Chapter 17, Problem 17.75QE

(a)

Interpretation Introduction

Interpretation:

The values of ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ has to be calculate for given two different temperatures.

  $\text{2NO}\left(\text{g}\right)\text{+}{\text{Br}}_{\text{2}}\left(\text{g}\right)\to \text{2NOBr}\left(\text{g}\right)$

Answer to Problem 17.75QE

The calculated Standard Gibbs free energy at $\text{400}\text{K}$ is $0.7kJ$ and the calculated Standard Gibbs free energy at $\text{600K}$ is $24.6kJ.$

Explanation of Solution

The given reaction as follows,

  $\text{2NO}\left(\text{g}\right)\text{+}{\text{Br}}_{\text{2}}\left(\text{g}\right)\to \text{2NOBr}\left(\text{g}\right)$

• The value of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{2NO}\left(\text{g}\right)\text{+}{\text{Br}}_{\text{2}}\left(\text{g}\right)\to \text{2NOBr}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)90.2530.9182.1\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}82.1\text{kJ/mol}\right)-\left(2\text{×}90.25\text{kJ/mol}\right)-\left(1\text{×}30.91\text{kJ/mol}\right)\\ \\ =-47.21kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $-47.21kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{2NO}\left(\text{g}\right)\text{+}{\text{Br}}_{\text{2}}\left(\text{g}\right)\to \text{2NOBr}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)210.65245.35273.5\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(2\text{×}273.5\text{J/mol}\text{.K}\right)-\left(2\text{×}210.65\text{J/mol}\text{.K}\right)-\left(1\text{×}245.35\text{J/mol}\text{.K}\right)\\ \\ =-119.65J/mol.K.\end{array}$

Hence, value of $\Delta {\text{S}}^{\circ }$ for the process is $-119.65J/mol.K.$

Using the equation as follows, the Standard Gibbs free energy can be determined as,

  $\begin{array}{c}At,T=400K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}-47.21\text{kJ}-\left(400K\right)\left(-\text{119}\text{.65}\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\left(\text{at}\text{equilibrium}\right)\\ \\ \text{=}\text{0}\text{.65kJ}\simeq 0.7kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{400}\text{K}$ is $0.7kJ$

  $\begin{array}{c}At,T=600K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}-47.21\text{kJ}-\left(600K\right)\left(-\text{119}\text{.65}\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}\text{24}\text{.58kJ}\simeq 24.6kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{600K}$ is $24.6kJ.$

(b)

Interpretation Introduction

Interpretation:

The values of ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ has to be calculate for given two different temperatures.

  ${\text{2NH}}_{\text{3}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)$

Answer to Problem 17.75QE

The calculated Standard Gibbs free energy at $\text{400}\text{K}$ is $196.0kJ$ and the calculated Standard Gibbs free energy at $\text{600K}$ is $222.6kJ.$

Explanation of Solution

The given reaction as follows,

  ${\text{2NH}}_{\text{3}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)$

• The value of $\Delta {\text{H}}_{rxn}^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}{\text{2NH}}_{\text{3}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-46.1150.630\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}50.63\text{kJ/mol}\right)-\left(2\text{×}-46.11\text{kJ/mol}\right)\\ \\ =142.85kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $142.85kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}{\text{2NH}}_{\text{3}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)192.34121.21130.57\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}130.57\text{J/mol}\text{.K}\right)+\left(1\text{×}121.21\text{J/mol}\text{.K}\right)-\left(2\text{×}192.34\text{J/mol}\text{.K}\right)\\ \\ =-132.9J/mol.K.\end{array}$

Hence, value of $\Delta {\text{S}}^{\circ }$ for the process is $-132.9J/mol.K.$

Using the equation as follows, the Standard Gibbs free energy can be determined as,

  $\begin{array}{c}At,T=400K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}142.85\text{kJ}-\left(400K\right)\left(-\text{132}\text{.9}\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}196.0kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{400}\text{K}$ is $196.0kJ.$

  $\begin{array}{c}At,T=600K,\\ \\ {\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{T}{\text{ΔS}}^{\text{o}}\\ \\ \text{=}142.85\text{kJ}-\left(600K\right)\left(-1\text{32}\text{.9}\frac{\text{J}}{\text{K}}\text{×}\frac{\text{1}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \\ \text{=}\text{222}\text{.59kJ}\simeq 222.6kJ.\end{array}$

Therefore, the calculated Standard Gibbs free energy at $\text{600K}$ is $222.6kJ.$