(D) For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$, the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$. Initial moles

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(D) For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$, the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$. Initial moles of $A$ in $1 \text{ L}$ flask $= 0.5 \text{ mol}$. Initial moles of $B$ in $1 \text{ L}$ flask $= 0.375 \text{ mol}$. After adding $0.1 \text{ mol}$ of $A$, total moles of $A = 0.5 + 0.1 = 0.6 \text{ mol}$. Let $x$ be the amount of $A$ reacted to reach new equilibrium. New $[A] = 0.6 - x$ and New $[B] = 0.375 + x$. $K_c = \frac{0.375 + x}{0.6 - x} = 0.75$. $0.375 + x = 0.45 - 0.75x$. $1.75x = 0.075 \Rightarrow x = 0.0428$. New $[A] = 0.6 - 0.0428 = 0.557 \text{ M}$ and New $[B] = 0.375 + 0.0428 = 0.418 \text{ M}$.

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

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Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$, the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.

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A
$0.367, 0.275$
B
$0.53, 0.4$
C
$0.742, 0.557$
D
$0.557, 0.418$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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Medium

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The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:

MediumAIIMS 1984

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MediumTS EAMCET 2024

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For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the correct expression of equilibrium constant $K$ is

Medium

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For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)} + D_{(s)}$,$2 \, \text{moles}$ of $A$,$3 \, \text{moles}$ of $B$ and $1 \, \text{mole}$ of $C$ are present in a $10 \, L$ vessel. If $K_c$ for the reaction is $3.6$,the reaction will proceed in:

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