Chapter 16, Problem 16.134MP

Interpretation Introduction

Interpretation:

For the given reaction molar concentration of ${\text{N}}_{\text{2}}$ $H_2$ and $N{H}_3$ at equilibrium has to be calculated.

Concept Introduction:

Standard free energy change of a reaction is related to equilibrium constant of a reaction involving gases as,

$\Delta G°=-{\text{ RT ln K}}_p$

Where,

$\begin{array}{l}\Delta G°=\text{ standard free energy change}\\ R=\text{ Universal gas constant}\\ \text{T}=\text{ Temperature}\\ {\text{K}}_b\text{= equilibrium constant, also equivalent to dissociation constant of base}\end{array}$

Standard free energy change of a reaction is also related to standard enthalpy change and entropy change as follows,

$\Delta G°=\Delta H°-T\Delta S°$

$\Delta H°$ refers to standard enthalpy change.  Change in enthalpy in a reaction and heat of formation are related by the formula,

$\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)$

Where,

$\begin{array}{l}\Delta H_{rxn}^{°}=\text{ enthalpy change in reaction}\\ \Delta H_f^{°}(product)=\text{ enthalpy or heat of formation of product}\\ \Delta H_f^{°}(reac\tan t)=\text{ enthalpy or heat of formation of }reac\tan t\end{array}$

$\Delta S°$ refers to standard entropy of the reaction which is termed as entropy change in a reaction under standard conditions of $298K$ and $1\text{ atm}$ pressure.  It is represented as,

$\Delta S°=S°\left(products\right)-S°\left(reac\tan ts\right)$