(C) Given reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ with $K_{c1} = 4$.For the target reaction: $N_{2(g)} +

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$\left( \frac{1}{4 \times 800R} \right)^2$

(C) Given reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ with $K_{c1} = 4$.For the target reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,we reverse the first reaction and multiply by $2$.Thus,$K_{c2} = (1/K_{c1})^2 = (1/4)^2 = 1/16$.The relationship between $K_P$ and $K_c$ is $K_P = K_c(RT)^{\Delta n}$.For the target reaction,$\Delta n = 2 - (1 + 3) = -2$.Temperature $T = 527 + 273 = 800 \ K$.$K_P = (1/16) \times (R \times 800)^{-2} = (1/4)^2 \times (800R)^{-2} = \left( \frac{1}{4 \times 800R} \right)^2$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

Medium

At $527 \ ^oC$,the reaction given below has $K_c = 4$.
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

A
$16 \times (800 \ R)^2$
B
$\left( \frac{800R}{4} \right)^{-2}$
C
$\left( \frac{1}{4 \times 800R} \right)^2$
D
None of these

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

In the reaction,$A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,equilibrium is established. If the pressure of $B$ is doubled,then to reestablish the equilibrium,how many times does the pressure of $D$ become as compared to its initial pressure?

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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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In a $500 \ mL$ capacity vessel,$CO$ and $Cl_2$ are mixed to form $COCl_2$. At equilibrium,it contains $0.2 \ mol$ of $COCl_2$ and $0.1 \ mol$ of each of $CO$ and $Cl_2$. The equilibrium constant $K_c$ for the reaction $CO + Cl_2 \rightleftharpoons COCl_2$ is:

MediumAIPMT 1998

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If the equilibrium constant of a reaction is $20.0$ at equilibrium,and the rate constant of the forward reaction is $10.0$,then the rate constant for the backward reaction is:

Medium

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For the reaction $3A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$ at a given temperature,$K_c$ is $9.0$. What must be the volume of the flask if a mixture of $2.0 \ mol$ each of $A, B$ and $C$ is obtained at equilibrium? .....$L$

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At $527 \ ^oC$,the reaction given below has $K_c = 4$.$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$What is the $K_P$ for the following reaction?$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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In the reaction,$A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,equilibrium is established. If the pressure of $B$ is doubled,then to reestablish the equilibrium,how many times does the pressure of $D$ become as compared to its initial pressure?

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$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).

In a $500 \ mL$ capacity vessel,$CO$ and $Cl_2$ are mixed to form $COCl_2$. At equilibrium,it contains $0.2 \ mol$ of $COCl_2$ and $0.1 \ mol$ of each of $CO$ and $Cl_2$. The equilibrium constant $K_c$ for the reaction $CO + Cl_2 \rightleftharpoons COCl_2$ is:

If the equilibrium constant of a reaction is $20.0$ at equilibrium,and the rate constant of the forward reaction is $10.0$,then the rate constant for the backward reaction is:

For the reaction $3A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$ at a given temperature,$K_c$ is $9.0$. What must be the volume of the flask if a mixture of $2.0 \ mol$ each of $A, B$ and $C$ is obtained at equilibrium? .....$L$

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At $527 \ ^oC$,the reaction given below has $K_c = 4$.
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
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).