N_2)^3 [∵ 3p_N_2 = p_H_2] ...

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Standard XII

Chemistry

Effective Equilibrium Constant

For the equil...

Question

For the equilibrium, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g),$ the equilibrium constant, $K_{p}$ is expressed as:

K_{p} = \frac{3^{3} \times p_{N_{2}}^{2}}{p_{N H_{3}}}

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K_{p} = \frac{3^{3} \times p_{N_{2}} p_{H_{2}}^{3}}{p_{N H_{3}}^{2}}

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K_{p} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}

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K_{p} = \frac{3^{(3 / 2)} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}

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Solution

The correct option is C $K_{p} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3} K_{p} = \frac{p_{N H_{3}}^{2}}{p_{N_{2}} . p_{H_{2}}^{3}} \Rightarrow K_{p} = \frac{p_{N H_{3}}^{2}}{p_{N_{2}} \times (3 p_{N_{2}})^{3}} [∵ 3 p_{N_{2}} = p_{H_{2}}] = \frac{p_{N H_{3}}^{2}}{p_{N_{2}} \times 3^{3} . p_{N_{2}}^{3}} = \frac{p_{N H_{3}}^{2}}{3^{3} \times p_{N_{2}}^{4}} \Rightarrow K_{p} = \frac{p_{N H_{3}}^{2}}{3^{3} \times p_{N_{2}}^{4}}$

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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)

K_{p}

and

K_{p}^{'}

are the equilibrium constants of the two reaction given below:

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {N H}_{3} (g)

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

Therefore,

K_{p}

and

K_{p}^{'}

are related by

Q. The equilibrium constant

K_{c}

for the reaction

N_{2 (g)} + 3 H_{2} ⇌ {2 N H}_{3 (g)}

0.286

500^{o} C

. Calculate

K_{c}

and

K_{p}

for the reaction

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

Q. A tenfold increase in pressure on the reaction

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

at equilibrium constant

K_{p}

will :

For $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) + h e a t$

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For the equilibrium, N2(g)+3H2(g)⇌2NH3(g), the equilibrium constant, Kp is expressed as:

The correct option is C Kp=33×p4N2p2NH3N2+3H2⇌2NH3Kp=p2NH3pN2.p3H2⇒Kp=p2NH3pN2×(3pN2)3[∵3pN2=pH2] =p2NH3pN2×33.p3N2 =p2NH333×p4N2⇒Kp=p2NH333×p4N2

For the equilibrium, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g),$ the equilibrium constant, $K_{p}$ is expressed as: