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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
Quick Facts
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50 Questions around this concept.
Solve by difficulty
For the reaction, the value of
is equal to
For the reaction, if
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 ?
What is the equilibrium expression for the reaction
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)
$
Concepts Covered - 2
Relation between Kp and Kc
Relation between Kp and Kc
Let us suppose we have a reaction :
The equilibrium constant Kc for this reaction is given as:
For the reaction:
The equilibrium constant Kp is given as:
Now, from Ideal gas Equation
Putting the value of P in terms of C in the expression for KP
It can be seen that:
Characteristics of Equilibrium Constant
The equilibrium constant has the following characteristics:
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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.
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If the reaction is reversed. The value of the equilibrium constant is inversed.
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If the equation is divided by 2, the equilibrium constant for the new equation is the square root of K, i.e
.
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If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of K, i.e K2.
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If the equation is written in two steps, then K = K1 x K2.
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The magnitude of the equilibrium constant gives an idea of the relative amounts of reactants and the products.
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The value of the equilibrium constant is not affected by the addition of a catalyst to the reaction.
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