Answer
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Hint: The constant present in the relation between reactants and products at chemical equilibrium is known as equilibrium constant, it is basically the ratio of rate constants defined for forward and backward reactions at equilibrium.
Complete step by step answer:
The constant for relation between the reactants and products at equilibrium state is known as equilibrium constant for the reaction, it is represented as $K$ . Its denotation changes as per the type of reactions and the basis on which we are finding the equilibrium constant. If we are using concentration in equilibrium equations then the equilibrium constant is represented as ${K_c}$ and if the partial pressure is used in equilibrium equation then the equilibrium constant is represented as ${K_p}$.
As we all know that at equilibrium the rate of forward reaction is equal to rate of backward reaction so we can say that the equilibrium is nothing but the ratio of the amount of reactant and the amount of product. The equilibrium constant is basically used to define the chemical behavior of the reaction.
So from all the above discussion we can say that; ${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}}$
Where, ${K_f}$ is the constant for forward reaction,${K_b}$ is the constant for backward reaction and ${K_{eq}}$ is the equilibrium constant.
Let us suppose a reaction
${\text{aA + bB}} \to {\text{cC + dD}}$
${\text{A and B}}$ are the reactants and ${\text{a and b}}$ are their respective coefficients.
$ \Rightarrow {K_f} = {\left[ {\text{A}} \right]^a}{\left[ {\text{B}} \right]^b}$
${\text{C and D}}$ are the products and ${\text{c and d}}$ are their respective coefficients.
$ \Rightarrow {K_b} = {\left[ {\text{C}} \right]^c}{\left[ {\text{D}} \right]^d}$
Therefore the equilibrium constant is:
${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$
And $\Delta n = (c + d) - (a + b)$, ($\Delta n$ =sum of coefficients of products-sum of coefficients of reactants)
And we know that concentration is always in mole per liter.
So by combining both of the above statements we get the unit of equilibrium constant which is as follows:
${\text{unit of Kc = (mol/lit}}{{\text{)}}^{\Delta {\text{n}}}}$
So, the correct answer is Option C .
Note:
The equilibrium constant calculated with the help of moles per liter is known as ${K_c}$ and the chemical equilibrium calculated with the help of partial pressures of the gaseous reactants and products is known as ${K_p}$ . The relation between both of these equilibrium constants is as follows:
${K_p} = {K_c}{(RT)^{\Delta n}}$
Complete step by step answer:
The constant for relation between the reactants and products at equilibrium state is known as equilibrium constant for the reaction, it is represented as $K$ . Its denotation changes as per the type of reactions and the basis on which we are finding the equilibrium constant. If we are using concentration in equilibrium equations then the equilibrium constant is represented as ${K_c}$ and if the partial pressure is used in equilibrium equation then the equilibrium constant is represented as ${K_p}$.
As we all know that at equilibrium the rate of forward reaction is equal to rate of backward reaction so we can say that the equilibrium is nothing but the ratio of the amount of reactant and the amount of product. The equilibrium constant is basically used to define the chemical behavior of the reaction.
So from all the above discussion we can say that; ${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}}$
Where, ${K_f}$ is the constant for forward reaction,${K_b}$ is the constant for backward reaction and ${K_{eq}}$ is the equilibrium constant.
Let us suppose a reaction
${\text{aA + bB}} \to {\text{cC + dD}}$
${\text{A and B}}$ are the reactants and ${\text{a and b}}$ are their respective coefficients.
$ \Rightarrow {K_f} = {\left[ {\text{A}} \right]^a}{\left[ {\text{B}} \right]^b}$
${\text{C and D}}$ are the products and ${\text{c and d}}$ are their respective coefficients.
$ \Rightarrow {K_b} = {\left[ {\text{C}} \right]^c}{\left[ {\text{D}} \right]^d}$
Therefore the equilibrium constant is:
${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$
And $\Delta n = (c + d) - (a + b)$, ($\Delta n$ =sum of coefficients of products-sum of coefficients of reactants)
And we know that concentration is always in mole per liter.
So by combining both of the above statements we get the unit of equilibrium constant which is as follows:
${\text{unit of Kc = (mol/lit}}{{\text{)}}^{\Delta {\text{n}}}}$
So, the correct answer is Option C .
Note:
The equilibrium constant calculated with the help of moles per liter is known as ${K_c}$ and the chemical equilibrium calculated with the help of partial pressures of the gaseous reactants and products is known as ${K_p}$ . The relation between both of these equilibrium constants is as follows:
${K_p} = {K_c}{(RT)^{\Delta n}}$
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