Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
Medium$0.5 \ mol$ of $CaCO_3(s)$ is heated in a $500 \ mL$ closed vessel at $400 \ K$. The reaction is $CaCO_{3(s)} \rightleftharpoons CaO(s) + CO_{2(g)}$. If the equilibrium constant $K_c = 0.9 \ mol \ L^{-1}$,calculate the amount of $CO_2$ in $mol$ at equilibrium and the percentage of the reaction completed.
(A) The equilibrium constant expression for the reaction is $K_c = [CO_2]$.
Given $K_c = 0.9 \ mol \ L^{-1}$,therefore at equilibrium,$[CO_2] = 0.9 \ mol \ L^{-1}$.
The volume of the vessel is $500 \ mL = 0.5 \ L$.
The amount of $CO_2$ at equilibrium is $n(CO_2) = [CO_2] \times V = 0.9 \ mol \ L^{-1} \times 0.5 \ L = 0.45 \ mol$.
Since $1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$,the amount of $CaCO_3$ decomposed is $0.45 \ mol$.
The initial amount of $CaCO_3$ is $0.5 \ mol$.
Percentage of reaction completed = $\frac{0.45 \ mol}{0.5 \ mol} \times 100 = 90 \%$.
Given $K_c = 0.9 \ mol \ L^{-1}$,therefore at equilibrium,$[CO_2] = 0.9 \ mol \ L^{-1}$.
The volume of the vessel is $500 \ mL = 0.5 \ L$.
The amount of $CO_2$ at equilibrium is $n(CO_2) = [CO_2] \times V = 0.9 \ mol \ L^{-1} \times 0.5 \ L = 0.45 \ mol$.
Since $1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$,the amount of $CaCO_3$ decomposed is $0.45 \ mol$.
The initial amount of $CaCO_3$ is $0.5 \ mol$.
Percentage of reaction completed = $\frac{0.45 \ mol}{0.5 \ mol} \times 100 = 90 \%$.
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$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,find the value of $K_c$ for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
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View SolutionNitric oxide reacts with $Br_{2}$ and gives nitrosyl bromide as per the reaction given below:
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When $0.087 \ mol$ of $NO$ and $0.0437 \ mol$ of $Br_{2}$ are mixed in a closed container at constant temperature,$0.0518 \ mol$ of $NOBr$ is obtained at equilibrium. Calculate the equilibrium amount of $NO$ and $Br_{2}$.
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View SolutionIn a $0.25 \ L$ tube,$4 \ mol$ of $NO$ undergoes dissociation. If the degree of dissociation is $10\%$,then the value of $K_c$ for the reaction $2NO \rightleftharpoons N_2 + O_2$ will be:
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$\Delta H = -57.2 \ kJ \ mol^{-1}$ and $K_C = 1.7 \times 10^{16}$
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Which of the following statement is $INCORRECT$?
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