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Question

Given:
N2 (g)+3H2 (g)2NH3 (g); K1
N2 (g)+O2 (g)2NO (g); K2
H2 (g)+12O2 (g)H2O (g); K3
where K1,K2,K3 are equilibrium constants.
The equilibrium constant for the reaction,
2NH3 (g)+52O2 (g)2NO (g)+3H2O (g) will be:

A
K1K2K3
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B
K1K2K3
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C
K1K23K2
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D
K2K33K1
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Solution

The correct option is D K2K33K1
For the given Equations:
eq.1,
N2 (g)+3H2 (g)2NH3 (g); K1

eq.2,
N2 (g)+O2 (g)2NO (g); K2

eq.3,
H2 (g)+12O2 (g)H2O (g); K3

We have to fine the equilibrium constant for
2NH3(g)+52O2 (g)2NO (g)+3H2O (g) will be:

Reversing the eq. 1,
we get , eq. 4
2NH3 (g)N2 (g)+3H2 (g); K4=1K1

Multiplying eq. 3 by 3,
we get , eq. 5
3H2 (g)+32O2 (g)3H2O (g); K5=(K3)3

Adding equation 2, 4, and 5, we get the required equation,
2NH3(g)+52O2 (g)2NO (g)+3H2O (g)

K=K2×K4×K5

putting values,

K=K2×1K1×(K3)3=K2×K33K1



Theory:

Characteristics of K (Equilibrium constant) :
Equilibrium constant (K) for the backward reaction is calculated by taking the reciprocal of equilibrium constant (K) for the forward reaction and vice versa .

Suppose for the given reaction :

A+BC then Kc=[C][A][B]
CA+B then K c=[A][B][C]=1K
If the equilibrium reaction stoichiometry is changed, the power of the equilibrium constant also gets changed by the same quantity.

Suppose for the given reaction :

A+BC ; Keq=K
2A+2B2C; K eq=K2

if,
A+BC ; Keq=K
12A+12B12C then K′′eq=K12

In case stepwise multiple equilibria leading to the final products, the equilibrium constant of the net equilibrium = product of each stepwise equilibrium constants.
Note:- whenever we add the reactions we multiply the K values and when we subtract the reaction we need to divide the K values.

Example 1 :
A+BC; K1
CD; K2
for A+BD; K=K1×K2

Example 2 :
H2(g)+2C(s)+52O2(g)2CO2(g)+H2O(l)K1

C2H2(g)+52O2(g)2CO2(g)+H2O(l)K2

H2(g)+2C(s)C2H2(g)K=K1K2


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