The following are the equilibria N2-3H2- 2NH3 K1 -i N2-O2- 2NO K2 -ii H2-1-2O2- H2O K3 -iiiThen the equilibrium constant K4 for the reaction 2NH3-5-2- 2NO-3H2O is-

The correct option is C K_2× K_3^3/K_1 Equation (III) is multiplied by 3 3H_2+3/2O_2⇌ 3H_2O nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard XII

Chemistry

Arrhenius Acid

The following...

Question

The following are the equilibria
$N_{2} + 3 H_{2} ⇌ 2 N H_{3} K_{1}$ ...(i)
$N_{2} + O_{2} ⇌ 2 N O K_{2}$ ...(ii)
$H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O K_{3}$ ...(iii)
Then the equilibrium constant $K_{4}$ for the reaction
$2 N H_{3} + \frac{5}{2} ⇌ 2 N O + 3 H_{2} O$ is:

\frac{K_{2} \times K_{3}}{K_{1}}

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\frac{K_{2} \times K_{1}^{3}}{K_{3}}

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\frac{K_{2} \times K_{3}^{3}}{K_{1}}

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\frac{K_{2} \times K_{1}}{K_{3}}

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Solution

The correct option is C $\frac{K_{2} \times K_{3}^{3}}{K_{1}}$
Equation (III) is multiplied by 3
$3 H_{2} + \frac{3}{2} O_{2} ⇌ 3 H_{2} O K_{3}^{3}$
It is then added to equation (II).
$3 H_{2} + N_{2} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O; K_{2} \times K_{3}^{3}$
The result is then added to reverse of equation (I) to obtain equation (4)
$2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O (K_{4})$
Hence, $K_{4} = \frac{K_{2} \times K_{3}^{3}}{K_{1}}$ .

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Similar questions

Q. The following equilibria are given

N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{1}

N_{2} + O_{2} ⇌ 2 N O, K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O, K_{3}

The equilibrium constant of the reaction, in terms of

K_{1}, K_{2}

and

K_{3}

is:

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

Q. The following equilibria are given :

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K_{1}

N_{2} + O_{2} ⇌ 2 N O

K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O

K_{3}

The equilibrium constant of the reaction :

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

in terms of

K_{1}, K_{2}

and

K_{3}

is:

The following equilibria are given: $N_{2}$ +3 $H_{2}$ $⇋$ $2 N H_{3}$ $K_{1}$ ; $N_{2}$ + $O_{2}$ $⇋$ 2NO $K_{2}$ ;
$H_{2}$ + $\frac{1}{2}$ $O_{2}$ $⇋$ $H_{2} O K_{3}$ ;

The Equilibrium constant for the reaction $2 N H_{3}$ + $\frac{5}{2}$ $O_{2}$ $⇋$ 2NO + 3 $H_{2}$ O in terms of $K_{1}$ , $K_{2}$ and $K_{3}$ is:

Q. The following equilibrium are given

N_{2} + 3 H_{2} \to 2 N H_{3} - K_{1}

N_{2} + O_{2} \to 2 N O - K_{2}

H_{2} + 1 / 2 O_{2} \to H_{2} O - K_{3}

The equilibrium constant of reaction,

3 H_{2} + 5 / 2 O_{2} \to 2 N O + 3 H_{2} O

, in terms of

K_{1}, K_{2}, K_{3}

Q. The following equilibrium are given

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

--------------

K_{1}

N_{2} + O_{2} ⇌ 2 N O

---------------

K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O

---------------

K_{3}

The equilibrium constant of the reaction

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

, in terms of

K_{1}, K_{2}

and

K_{3}

is :

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The following are the equilibriaN2+3H2⇌2NH3K1 ...(i)N2+O2⇌2NOK2 ...(ii)H2+12O2⇌H2OK3 ...(iii)Then the equilibrium constant K4 for the reaction 2NH3+52⇌2NO+3H2O is:

The correct option is C K2×K33K1Equation (III) is multiplied by 3 3H2+32O2⇌3H2OK33It is then added to equation (II). 3H2+N2+52O2⇌2NO+3H2O;K2×K33The result is then added to reverse of equation (I) to obtain equation (4)2NH3+52O2⇌2NO+3H2O(K4)Hence, K4=K2×K33K1.

The following are the equilibria
$N_{2} + 3 H_{2} ⇌ 2 N H_{3} K_{1}$ ...(i)
$N_{2} + O_{2} ⇌ 2 N O K_{2}$ ...(ii)
$H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O K_{3}$ ...(iii)
Then the equilibrium constant $K_{4}$ for the reaction
$2 N H_{3} + \frac{5}{2} ⇌ 2 N O + 3 H_{2} O$ is: