Question

Relationship between $ {K_p}\; $ and $ {K_c}\; $ is as follows:
 $ {K_p}\; = {{\text{K}}_{\text{c}}}{\left( {RT} \right)^{ - \vartriangle n}} $
if true enter $ 1 $, else enter $ 0 $.

Answer 1

Hint: $ {K_p}\; $ and $ {K_c}\; $ are the equilibrium constants for ideal type gas mixtures for reversible reactions. When concentration is expressed in terms of atmospheric pressure $ {K_p} $ is used. When concentration is expressed in terms of molarity $ {K_c} $ is used. We will derive the relation between $ {K_p}\; $ and $ {K_c}\; $ to see how they are related to each other.

Complete answer:
For deriving the relation between $ {K_p}\; $ and $ {K_c}\; $ , consider a simple reaction given below:
 $ pP + qQ \to rR + sS $
In above reaction, ‘p’ mole of reactant ‘P’ reacts with ‘q’ mole of reactant ‘Q’ to give product ‘r’ mole of R and ‘s’ mole of S. here p, q, r, s are stoichiometric coefficients of P, Q, R, and S respectively.
 $ {K_c}\; $ is the equilibrium constant associated with concentration. it is given by
 $ {{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {\text{R}} \right]}^{\text{r}}}{{\left[ {\text{S}} \right]}^{\text{s}}}}}{{{{\left[ {\text{P}} \right]}^{\text{p}}}{{\left[ {\text{Q}} \right]}^{\text{q}}}}} $ …… ( $ 1 $ )
where, R is the molar concentration of product ‘R’, S is the molar concentration of product ‘S’, P is the molar concentration of reactant ‘P’, Q is the molar concentration of reactant ‘Q’.
Similarly,
 $ {{\text{K}}_{\text{p}}} = \dfrac{{{{\left[ {{{\text{P}}_{\text{R}}}} \right]}^{\text{r}}}{{\left[ {{{\text{P}}_{\text{S}}}} \right]}^{\text{s}}}}}{{{{\left[ {{{\text{P}}_{\text{P}}}} \right]}^{\text{p}}}{{\left[ {{{\text{P}}_{\text{Q}}}} \right]}^{\text{q}}}}} $ ……… ( $ 2 $ )
where, $ {{\text{P}}_{\text{R}}} $ is the Partial pressure of product R, $ {{\text{P}}_{\text{S}}} $ is the Partial pressure of product S, $ {{\text{P}}_{\text{P}}} $ is the Partial pressure of reactant P, $ {{\text{P}}_{\text{Q}}} $ is the Partial pressure of reactant Q.
 Ideal Gas Equation,
 $ pV = nRT $ or $ {\text{p}} = \dfrac{{nRT}}{{\text{V}}} $ …….. ( $ 3 $ )
we know $ \dfrac{{\text{n}}}{{\text{V}}} $ is the formula of molarity now arranging equation ( $ 1 $ ) and ( $ 2 $ ) in ( $ 3 $ ) we get,
 $ {{\text{P}}_{\text{R}}} = \left[ {\text{R}} \right]RT $
 $ {{\text{P}}_{\text{S}}} = \left[ {\text{S}} \right]RT $
now for reactants P and Q
 $ {{\text{P}}_{\text{P}}} = \left[ {\text{P}} \right]RT $
 $ {{\text{P}}_{\text{Q}}} = \left[ {\text{Q}} \right]RT $
Substituting above $ 4 $ equations in equation ( $ 2 $ ) we get
 $ {{\text{K}}_{\text{p}}} = \dfrac{{{{\left[ {\text{R}} \right]}^{\text{r}}}{{\left[ {\text{S}} \right]}^{\text{s}}}{{\left( {RT} \right)}^{\left( {{\text{r}} + {\text{s}}} \right)}}}}{{{{\left[ {\text{P}} \right]}^{\text{p}}}{{\left[ {\text{S}} \right]}^{\text{q}}}{{\left( {RT} \right)}^{\left( {{\text{p}} + {\text{q}}} \right)}}}} $
 $ {{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{\left[ {RT} \right]^{\left( {{\text{r}} + {\text{s}}} \right) - \left( {{\text{p}} + {\text{q}}} \right)}} $
 $ \left( {{\text{r}} + {\text{s}}} \right) - \left( {p + q} \right) $ signify change in moles of product and reactant, which can be represented by $ \vartriangle n $
so our relation becomes
 $ {K_p}\; = {{\text{K}}_{\text{c}}}{\left( {RT} \right)^{\vartriangle n}} $
So given formula $ {K_p}\; = {{\text{K}}_{\text{c}}}{\left( {RT} \right)^{ - \vartriangle n}} $ is wrong.

Note:
Closely observe the variables representing concentration and partial pressure of each reactant and product as it could get very confusing. Also assign variables in a manner that did not confuse with some other quantity.