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Relation between Kp and Kc - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

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  • 66 Questions around this concept.

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For the reaction, \mathrm{CO_{(g)}+Cl_{2(g)}\rightleftharpoons COCl_{2(g)}}    the value of \mathrm {\frac{K_{p}}{K_{c} }}   is equal to

For the reaction, \mathrm{SO_{2(g)}+\frac{1}{2}O_{2(g)}\rightleftharpoons SO_{3(g)}}  if \mathrm{K_P=K_C(RT)^x} where the symbols have usual meaning then the value of x is 

Change in volume of the system does not alter the number of moles in which of the following equilibria?

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In which of the following reactions, increase in the volume at constant temperature does not affect the number of moles at equilibrium ?

$\mathrm{PCl}_5, \mathrm{PCl}_3$ and $\mathrm{Cl}_2$ are at equilibrium at 500 K in a closed container and their concentrations are $0.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}{ }^{-1}, 1.2 \times 10^{-3} \mathrm{mol} L^{-1}$ and $1.2 \times 10^{-3} \mathrm{molL}^{-1}$ respectively. The value of $K c$ for the reaction $\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ will be

For which of the following equilibrium, the value of $K_P$  is not greater than the value of $K_C$  ? 

 

What is the equilibrium expression for the reaction

P_{4(s)}+5O_{2(g)}\rightleftharpoons P_{4}O_{10(s)}?

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At $1200 \mathrm{~K}, K_c=36$ for the reaction: $\mathrm{H}_2(g)+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$. What is $K_c$ for the following reaction at $1200 \mathrm{~K}$ ?

$
HI(g) \rightleftharpoons \frac{1}{2} H_2(g)+\frac{1}{2} \mathrm{I}_2(g)
$

For a given liquid equlibrium vapour pressure depends upon :-

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For the given hypothetical reactions, the equilibrium constants are as follows:

$\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0\end{aligned}$

$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0$

The equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is

Concepts Covered - 2

Relation between Kp and Kc

Relation between Kp and Kc

Let us suppose we have a reaction :

\mathrm{n}_{1} \mathrm{A}(\mathrm{g})+\mathrm{n}_{2} \mathrm{B}(\mathrm{g}) \leftrightharpoons \mathrm{n}_{3} \mathrm{C}(\mathrm{g})+\mathrm{n}_{4} \mathrm{D}(\mathrm{g})

The equilibrium constant Kc for this reaction is given as:

\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{C}]^{\mathrm{n}_{3}}[\mathrm{D}]^{\mathrm{n}_{4}}}{[\mathrm{A}]^{\mathrm{n}_{1}}[\mathrm{B}]^{\mathrm{n}_{2}}}

For the reaction:

\mathrm{n}_{1} \mathrm{A}(\mathrm{g})+\mathrm{n}_{2} \mathrm{B}(\mathrm{g}) \leftrightharpoons \mathrm{n}_{3} \mathrm{C}(\mathrm{g})+\mathrm{n}_{4} \mathrm{D}(\mathrm{g})

The equilibrium constant Kp is given as:

\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{C}}\right)^{\mathrm{n}_{3}}\left(\mathrm{P}_{\mathrm{D}}\right)^{\mathrm{n}_{4}}}{\left(\mathrm{P}_{\mathrm{A}}\right)^{\mathrm{n}_{1}}\left(\mathrm{P}_{\mathrm{B}}\right)^{\mathrm{n}_{2}}}

Now, from Ideal gas Equation

\mathrm{PV = nRT}

\mathrm{P =\frac{n}{V} RT}

\mathrm{P =C RT}

Putting the value of P in terms of C in the expression for KP

\mathrm{K _{ p }=\frac{[ C ]^{ n _{3}}( RT )^{ n _{3}}[ D ]^{ n _{4}}( RT )^{ n _{4}}}{[ A ]^{ n _{1}}( RT )^{ n _{1}}[ B ]^{ n _{2}}( RT )^{ n _{2}}}}

\mathrm{K _{ P }=\frac{[ C ]^{ n _{3}}[ D ]^{ n _{4}}}{[ A ]^{ n _{1}}[ B ]^{ n _{2}}}[ RT ]^{\left( n _{3}+ n _{4}\right)-\left( n _{1}+ n _{2}\right)}}

\mathrm{Putting \ \Delta n_g = \left (n_3 + n_4 \right ) - \left (n_1+n_2 \right ), we\ have}

\mathrm{K_P = K_C \left (RT \right )^{\Delta n_g}}

It can be seen that: 

  • \mathrm{When\ \Delta n_g = 0, then\ K_P= K_C}
  • \mathrm{When\ \Delta n_g > 0, then\ K_P> K_C}
  • \mathrm{When\ \Delta n_g < 0, then\ K_P< K_C}
Characteristics of Equilibrium Constant

The equilibrium constant has the following characteristics:

  • The value of the equilibrium constant for a particular reaction is always constant depending only upon the temperature of the reaction and is independent of the concentrations of the reactants with which we start of the direction from which the equilibrium is approached.

  • If the reaction is reversed. The value of the equilibrium constant is inversed.

  • If the equation is divided by 2, the equilibrium constant for the new equation is the square root of K, i.e \sqrt{K}.

  • If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of K, i.e K2.

  • If the equation is written in two steps, then K = K1 x K2.

  • The magnitude of the equilibrium constant gives an idea of the relative amounts of reactants and the products.

  • The value of the equilibrium constant is not affected by the addition of a catalyst to the reaction.

Study it with Videos

Relation between Kp and Kc
Characteristics of Equilibrium Constant

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Books

Reference Books

Relation between Kp and Kc

Chemistry Part I Textbook for Class XI

Page No. : 202

Line : 1

Characteristics of Equilibrium Constant

Chemistry Part I Textbook for Class XI

Page No. : 200

Line : 1

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