Chapter 20.3, Problem 20.4BFP

Interpretation Introduction

Interpretation:

The standard free energy change $\Delta {\text{G}}_{\text{rxn}}^{\text{o}}$ has to be calculated for given reaction at $298K$.

    $4N{H}_3\left(g\right)+5O_2\left(g\right)\stackrel{}{\to }2NO\left(g\right)+6H_{2}O\left(g\right)$

Concept introduction:

• Enthalpy $\left(\text{H}\right)$: it is the total amount of heat in a particular system.
• Entropy $\left(\text{S}\right)$ : it is used to describe the disorder. It is the amount of arrangements possible in a system at a particular state. ${\text{ΔS}}_{\text{univ}}={\text{ΔS}}_{\text{sys}}+{\text{ΔS}}_{\text{surr}}$
• Free energy change $\left({\text{ΔG}}^{\text{o}}\right)$: change in the free energy takes place while reactant converts to product where both are in standard state.

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

  ${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ can be calculated by the following equation.

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}{\text{ = S}}^{\text{°}}{}_{\text{Products}}-{\text{ S}}^{\text{°}}{}_{\text{reactants}}$

Where,

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard 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 $G^o$.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Where,

${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ is the change in standard free energy of the system

$\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ is the change in standard enthalpy of the system

$\text{T}$ is the absolute value of the temperature

$\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$ is the change in entropy in the system

Answer to Problem 20.4BFP

The standard free-energy change of given reaction is $\underset{\_}{-959kJ}$.

Explanation of Solution

Given,

Standared free energy equation is,

$4N{H}_3\left(g\right)+5O_2\left(g\right)\stackrel{}{\to }2NO\left(g\right)+6H_{2}O\left(g\right)$

The balanced equation rightfully, multiplying the reactant $N{H}_3,O_2$ and the products $NOand{H}_{2}O$ with the coefficents $4,5,2and6$ respectively.

Calculation of enthalpy ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ value at $298K$ 

Standard enthalpy change equation is,

${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

$\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(4\text{mol}NO\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofNO\right)+\left(6mol{H}_{2}O\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{H}_{2}O\right)\right]\\ -\left[\left(4molN{H}_3\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofN{H}_3\right)+\left(5mol{O}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{O}_2\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(4\text{mol}NO\right)\left(90.29kJ/mol\right)+\left(6mol{H}_{2}O\right)\left(-241.826kJ/mol\right)\right]-\\ \left[\left(4molN{H}_3\right)\left(-45.9kJ/mol\right)+\left(5mol{O}_2\right)\left(0kJ/mol\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-906.196kJ\end{array}$

The enthalpy change is expected to be negative because in gas mole is negative.

Therefore, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{-906.196kJ}$ 

Calculation of entropy ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ value at $298K$

Standard entropy change equation is,

${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(4\text{mol}NO\right)\left(S^{o}ofNO\right)+\left(6mol{H}_{2}O\right)\left(S^{o}of{H}_{2}O\right)\right]\\ -\left[\left(4molN{H}_3\right)\left(S^{o}ofN{H}_3\right)+\left(5mol{O}_2\right)\left(S^{o}of{O}_2\right)\right]\\ \\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(4\text{mol}NO\right)\left(210.65\text{J/mol×K}\right)+\left(6mol{H}_{2}O\right)\left(\text{188}\text{.72}\text{J/mol×K}\right)\right]-\\ \left[\left(4molN{H}_3\right)\left(193.0J/mol\times K\right)+\left(5mol{O}_2\right)\left(205.0J/mol\times K\right)\right]\\ \\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}177.92\text{J/K}\end{array}$

Hence, the entropy ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{177.92\text{J/K}}$ 

Calculation of standard entropy value $\Delta {\text{G}}_{\text{rxn}}^{\text{o}}$

The standard entropy equation is,

${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Enthalpy and entropy values are substituted in above equation.

$\begin{array}{l}\Delta G_{rxn}^o=-906.196kJ-\left[\left(298K\right)\left(177.92\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\right]\\ \\ \Delta G_{rxn}^o=-959.21616kJ\\ \\ \Delta G_{rxn}^o=-959kJ(Roundedvalue)\end{array}$

Hence, the standard free-energy change of the reaction is $\Delta G_{rxn}^o=\underset{\_}{-959kJ}$

The standard free energy of the reaction is negative. So, the reaction is spontaneous.