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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$10 \%$ of $PCl_5$ decomposes at a definite temperature and $4 \ atm$ pressure. At the same temperature,if $20 \%$ of $PCl_5$ decomposes,find the pressure. (Temperature remains constant.) (in $atm$)
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View SolutionIn a closed vessel at $448^{\circ} C$,$0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ react to form hydrogen iodide.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.
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View SolutionThe number of correct statement$(s)$ involving equilibria in physical processes from the following is:
$A$. Equilibrium is possible only in a closed system at a given temperature.
$B$. Both the opposing processes occur at the same rate.
$C$. When equilibrium is attained at a given temperature,the value of all its parameters becomes constant.
$D$. For dissolution of solids in liquids,the solubility is constant at a given temperature.
$A$. Equilibrium is possible only in a closed system at a given temperature.
$B$. Both the opposing processes occur at the same rate.
$C$. When equilibrium is attained at a given temperature,the value of all its parameters becomes constant.
$D$. For dissolution of solids in liquids,the solubility is constant at a given temperature.
Given the equilibrium constants for the following three reactions:
$(1) N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$
$(2) N_2 + O_2 \rightleftharpoons 2NO; K_2$
$(3) H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$
The equilibrium constant for the reaction of $NH_3$ with oxygen to form $NO$ and $H_2O$ is:
$(1) N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$
$(2) N_2 + O_2 \rightleftharpoons 2NO; K_2$
$(3) H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$
The equilibrium constant for the reaction of $NH_3$ with oxygen to form $NO$ and $H_2O$ is:
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View SolutionAt $717 \ K$,$3.2 \ mol$ of $HI$ is heated in a closed tube. $20\%$ of $HI$ decomposes at equilibrium according to the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$. Find $K_c$ and the moles of $HI$,$H_2$,and $I_2$ at equilibrium.
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