(C) The reaction is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $8, 0, 0$At equilibrium,$n(A_2) = 2 \ mol$. Since the stoichio

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(C) The reaction is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $8, 0, 0$At equilibrium,$n(A_2) = 2 \ mol$. Since the stoichiometry of $A_2$ is $1$,the amount of $AB_3$ reacted is $2 \times 2 = 4 \ mol$.Remaining moles at equilibrium:$n(AB_3) = 8 - 4 = 4 \ mol$$n(A_2) = 2 \ mol$$n(B_2) = 3 \times 2 = 6 \ mol$Since the volume is $1.0 \ L$,the concentrations are $[AB_3] = 4 \ M, [A_2] = 2 \ M, [B_2] = 6 \ M$.$K_c = \frac{[A_2][B_2]^3}{[AB_3]^2} = \frac{2 \times (6)^3}{(4)^2} = \frac{2 \times 216}{16} = \frac{432}{16} = 27$.

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

Medium

$8 \ mol$ of $AB_3$ is taken in a $1.0 \ dm^3$ container. It dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. If $2 \ mol$ of $A_2$ are present at equilibrium,the equilibrium constant $K_c$ of the reaction is ....... $mol^2 \ L^{-2}$.

A
$12$
B
$3$
C
$27$
D
$36$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

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For the reaction $CO_{(g)} + Cl_{2_{(g)}} \rightleftharpoons COCl_{2_{(g)}}$,the value of $K_p/K_c$ is ....

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$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
In an equilibrium mixture,the partial pressures are
$P_{SO_{3}} = 43 \ kPa$,$P_{O_{2}} = 530 \ Pa = 0.53 \ kPa$,and
$P_{SO_{2}} = 45 \ kPa$. The equilibrium constant $K_{p} = ...... \times 10^{-2} \ kPa^{-1}$. (Nearest integer)

MediumJEE MAIN 2021

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At $700 \, K$,the equilibrium constant $K_p$ for the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ is $1.80 \times 10^{-3}$. The numerical value in $mol \, L^{-1}$ of $K_c$ for this reaction at the same temperature will be $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

Medium

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At $1000\, K$ and $2\, atm$ pressure,a gaseous mixture of $CO$ and $CO_{2}$ in equilibrium with solid carbon has $84\%$ $CO_{(g)}$ by mass. Calculate $K_{p}$ for the reaction: $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$ at this temperature.

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$8 \ mol$ of $AB_3$ is taken in a $1.0 \ dm^3$ container. It dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. If $2 \ mol$ of $A_2$ are present at equilibrium,the equilibrium constant $K_c$ of the reaction is ....... $mol^2 \ L^{-2}$.

Similar Questions

For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

For the reaction $CO_{(g)} + Cl_{2_{(g)}} \rightleftharpoons COCl_{2_{(g)}}$,the value of $K_p/K_c$ is ....

$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$In an equilibrium mixture,the partial pressures are$P_{SO_{3}} = 43 \ kPa$,$P_{O_{2}} = 530 \ Pa = 0.53 \ kPa$,and$P_{SO_{2}} = 45 \ kPa$. The equilibrium constant $K_{p} = ...... \times 10^{-2} \ kPa^{-1}$. (Nearest integer)

At $700 \, K$,the equilibrium constant $K_p$ for the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ is $1.80 \times 10^{-3}$. The numerical value in $mol \, L^{-1}$ of $K_c$ for this reaction at the same temperature will be $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

At $1000\, K$ and $2\, atm$ pressure,a gaseous mixture of $CO$ and $CO_{2}$ in equilibrium with solid carbon has $84\%$ $CO_{(g)}$ by mass. Calculate $K_{p}$ for the reaction: $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$ at this temperature.

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$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
In an equilibrium mixture,the partial pressures are
$P_{SO_{3}} = 43 \ kPa$,$P_{O_{2}} = 530 \ Pa = 0.53 \ kPa$,and
$P_{SO_{2}} = 45 \ kPa$. The equilibrium constant $K_{p} = ...... \times 10^{-2} \ kPa^{-1}$. (Nearest integer)