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

${K_p}\;$ 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.