Chapter 6, Problem 6.15CP

Interpretation Introduction

Interpretation:For the given reaction, the change in partial pressure ${\text{CO}}_{\text{2}}$ needs to be explained ifXe gas is added to the reaction at constant pressure andtemperature.

  $\text{CO}\left(g\right)\text{ + }\text{1}/\text{2}{\text{O}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{CO}}_{\text{2}}\left(g\right){\text{ ΔH}}_{\text{R}}^{\text{0}}\text{ = -283}\text{.0 kJ/mol}$

Concept Introduction: A reversible chemical reaction can move in either forward or backward direction. The stage of a reversible reaction at which the rate of forward and backward reaction become equal to each other is called an equilibrium stage. The equilibrium constant for a reaction is the ratio of equilibrium concentration of product and reactant. If the reaction is given in gaseous state, then the equilibrium pressure must be considered. The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ can be written as:

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{=K}}_{\text{c}}{\text{(R×T)}}^{\text{Δn}}\\ \begin{array}{l}\text{Here:}\hfill \\ {\text{K}}_{\text{p}}{\text{and K}}_{\text{c}}\text{= equilibrium constant}\hfill \\ \text{R = Gas constant}\hfill \\ \text{T = temperature}\hfill \\ \Delta \text{n=number of gaseous product – number of gaseous reactants}\hfill \end{array}\end{array}$