Chapter 19, Problem 19.34P

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{p}}$ for the given reaction equation, ${\text{CaCO}}_{\text{3(s)}}{\text{ + C}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ CaO}}_{\text{(s)}}{\text{ + 2CO}}_{\text{(g)}}$ has to be calculated.

Concept Introduction:

${\text{K}}_{\text{c}}$ or ${\text{K}}_{\text{p}}$ for forward and reverse reaction:

A equilibrium constant (or reaction quotient) for a forward reaction is the reciprocal of the equilibrium constant (or reaction quotient) for the reverse reaction.

  ${\text{K}}_{\text{c(fwd)}}=\frac{1}{{\text{K}}_{\text{c(rev)}}}{\text{ and Q}}_{\text{c(fwd)}}=\frac{1}{{\text{Q}}_{\text{c(rev)}}}$

When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. When a reaction is reversed, the reciprocal of the equilibrium constant is used as the new equilibrium constant.

If an overall reaction is the sum of two or more reactions, the overall equilibrium constant (or reaction quotients) is the product of the equilibrium constants (or reaction quotients) for the steps:

  $\begin{array}{l}{\text{Q}}_{\text{overall}}={\text{ Q}}_{\text{1}}\times {\text{Q}}_{\text{2}}\times {\text{Q}}_{\text{3}}\times \mathrm{........}\\ {\text{K}}_{\text{overall}}={\text{ K}}_{\text{1}}\times {\text{K}}_{\text{2}}\times {\text{K}}_{\text{3}}\times \mathrm{........}\end{array}$