(B) Given reactions:$1) A + D \rightleftharpoons AD, K_1 = \frac{[AD]}{[A][D]}$$2) AD + D \rightleftharpoons AD_2, K_2 = \frac{[AD_2]}{[AD][D]}$$3) AD

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$\log K_1 + \log K_2 + \log K_3 = \log K$

(B) Given reactions:$1) A + D \rightleftharpoons AD, K_1 = \frac{[AD]}{[A][D]}$$2) AD + D \rightleftharpoons AD_2, K_2 = \frac{[AD_2]}{[AD][D]}$$3) AD_2 + D \rightleftharpoons AD_3, K_3 = \frac{[AD_3]}{[AD_2][D]}$Adding these three equations gives the net reaction:$A + 3D \rightleftharpoons AD_3$The equilibrium constant $K$ for the net reaction is the product of the individual equilibrium constants:$K = K_1 \times K_2 \times K_3$Taking the logarithm on both sides:$\log K = \log(K_1 \times K_2 \times K_3)$$\log K = \log K_1 + \log K_2 + \log K_3$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

A
$K_1 + K_2 + K_3 = K$
B
$\log K_1 + \log K_2 + \log K_3 = \log K$
C
$K_1 + K_2 = K_3 + K$
D
$\log K_1 + \log K_2 = \log K_3 + \log K$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

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The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

DifficultJEE MAIN 2024

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For the reaction ${N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}}$,the reaction quotient is given by ${Q = [NH_3]^2 / ([N_2][H_2]^3)}$. The reaction will proceed from left to right when ........

Medium

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Consider the following equilibrium reaction in gaseous state at $T(K)$.
$A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g)$
The initial concentration of $B$ is $1.5$ times that of $A$. At equilibrium,the concentrations of $A$ and $B$ are equal. The equilibrium constant for the reaction is

MediumAP EAMCET 2025

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For ammonia formation from constituent elements,the expression for $K_{C}$ is

EasyAP EAMCET 2022

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If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

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For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

For the reaction ${N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}}$,the reaction quotient is given by ${Q = [NH_3]^2 / ([N_2][H_2]^3)}$. The reaction will proceed from left to right when ........

Consider the following equilibrium reaction in gaseous state at $T(K)$.$A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g)$The initial concentration of $B$ is $1.5$ times that of $A$. At equilibrium,the concentrations of $A$ and $B$ are equal. The equilibrium constant for the reaction is

For ammonia formation from constituent elements,the expression for $K_{C}$ is

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Consider the following equilibrium reaction in gaseous state at $T(K)$.
$A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g)$
The initial concentration of $B$ is $1.5$ times that of $A$. At equilibrium,the concentrations of $A$ and $B$ are equal. The equilibrium constant for the reaction is