(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,so $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$.For $X: \Delta n_g = 2

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(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,so $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$.For $X: \Delta n_g = 2 - (2 + 1) = -1$. Thus,$\frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT}$.For $Y: \Delta n_g = (1 + 1) - 1 = 1$. Thus,$\frac{K_p}{K_c} = (RT)^1 = RT$.For $Z: \Delta n_g = (1 + 1) - 2 = 0$. Thus,$\frac{K_p}{K_c} = (RT)^0 = 1$.Since $T = 300 \ K$ $(RT > 1)$,the values are $\frac{1}{RT} Therefore,the increasing order is $X < Z < Y$.

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

Easy

For the following gaseous equilibria at $300 \ K$,find the increasing order of the ratio $\frac{K_p}{K_c}$ for $X, Y,$ and $Z$:
$X: 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$Y: PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
$Z: 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$

A
$X = Y = Z$
B
$X < Y < Z$
C
$X < Z < Y$
D
$Z < Y < X$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

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For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

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At $T(K)$,the following gaseous equilibrium is established: $W + X \rightleftharpoons Y + Z$. The initial concentration of $W$ is two times the initial concentration of $X$. The system is heated to $T(K)$ to establish equilibrium. At equilibrium,the concentration of $Y$ is four times the concentration of $X$. What is the value of $K_c$?

MediumAP EAMCET 2025

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The following equilibrium is established at $STP$. $B_{2(g)} \rightleftharpoons 2B_{(g)}$. Atoms of $B$ occupy $20 \%$ of total volume at $STP$. The total pressure of the system is $1 \ bar$. What is its $K_p$?

MediumAP EAMCET 2025

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For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

MediumTS EAMCET 2018

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For the following gaseous equilibria at $300 \ K$,find the increasing order of the ratio $\frac{K_p}{K_c}$ for $X, Y,$ and $Z$:$X: 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$$Y: PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$$Z: 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$

Similar Questions

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

The following equilibrium is established at $STP$. $B_{2(g)} \rightleftharpoons 2B_{(g)}$. Atoms of $B$ occupy $20 \%$ of total volume at $STP$. The total pressure of the system is $1 \ bar$. What is its $K_p$?

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

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For the following gaseous equilibria at $300 \ K$,find the increasing order of the ratio $\frac{K_p}{K_c}$ for $X, Y,$ and $Z$:
$X: 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$Y: PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
$Z: 2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$