Chapter 15.2, Problem 1RC

Interpretation Introduction

Interpretation:

For the given reactions, value of equilibrium constant ${\text{K}}_{\text{c}}$ is equal to ${\text{K}}_{\text{P}}$ has to be predicted.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Relation between $K_c$ and$K_P$:

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{\text{Δn}}\\ \text{where,}\\ \text{Δn}\text{is}\text{moles}\text{of}\text{products}\text{-}\text{moles}\text{of}\text{rectants}\\ \text{R}\text{is}\text{Gas}\text{Constant}\\ \text{T}\text{is}\text{Temperature}\text{in}\text{Kelvin}\end{array}$