
Consider the following reversible chemical reactions :
${A_2}(g) + B{r_2}(g) \rightleftharpoons 2AB(g)...(1){K_1}$
$6AB(g) \rightleftharpoons 3{A_2}(g) + 3{B_2}...(2){K_2}$
a.) ${K_2}$= $K_1^3$
b.) ${K_2}$= $K_1^{ - 3}$
c.) ${K_1}$${K_2}$= 3
d.) ${K_1}$${K_2}$= $\dfrac{1}{3}$
Answer
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Hint: The Equilibrium is a situation when the rate of forward reaction is equal to the rate of backward reaction. The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction. Further, we know that if we multiply an equation by n factor then its equilibrium constant can be given by ${K^n}$. By following these, we can get our answer.
Complete step by step answer :
The Equilibrium is a situation when the rate of forward reaction is equal to the rate of backward reaction.
The Equilibrium constant of a reaction is the constant measured at the equilibrium.
The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.
Now, let us see the reactions as-
${A_2}(g) + B{r_2}(g) \rightleftharpoons 2AB(g)...(1){K_1}$
$6AB(g) \rightleftharpoons 3{A_2}(g) + 3{B_2}...(2){K_2}$
For the first reaction; we have the equilibrium constant is ${K_1}$.
And for the second reaction, it is ${K_2}$.
If we reverse the second reaction and let its equilibrium constant be K and be its third reaction.
Then, it can be written as-
$3{A_2}(g) + 3{B_2} \rightleftharpoons 6AB(g)$
The equilibrium constant of this reaction i.e. K will be the reverse of the equilibrium constant of the second reaction.
Further, we know that if we multiply equation 1 with 3; we get a third equation.
And we know that if we multiply an equation by n factor then its equilibrium constant can be given by ${K^n}$.
So, for the third equation, we can write K = $K_1^3$
Also, we have K = $K_2^{ - 1}$
Thus, $K_1^3$= $K_2^{ - 1}$
So, this is our answer. But, this is not in any options.
Note : It must be noted that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. The equilibrium constant describes the relationship between products and reactants of a reaction at equilibrium with respect to specific units.
Complete step by step answer :
The Equilibrium is a situation when the rate of forward reaction is equal to the rate of backward reaction.
The Equilibrium constant of a reaction is the constant measured at the equilibrium.
The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.
Now, let us see the reactions as-
${A_2}(g) + B{r_2}(g) \rightleftharpoons 2AB(g)...(1){K_1}$
$6AB(g) \rightleftharpoons 3{A_2}(g) + 3{B_2}...(2){K_2}$
For the first reaction; we have the equilibrium constant is ${K_1}$.
And for the second reaction, it is ${K_2}$.
If we reverse the second reaction and let its equilibrium constant be K and be its third reaction.
Then, it can be written as-
$3{A_2}(g) + 3{B_2} \rightleftharpoons 6AB(g)$
The equilibrium constant of this reaction i.e. K will be the reverse of the equilibrium constant of the second reaction.
Further, we know that if we multiply equation 1 with 3; we get a third equation.
And we know that if we multiply an equation by n factor then its equilibrium constant can be given by ${K^n}$.
So, for the third equation, we can write K = $K_1^3$
Also, we have K = $K_2^{ - 1}$
Thus, $K_1^3$= $K_2^{ - 1}$
So, this is our answer. But, this is not in any options.
Note : It must be noted that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. The equilibrium constant describes the relationship between products and reactants of a reaction at equilibrium with respect to specific units.
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