Chapter 20, Problem 20.76P

Interpretation Introduction

Interpretation:

For the following given reaction the standard equilibrium ${\text{K}}_p$ value has to be calculated at ${25}^{o}C$.

    $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

Concept introduction:

Free energy change$\text{ΔG}$: change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^{\text{o}}\text{+}\text{RT}\ln \left(\text{K}\right)\\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}-{\text{TΔS}}^{\text{o}}\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{o}}$, ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ is standard free energy, enthalpy and entropy values.

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{mΔG}}_{\text{f}}{}^{\text{°}}}\text{(Products)-}{\displaystyle \sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}}\text{(Reactants)}$

Where,

  ${\text{nΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{Reactants}\right)}$ is the standard entropy of the reactants

  ${\text{mΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{products}\right)}$ is the standard free energy of the products