
S:${{K}_{c}},{{K}_{p}}\text{ and }{{K}_{x}}$ are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively.
E: ${{K}_{c}}\text{ and }{{\text{K}}_{p}}$ do not depend on equilibrium but ${{K}_{x}}$ depends upon equilibrium pressure if $\Delta n\ne 0$.
(a) S is correct but E is wrong
(b) S is wrong but E is correct
(c) Both S and E are correct and E is correct explanation of S
(d) Both S and E are correct but E is not correct explanation of S
Answer
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Hint: Equilibrium constants is simply the ratio of products to reactants raise to their power of molar coefficient and is represented as ${{K}_{c}},{{K}_{p}}\text{ and }{{K}_{x}}$ depending upon whether the reactants and the products are given in concentration, pressure or in molar fraction and equilibrium constant in terms of molar fraction is as ${{K}_{x}}={{K}_{p}}{{({{P}_{T}})}^{-\Delta n}}$. Now you can easily answer the statement.
Complete Solution :
First of all, let’s discuss the equilibrium constant. Equilibrium constant may be defined as the ratio of the concentration of products to the reactants raised to the power of their molar coefficient.
Consider the general reaction as:
$2A+2B\to 2AB$
Equilibrium constant for the reaction is as:
${{K}_{c}}=\dfrac{{{[AB]}^{2}}}{{{[A]}^{2}}{{[B]}^{2}}}$
Here, ${{K}_{c}}$ is the equilibrium constant in terms of concentration and square brackets represents the concentration of reactants and products.
In terms of partial pressure;
Consider the reaction as;
$aA\to cC$
The equilibrium constant for the reaction is as;
${{K}_{p}}=\dfrac{P_{C}^{c}}{P_{A}^{a}}$
Here, ${{K}_{p}}$ is the equilibrium constant in terms of pressure and ${{P}_{A}}\text{ and }{{P}_{C}}$ represents the partial pressure of the reactants and products.
In terms of molar fraction;
Considering the same reaction;
$aA\to cC$
The equilibrium constant for the reaction is as;
$\begin{align}
& {{K}_{p}}=\dfrac{P_{C}^{c}X_{C}^{c}}{P_{A}^{a}X_{A}^{a}}=\dfrac{X_{C}^{c}}{X_{A}^{a}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{p}}={{K}_{x}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{x}}={{K}_{p}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{x}}={{K}_{p}}{{({{P}_{T}})}^{-\Delta n}} \\
\end{align}$
Here, ${{K}_{x}}$ and ${{K}_{p}}$ are the equilibrium constant in terms of molar fraction and pressure respectively and ${{P}_{T}}$ represents the partial pressure of the reactants and products and $\Delta n$ represents the no. of moles of the reactants minus the no.of moles of the products.
Now considering the statement as:
S is correct as ${{K}_{c}},{{K}_{p}}\text{ and }{{K}_{x}}$ are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively.
E is also correct as ${{K}_{c}}\text{ and }{{\text{K}}_{p}}$ do not depend on equilibrium but ${{K}_{x}}$ depends upon equilibrium pressure if $\Delta n\ne 0$ because ${{K}_{x}}={{K}_{p}}{{({{P}_{T}})}^{-\Delta n}}$ .
Hence, Both S and E are correct but E is not the correct explanation of S.
So, the correct answer is “Option D”.
Note: Always keep in mind that equilibrium constants change with the change in the temperature and as the temperature is increased, the value of equilibrium constant also increases and vice-versa.
Complete Solution :
First of all, let’s discuss the equilibrium constant. Equilibrium constant may be defined as the ratio of the concentration of products to the reactants raised to the power of their molar coefficient.
Consider the general reaction as:
$2A+2B\to 2AB$
Equilibrium constant for the reaction is as:
${{K}_{c}}=\dfrac{{{[AB]}^{2}}}{{{[A]}^{2}}{{[B]}^{2}}}$
Here, ${{K}_{c}}$ is the equilibrium constant in terms of concentration and square brackets represents the concentration of reactants and products.
In terms of partial pressure;
Consider the reaction as;
$aA\to cC$
The equilibrium constant for the reaction is as;
${{K}_{p}}=\dfrac{P_{C}^{c}}{P_{A}^{a}}$
Here, ${{K}_{p}}$ is the equilibrium constant in terms of pressure and ${{P}_{A}}\text{ and }{{P}_{C}}$ represents the partial pressure of the reactants and products.
In terms of molar fraction;
Considering the same reaction;
$aA\to cC$
The equilibrium constant for the reaction is as;
$\begin{align}
& {{K}_{p}}=\dfrac{P_{C}^{c}X_{C}^{c}}{P_{A}^{a}X_{A}^{a}}=\dfrac{X_{C}^{c}}{X_{A}^{a}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{p}}={{K}_{x}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{x}}={{K}_{p}}\times {{({{P}_{T}})}^{b-a}} \\
& {{K}_{x}}={{K}_{p}}{{({{P}_{T}})}^{-\Delta n}} \\
\end{align}$
Here, ${{K}_{x}}$ and ${{K}_{p}}$ are the equilibrium constant in terms of molar fraction and pressure respectively and ${{P}_{T}}$ represents the partial pressure of the reactants and products and $\Delta n$ represents the no. of moles of the reactants minus the no.of moles of the products.
Now considering the statement as:
S is correct as ${{K}_{c}},{{K}_{p}}\text{ and }{{K}_{x}}$ are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively.
E is also correct as ${{K}_{c}}\text{ and }{{\text{K}}_{p}}$ do not depend on equilibrium but ${{K}_{x}}$ depends upon equilibrium pressure if $\Delta n\ne 0$ because ${{K}_{x}}={{K}_{p}}{{({{P}_{T}})}^{-\Delta n}}$ .
Hence, Both S and E are correct but E is not the correct explanation of S.
So, the correct answer is “Option D”.
Note: Always keep in mind that equilibrium constants change with the change in the temperature and as the temperature is increased, the value of equilibrium constant also increases and vice-versa.
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