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Relation between Kp and Kc is considered one the most difficult concept.
48 Questions around this concept.
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?
NEET 2025: Syllabus | Most Scoring concepts | NEET PYQ's (2015-24)
In which of the following reactions, increase in the volume at constant temperature does not affect the number of moles at equilibrium ?
In which of the following equilibrium $K_c$ and $K_p$ are not equal?
The equilibrium constants of the following are:
$\begin{aligned} & \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \quad \mathrm{~K}_1 \\ & \mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \quad \mathrm{K}_2 \\ & \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \quad \mathrm{K}_3\end{aligned}$
The equilibrium constant (K) of the reaction:
$2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{\mathrm{~K}}{=} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}$
If the equilibrium constant for
the equilibrium constant for
will be:
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What is the equilibrium expression for the reaction
Relation between Kp and Kc
The reaction we have:
The equilibrium constant Kc for this reaction is given as:
For the reaction:
The equilibrium constant Kp is given as:
As we know:
PV = nRT
P = (n/V)RT
Also, n/V = C
Thus, P = CRT
Therefore, ?n = (n3 + n4) - (n1 + n2)]
Hence, Kp = Kc(RT)?n
Expressions of Kc and Kp for some reactions
Evaluation of Kc and Kp for HI formation
For the reaction:
Initially: a mole b mole 0 mole
At equil: (a - X) (b - X) 2X moles
Thus, according to the equilibrium constant equation we have:
Similarly, for Kp we have;
Partial pressure of individual species are given as follows:
On substituting these values in the above equation, we get:
Evaluation of Kc and Kp for PCl5 dissociation
Similarly for PCl5 dissociation reaction we have:
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 .
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.
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