(B) The dissociation reaction is: $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$Initial pressure: $P_0 = 600 \ mm \ Hg$,$P_{XY} = 0$,$P_Y = 0$At eq

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(B) The dissociation reaction is: $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$Initial pressure: $P_0 = 600 \ mm \ Hg$,$P_{XY} = 0$,$P_Y = 0$At equilibrium: $P_{XY_2} = 600 - x$,$P_{XY} = x$,$P_Y = x$Total pressure at equilibrium: $P_{total} = (600 - x) + x + x = 600 + x$Given $P_{total} = 800 \ mm \ Hg$,so $600 + x = 800$,which gives $x = 200 \ mm \ Hg$Equilibrium partial pressures are: $P_{XY_2} = 600 - 200 = 400 \ mm \ Hg$,$P_{XY} = 200 \ mm \ Hg$,$P_Y = 200 \ mm \ Hg$$K_p = \frac{P_{XY} \times P_Y}{P_{XY_2}} = \frac{200 \times 200}{400} = 100 \ mm \ Hg$

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

Medium

$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

A
$50$
B
$100$
C
$166.6$
D
$400$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

Medium

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The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

DifficultAIEEE 2012

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The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

DifficultJEE MAIN 2026

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For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$K_p = 0.492 \ atm$ at $300 \ K$. $K_c$ for the reaction at same temperature is . . . . . . $\times 10^{-2}$. (Given: $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

DifficultJEE MAIN 2024

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At $T(K)$,when $1 \ mol$ of $X$ and $1 \ mol$ of $Y$ are heated in a $1 \ L$ flask,$0.5 \ mol$ of $Z$ is formed at equilibrium. The $K_c$ value for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)} + A_{(g)}$ is:

MediumTS EAMCET 2023

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$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

Similar Questions

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$K_p = 0.492 \ atm$ at $300 \ K$. $K_c$ for the reaction at same temperature is . . . . . . $\times 10^{-2}$. (Given: $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

At $T(K)$,when $1 \ mol$ of $X$ and $1 \ mol$ of $Y$ are heated in a $1 \ L$ flask,$0.5 \ mol$ of $Z$ is formed at equilibrium. The $K_c$ value for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)} + A_{(g)}$ is:

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