Chapter 14, Problem 14.22P

Interpretation Introduction

Interpretation:

The temperature at which the decomposition pressure of limestone is 1 atm should be determined

Concept Introduction:

• For a given reaction at a certain temperature (T) and pressure (P), the equilibrium constant K is related to the free energy change ?G_T by the relation:

${\text{lnK}}_{\text{eq}}\text{ = }\frac{{\text{ΔG}}_{\text{T}}}{\text{RT}}\text{ -------(1)}$

• The equilibrium constant can also be expressed in terms of the coefficients of heat capacity C_p as follows:

  $\begin{array}{l}{\text{lnK}}_{\text{eq}}\text{ = }\frac{{\text{ΔG}}_{{}_{\text{T}}}^0}{\text{RT}}\text{ = }\frac{{\text{ΔG}}_{298}^0{\text{-ΔH}}_{{}_{298}}^0}{{\text{RT}}_{\text{0}}}\text{ + }\frac{{\text{ΔH}}_{298}^0}{\text{RT}}\text{ + ΔAlnT - }\frac{\text{ΔB}}{\text{2}}{\text{T}}^{\text{2}}\text{ + }\frac{\text{ΔC}}{\text{6}}{\text{T}}^{\text{3}}\text{ -----(2)}\\ \text{where:}\\ {\text{ΔG}}_{298}^0{\text{ and ΔH}}_{{}_{298}}^0\text{ are free energy and enthalpy of the reaction at a reference temperature of 298 K}\end{array}$