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Question

# The following equilibria are givenN2+3H2⇌2NH3,K1N2+O2⇌2NO,K2H2+12O2⇌H2O,K3The equilibrium constant of the reaction, in terms of K1,K2 and K3 is:2NH3+52O2⇌2NO+3H2O

A
K=K2×K23K1
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B
K=K22×K3K1
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C
K=K1×K2K3
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D
K=K2×K33K1
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Solution

## The correct option is D K=K2×K33K1If we reverse a reaction, then the new equilibrium constant becomes reciprocal to the older.If we add two reactions, then equilibrium constants multiply to get a new equilibrium constant.If we multiply the reaction by '2', then the equilibrium constant gets squared.N2+3H2⇌2NH3⟶K12NH3⇌N2+3H2⟶1K1−−−−−−(1)Multiplying equation (1) by 2.N2+O2⇌2NO⟶K2−−−−−−(2)H2+12O2⇌H2O⟶K3Multiplying the above reaction by 3,3H2+32O2⇌3H2O⟶K33−−−−−−(3)Adding (1), (2) and (3) reactions,2NH3+52O2⇌2NO+3H2OResultant equilibrium constant=K′=(K3)3⋅K2K1

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