Chapter 14, Problem 14.7P

Interpretation Introduction

Interpretation:

To develop general equation of standard Gibbs energy of reaction $\Delta G_{}^0$

N₂ + 3H₂ → 2 NH₃

Concept Introduction :

Gibbs change of Energy is given as below,

  $\Delta G^o=\Delta H_0^0-\frac{T}{T_O}(\Delta H_0^0-\Delta G_0^0)+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT-RT{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{RT}}dT$

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

Where,

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}=A\ln \tau +\left[B{T}_0^{}+\left(C{T}_0^2+\frac{D}{T_0^2}\right)\left(\frac{\tau +1}{\tau }\right)\right](\tau -1)$

A, B, C, D are constants

T_o = Initial temperature = 25 ⁰C = 298 K

T = Final temperature

  $\tau =\frac{T}{T_o}=\frac{T}{298}$

C_p = Molar heat capacity

R = Universal Gas constant = 8.314 J/molK

  $\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

  $\Delta H_{298}^0$ = Heat of formation at 298 K

  $\Delta H_0^0$ = Standard Heat of Reaction

  $\Delta G_{298}^0$ = Standard Gibbs Energy of reaction at 298 K

  $\Delta G_{}^0$ = Standard Gibbs Energy of reaction