Question

Relation between kp and kc

Answer 1

Here is your answer.........
Relation between Kp and Kc : -
Kp = Kc (RT)^∆n

derivation:
let us consider the following reversible reaction:

aA + bB ⇌ cC + dD

the equilib. const. for the reaction expressed in terms of the concentration (mol / lit) may be expressed as:

K c = [C]^c [D]^d / [A]^a [B]^b

the equilib. const. in terms of partial pressures may be given as:

K p = p^cC p^dD / p^aA p^bB

here, pA, pB, pC and pD represents the partial pressures of the substance A, B, C and D resp. if gases are assumed to be ideal, then according to ideal gas equation:

pV = n*R*T
p = nRT / V
Where p = pressure in Pa
n = amount of gas in mol
V = Volume in m³
T = temperature (in K)

n/V = concentration, C

or

p = C*R*T or [gas] RT

if C is in mol dm⁻³ and p is in bar,
R = 0.0831 bar dm³ mol⁻¹ K⁻¹
so at constant temp,
pressure of the gas P is proportional to its concentration C
tht is,

 a general reaction:

aA + bB↔ cC + dD

The equilibrium constant will be given as:

Kp = (pC) c (pD) d / (pA) a (pB) b __________(1)

 p = CRT

hence

pA = [A] RT

similarly,
pB = [B] RT
pC = [C] RT
pD = [D] RT

where [B], [C] and [D] are the molar concentration

so,
Kp = [([C] RT) c ([D] RT) d]/[([A] RT) a ([B] RT) b]
= [C] c [D] d (RT) c+d/[A] a [B] b (RT) a+b
= [C] c [D] d (RT) c+d – a+b/[A] a [B] b
= Kc (RT)^c+d – a+b
= Kc (RT) ^∆n

here ∆n = (c + d) – (a + b)
that is no. of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.

finally, relation between Kp and Kc is given as:

Kp = Kc (RT)^∆n

I hope it helps you ^_^

Answer 2

Answer:

answer in attachment