(D) Given $K_c = 99.0$. Initial moles of $A_2 = 2 \ mol$,Volume = $1 \ L$,so $[A_2]_{initial} = 2 \ mol \ L^{-1}$. Let $x$ be the concentration of $B_

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(D) Given $K_c = 99.0$. Initial moles of $A_2 = 2 \ mol$,Volume = $1 \ L$,so $[A_2]_{initial} = 2 \ mol \ L^{-1}$. Let $x$ be the concentration of $B_2$ formed at equilibrium. Reaction: $A_{2(g)} \rightleftharpoons B_{2(g)}$ Initial: $2 \ 0$ Equilibrium: $(2-x) \ x$ $K_c = \frac{[B_2]}{[A_2]} = \frac{x}{2-x} = 99.0$. $x = 99(2-x) = 198 - 99x$. $100x = 198 \implies x = 1.98 \ mol \ L^{-1}$. Concentration of $B_2$ at equilibrium = $1.98 \ mol \ L^{-1}$. Concentration of $A_2$ at equilibrium = $2 - 1.98 = 0.02 \ mol \ L^{-1}$. Thus,the concentrations of $A_2$ and $B_2$ are $0.02 \ mol \ L^{-1}$ and $1.98 \ mol \ L^{-1}$ respectively.

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

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At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

AP EAMCET 2024 All AP EAMCET

A
$0.0187, 1.86$
B
$1.98, 0.02$
C
$0.0187, 1.86$
D
$0.02, 1.98$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ 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:

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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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If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

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When $2 \ mol$ of $HI$ is heated in a closed vessel at $440 \ ^\circ C$,$22\%$ of $HI$ dissociates until equilibrium is reached. The equilibrium constant $K_c$ for the reaction is ..........

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The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is

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At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

Similar Questions

The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ 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:

If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

When $2 \ mol$ of $HI$ is heated in a closed vessel at $440 \ ^\circ C$,$22\%$ of $HI$ dissociates until equilibrium is reached. The equilibrium constant $K_c$ for the reaction is ..........

The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is

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