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  • Tell me? - Consider the following reversible chemical reactions : The relation between is : - Equilibrium - JEE Main

Consider the following reversible chemical reactions : 

\\A_2(g)+B_2(g)\overset{K_1}{\rightleftharpoons }2AB(g)............1\\\\6AB(g)\overset{K_2}{\rightleftharpoons }3A_2(g)+3B_2(g).........2

The relation between K_{1}\: and\: K_{2}  is :

 

  • Option 1)

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

  • Option 2)

    K_{2}=K_{1}\: ^{3}

  • Option 3)

    K_{1}K_{2}=3

  • Option 4)

    K_{2}=K_{1}\: ^{-3}

Answers (1)

best_answer

 

Equilibrium constant for the reverse reaction -

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

- wherein

K'_{c}=\frac{1}{K_{c}}

{K'_{c}}   is  equilibrium constant for reverse direction.

 

  

 

Equilibrium constant for a net reaction -

K_{net}=K_{1}\times K_{2}.........

- wherein

The equilibrium constant for a net reaction obtained after adding two reactions equal the product of the equilibrium constant for individual reactions.

Rxn \left \{ A_{2}(g)+B_{2} (g)\rightleftharpoons 2AB\right \}\times 3

AB(g)\rightleftharpoons 3A_{2}+3B_{2}(g)

(\frac{1}{K_{1}})^{3}=K_{2}

K_{2}=(K_{1})^{-3}

 


Option 1)

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

Option 2)

K_{2}=K_{1}\: ^{3}

Option 3)

K_{1}K_{2}=3

Option 4)

K_{2}=K_{1}\: ^{-3}
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