(A) For the reaction $A \rightleftharpoons n B$ at equilibrium: Initial concentration of $A = 0.06 \ mol \ L^{-1}$,$B = 0 \ mol \ L^{-1}$. Equilibrium

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(A) For the reaction $A \rightleftharpoons n B$ at equilibrium: Initial concentration of $A = 0.06 \ mol \ L^{-1}$,$B = 0 \ mol \ L^{-1}$. Equilibrium concentration of $A = 0.03 \ mol \ L^{-1}$,$B = 0.06 \ mol \ L^{-1}$. Change in concentration of $A = 0.06 - 0.03 = 0.03 \ mol \ L^{-1}$. Change in concentration of $B = 0.06 - 0 = 0.06 \ mol \ L^{-1}$. According to stoichiometry,$n = \frac{\Delta [B]}{\Delta [A]} = \frac{0.06}{0.03} = 2$. The reaction is $A \rightleftharpoons 2 B$. Equilibrium constant $K_c = \frac{[B]^2}{[A]} = \frac{(0.06)^2}{0.03} = \frac{0.0036}{0.03} = 0.12$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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For the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are

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$2$ and $0.12$
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$2$ and $1.2$
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$3$ and $0.12$
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$3$ and $1.2$

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6-1.Equilibrium (Chemical Equilibrium)

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Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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For the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are

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At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are:

In a $1.0 \, L$ vessel at $90 \, ^\circ C$,$0.2 \, mol$ of $H_{2(g)}$ and $2.0 \, mol$ of $S_{(s)}$ are mixed. For the reaction $H_{2(g)} + S_{(s)} \rightleftharpoons H_2S_{(g)}$; $K_p = 6.8 \times 10^{-2}$,the partial pressure of $H_2S_{(g)}$ at equilibrium will be ............ $atm$.

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