(D) The balanced chemical equation is: $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$Initial moles: $1 \quad 1 \quad 0 \quad 0$Moles at

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(D) The balanced chemical equation is: $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$Initial moles: $1 \quad 1 \quad 0 \quad 0$Moles at equilibrium: $(1 - 3/4) \quad (1 - 3/4) \quad 3/4 \quad 3/4$Equilibrium moles: $1/4 \quad 1/4 \quad 3/4 \quad 3/4$Since the volume is constant,the concentration is proportional to the number of moles. The equilibrium constant $K_c$ is given by:$K_c = \frac{[AD][CB]}{[AB][CD]} = \frac{(3/4) \times (3/4)}{(1/4) \times (1/4)} = \frac{9/16}{1/16} = 9$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

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In the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction will be........

A
$9/16$
B
$1/9$
C
$16/9$
D
$9$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

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In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \ mol$ each of the two reactants are mixed,$0.6 \ mol$ each of the products are formed. The equilibrium constant calculated is

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At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

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At $400 \ K$ for the reaction $2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$,the concentration of $NO_{2}$ is $0.710 \ M$ and $N_{2}O_{4}$ is $0.145 \ M$. Find the equilibrium constant $K_{c}$. (in $M^{-1}$)

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For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?

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The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \ mol$ each of the two reactants are mixed,$0.6 \ mol$ each of the products are formed. The equilibrium constant calculated is

At $400 \ K$ for the reaction $2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$,the concentration of $NO_{2}$ is $0.710 \ M$ and $N_{2}O_{4}$ is $0.145 \ M$. Find the equilibrium constant $K_{c}$. (in $M^{-1}$)

For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?

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For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?