Difficult
Explain the use of the equilibrium constant to predict the extent of a reaction with an example.
(N/A) The numerical value of the equilibrium constant $(K_{c})$ for a reaction indicates the extent of the reaction,but it does not provide information about the rate of the reaction.
The magnitude of $K_{c}$ or $K_{p}$ is directly proportional to the concentrations of products and inversely proportional to the concentrations of the reactants. $A$ high value of $K$ indicates that the concentration of products is high,while a low value of $K$ indicates that the concentration of products is low.
Value of $K \propto \text{[Products]} \propto \frac{1}{\text{[Reactants]}}$
The following generalizations describe the composition of equilibrium mixtures:
$(a)$ If $K_{c} > 10^{3}$: Products predominate over reactants; i.e.,if $K_{c}$ is very large,the reaction proceeds nearly to completion. Examples:
$(i)$ The reaction $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ at $500 \ K$ has a very large equilibrium constant,$K_{c} = 2.4 \times 10^{47}$.
$(ii)$ $H_{2(g)} + Cl_{2(g)} \rightleftharpoons 2 HCl_{(g)}$ at $300 \ K$ has a very large $K_{c} = 4.0 \times 10^{31}$.
$(iii)$ $H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ at $300 \ K$ has a very large $K_{c} = 5.4 \times 10^{18}$.
The magnitude of $K_{c}$ or $K_{p}$ is directly proportional to the concentrations of products and inversely proportional to the concentrations of the reactants. $A$ high value of $K$ indicates that the concentration of products is high,while a low value of $K$ indicates that the concentration of products is low.
Value of $K \propto \text{[Products]} \propto \frac{1}{\text{[Reactants]}}$
The following generalizations describe the composition of equilibrium mixtures:
$(a)$ If $K_{c} > 10^{3}$: Products predominate over reactants; i.e.,if $K_{c}$ is very large,the reaction proceeds nearly to completion. Examples:
$(i)$ The reaction $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ at $500 \ K$ has a very large equilibrium constant,$K_{c} = 2.4 \times 10^{47}$.
$(ii)$ $H_{2(g)} + Cl_{2(g)} \rightleftharpoons 2 HCl_{(g)}$ at $300 \ K$ has a very large $K_{c} = 4.0 \times 10^{31}$.
$(iii)$ $H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ at $300 \ K$ has a very large $K_{c} = 5.4 \times 10^{18}$.
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For the reactions and their equilibrium constants given below,
$CuCl_4^{2-} + Br^{-} \rightleftharpoons CuCl_3Br^{2-} + Cl^{-}$; $K_1$
$CuCl_3Br^{2-} + Br^{-} \rightleftharpoons CuCl_2Br_2^{2-} + Cl^{-}$; $K_2$
$CuCl_2Br_2^{2-} + Br^{-} \rightleftharpoons CuClBr_3^{2-} + Cl^{-}$; $K_3$
$CuClBr_3^{2-} + Br^{-} \rightleftharpoons CuBr_4^{2-} + Cl^{-}$; $K_4$
The equilibrium constant,$K$ for the reaction $CuCl_4^{2-} + 3Br^{-} \rightleftharpoons CuClBr_3^{2-} + 3Cl^{-}$,is
$CuCl_4^{2-} + Br^{-} \rightleftharpoons CuCl_3Br^{2-} + Cl^{-}$; $K_1$
$CuCl_3Br^{2-} + Br^{-} \rightleftharpoons CuCl_2Br_2^{2-} + Cl^{-}$; $K_2$
$CuCl_2Br_2^{2-} + Br^{-} \rightleftharpoons CuClBr_3^{2-} + Cl^{-}$; $K_3$
$CuClBr_3^{2-} + Br^{-} \rightleftharpoons CuBr_4^{2-} + Cl^{-}$; $K_4$
The equilibrium constant,$K$ for the reaction $CuCl_4^{2-} + 3Br^{-} \rightleftharpoons CuClBr_3^{2-} + 3Cl^{-}$,is
MediumKVPY 2009
View SolutionIf the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?
$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:
Difficult
View SolutionIf $K_1$ and $K_2$ are respective equilibrium constants for the two reactions:
$XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)}$
$XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)}$
The equilibrium constant for the reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$ will be:
$XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)}$
$XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)}$
The equilibrium constant for the reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$ will be:
MediumAIIMS 2013
View Solution$K_{a_1}, K_{a_2}$ and $K_{a_3}$ are the respective ionization constants for the following reactions $(a), (b),$ and $(c)$.
$(a)$ $H_2C_2O_4 \rightleftharpoons H^{+} + HC_2O_4^-$
$(b)$ $HC_2O_4^- \rightleftharpoons H^{+} + C_2O_4^{2-}$
$(c)$ $H_2C_2O_4 \rightleftharpoons 2H^{+} + C_2O_4^{2-}$
The relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is given as
$(a)$ $H_2C_2O_4 \rightleftharpoons H^{+} + HC_2O_4^-$
$(b)$ $HC_2O_4^- \rightleftharpoons H^{+} + C_2O_4^{2-}$
$(c)$ $H_2C_2O_4 \rightleftharpoons 2H^{+} + C_2O_4^{2-}$
The relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is given as
DifficultJEE MAIN 2022
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