Chapter 18, Problem 18.50QP

Calculate ΔH° and ΔG° for the following reactions at 25°C, using thermodynamic data from Appendix C; interpret the signs of ΔH° and ΔG°.

1. a $2\text{PbO}(s)+{\text{N}}_2(g)\stackrel{}{\to }2\text{Pb}(s)+2\text{NO}(g)$\
2. b ${\text{CS}}_{\text{2}}(l)+2{\text{H}}_{\text{2}}\text{O(}l)\stackrel{}{\to }{\text{CO}}_2(g)+2{\text{H}}_{\text{2}}\text{S(}g\text{)}$

Interpretation Introduction

Interpretation:

For the given reactions the values and sign of ${\text{ΔH}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$ has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Standard free energy of formation:

The change in free energy for the formation of $\text{1}\text{mole}$ of any substance from its starting materials in their standard (stable) states (i.e. $\text{1}\text{atm}\text{and}\text{25}{}^{\text{o}}\text{C}$).

Answer to Problem 18.50QP

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}\text{619}\text{.4}\text{kJ}\text{. Given reaction is endothermic}\text{.}\\ {\text{ΔG}}^{\text{o}}\text{=}\text{551}\text{.8}\text{kJ}\text{.}\text{Given reaction is non spontaneous reaction}\text{.}\end{array}$

Explanation of Solution

To calculate: The values and sign of ${\text{ΔH}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$

Given reaction and information

$\begin{array}{l}{\text{2PbO}}_{\text{(s)}}\text{+ }{\text{N}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2Pb}}_{\text{(s)}}{\text{+ 2NO}}_{\text{(g)}}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{2}\text{×}\text{-219}\text{.4}\text{0}\text{0}\text{2}\text{×}\text{90}\text{.29}\text{kJ}\\ {\text{ΔG}}^{\text{o}}{}_{\text{f}}\text{:}\text{2}\text{×}\text{-189}\text{.3}\text{0}\text{0}\text{2}\text{×}\text{86}\text{.60}\text{kJ}\end{array}$

Calculate the value and sign of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{(2}\text{×90}\text{.29) - (2}\text{×}\text{-219}\text{.4)}\right]\text{kJ}\\ \text{=}\left[\text{180}\text{.58}\text{+}\text{438}\text{.8}\right]\text{kJ}\\ \text{=}\text{619}\text{.4}\text{kJ}\text{.}\end{array}$

The given reaction is endothermic. It absorbs $\text{619}\text{.4}\text{kJ}$ of heat.

Calculate the value and sign of ${\text{ΔG}}^{\text{o}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{(2}\text{×86}\text{.60) - (2}\text{×}\text{-189}\text{.3)}\right]\text{kJ}\\ \text{=}\left[\text{173}\text{.2}\text{+}\text{378}\text{.6}\right]\text{kJ}\\ \text{=}\text{551}\text{.8}\text{kJ}\text{.}\end{array}$

The larger positive value of ${\text{ΔG}}^{\text{o}}$ shows that the given reaction is non-spontaneous reaction (equilibrium composition is mainly reactants).

Interpretation Introduction

Interpretation:

For the given reactions the values and sign of ${\text{ΔH}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$ has to be calculated.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Standard free energy of formation:

The change in free energy for the formation of $\text{1}\text{mole}$ of any substance from its starting materials in their standard (stable) states (i.e. $\text{1}\text{atm}\text{and}\text{25}{}^{\text{o}}\text{C}$).

Answer to Problem 18.50QP

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}\text{47}\text{.4}\text{kJ}\text{. Given reaction is endothermic}\text{.}\\ {\text{ΔG}}^{\text{o}}\text{=}\text{-52}\text{.1}\text{kJ}\text{.}\text{Given reaction is spontaneous reaction}\text{.}\end{array}$

Explanation of Solution

To calculate: The values and sign of ${\text{ΔH}}^{\text{o}}\text{and}{\text{ΔG}}^{\text{o}}$

Given reaction and information

$\begin{array}{l}{\text{CS}}_{\text{2}}{}_{\text{(l)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ CO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2H}}_{\text{2}}{\text{S}}_{\text{(g)}}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{89}\text{.70}\text{2}\text{×}\text{-285}\text{.8}\text{-393}\text{.5}\text{2}\text{×}\text{-20}\text{.50}\text{kJ}\\ {\text{ΔG}}^{\text{o}}{}_{\text{f}}\text{:}\text{65}\text{.27}\text{2}\text{×}\text{-237}\text{.1}\text{-394}\text{.4}\text{2}\text{×}\text{-33}\text{.33}\text{kJ}\end{array}$

Calculate the value and sign of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{-}\text{393}\text{.5}\right)\text{+}\left(\text{2}\text{×}\text{-20}\text{.50}\right)\right]\text{-}\left[\left(\text{89}\text{.70}\right)\text{+}\left(\text{2}\text{×}\text{-285}\text{.8}\right)\right]\text{kJ}\\ \text{=}\left[\left(\text{-434}\text{.5}\right)\right]\text{-}\left[\left(\text{-481}\text{.9}\right)\right]\text{kJ}\\ \text{=}\text{47}\text{.4}\text{kJ}\text{.}\end{array}$

The given reaction is endothermic. It absorbs $\text{47}\text{.4}\text{kJ}$ of heat.

Calculate the value and sign of ${\text{ΔG}}^{\text{o}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔG}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{-}\text{394}\text{.4}\right)\text{+}\left(\text{2}\text{×}\text{-33}\text{.33}\right)\right]\text{-}\left[\left(\text{65}\text{.27}\right)\text{+}\left(\text{2}\text{×}\text{-237}\text{.1}\right)\right]\text{kJ}\\ \text{=}\left[\text{-461}\text{.06}\text{+}\text{408}\text{.93}\right]\text{kJ}\\ \text{=}\text{-}\text{52}\text{.1}\text{kJ}\text{.}\end{array}$

The negative value of ${\text{ΔG}}^{\text{o}}$ shows that the given reaction is spontaneous reaction (equilibrium composition is mainly products).