Question

Relation between Kc and Kp​

Answer 1

Answer:

Derive the relationship between Kp and Kc. Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e. Where ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.

Answer 2

Solution:

$\boxed{\sf{xA+yB⇌lC+mD}}$

$\tt \large \: K_{c}=\frac{[C]^{l}[D]^{m}}{[A]^{x} [B]^{y}}$

$\tt \large \: K_{p}=\frac{P_{C}^{l}×P_{D}^{m}}{P_{A}^{x}×P_{B}^{y}}$

Ideal Gas Equation:

$\boxed{\tt \:P = \frac{n}{V} RT}$

$P_{A}=[A]RT; \\ P_{B}=[B]RT; \\ P_{C}=[C]RT; \\ P_{D}=[D]RT$

$\tt \large \: K_{p}=\frac{P_{C}^{l}×P_{D}^{m}}{P_{A}^{x}×P_{B}^{y}}= \frac{[[C]RT]^{l}×[[D]RT]^{m}}{[[A]RT]^{x}×[[B]RT]^{y}}$

$\large \: \tt \: =\frac{[C]^{l}[RT]^{l}×[D]^{m}[RT]^{m}}{[A]^{x}[RT]^{x}×[B]^{y}[RT]^{y}}$

$\large \tt= \frac{[C]^{l} [D]^{m}\: [RT]^{l+m}}{[A]^{x}[B]^{y} \: [RT] ^{x+y}}$

$\tt K_{p}=K_{c} [RT]^{(l+m)-(x+y)}$

$\tt K_{p}=K_{c}[RT]^{∆ng} \: => \boxed{\sf ∆ng=0}$

$\tt K_{p}=K_{c}[RT]^0$

$\huge \boxed{\sf K_{p}=K_{c}}$