(A) For the first equilibrium: $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$, the equilibrium constant is $K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}}$.

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(A) For the first equilibrium: $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$, the equilibrium constant is $K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}}$.For the second equilibrium: $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$, the equilibrium constant is $K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$.Comparing the two expressions, we can see that $K_2 = \frac{1}{K_1^2} = (\frac{1}{K_1})^2$.Therefore, the correct relation is $K_2 = (\frac{1}{K_1})^2$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

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If the equilibrium constants of the following equilibria $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$ and $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$ are given by $K_1$ and $K_2$ respectively, which of the following relations is correct?

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A
$K_2 = (\frac{1}{K_1})^2$
B
$K_1 = (\frac{1}{K_2})^3$
C
$K_2 = \frac{1}{K_1}$
D
$K_2 = (K_1)^2$

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For the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO_2$ and $CO$ are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is ............ .

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

MediumJEE MAIN 2021

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For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

MediumAP EAMCET 2025

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One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:

MediumAP EAMCET 2023

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For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$,the equilibrium constants are $K_1$ and $K_2$ respectively. What is the relationship between $K_1$ and $K_2$?

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If the equilibrium constants of the following equilibria $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$ and $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$ are given by $K_1$ and $K_2$ respectively, which of the following relations is correct?

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For the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO_2$ and $CO$ are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is ............ .

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$Its $K_c$ value is:

For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$,the equilibrium constants are $K_1$ and $K_2$ respectively. What is the relationship between $K_1$ and $K_2$?

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is: