Chapter 3, Problem 3D.3E

(a)

Interpretation Introduction

Interpretation:

The standard reaction entropy for the chemical reaction given has to be calculated.

Concept Information:

Standard reaction entropy: Difference in molar entropy between the products and the reactants in their standard states is called the standard reaction entropy,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}{\displaystyle \sum _{\text{products}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{νS}}_{\text{m}}{}^{\text{o}}}$

Where,

$\text{ν}$ Are the stoichiometric coefficients in the chemical equation.

Explanation of Solution

Given reaction: ${\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}\text{3}{\text{H}}_{\text{2}}\left(\text{g}\right)\to \text{2}{\text{NH}}_{\text{3}}\left(\text{g}\right)$

Molar entropy of Ammonia gas: ${\text{S}}_{\text{m}}{}^{\text{o}}\left({\text{NH}}_{\text{3}}\right)\text{=}\text{192}\text{.45}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Molar entropy of Nitrogen gas: ${\text{S}}_{\text{m}}{}^{\text{o}}\left({\text{N}}_{\text{2}}\right)\text{=}\text{191}\text{.61}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Molar entropy of Hydrogen gas: ${\text{S}}_{\text{m}}{}^{\text{o}}\left({\text{H}}_{\text{2}}\right)\text{=}\text{130}\text{.684}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Hence, the standard entropy of the reaction as follows,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}{\displaystyle \sum _{\text{products}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\\ \\ =2\left(\text{192}\text{.45}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)-\left[\left(\text{191}\text{.61}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)+3\left(\text{130}\text{.684}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\right]\\ \\ =\left(384.9-583.662\right){\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ =-198.8J{K}^{-1}mo{l}^{-1}\end{array}$

Therefore, the change in entropy of the given reaction is $-198.8J{K}^{-1}mo{l}^{-1}$

(b)

Interpretation Introduction

Interpretation:

The change in entropy of the surroundings at $\text{298K}$ has to be calculated.

Concept Information:

Standard reaction entropy: Difference in molar entropy between the products and the reactants in their standard states is called the standard reaction entropy,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}{\displaystyle \sum _{\text{products}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{νS}}_{\text{m}}{}^{\text{o}}}$

Where,

$\text{ν}$ is the stoichiometric coefficients in the chemical equation

Total entropy: The sum of entropy of system and surrounding gives the total entropy.

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}{}_{\text{total}}\text{=}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}{}_{\text{sys}}\text{+}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}{}_{\text{surr}}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}{}_{\text{total}}\text{=}\left[{\displaystyle \sum _{\text{products}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\right]\text{+}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}{}_{\text{surr}}\end{array}$

Explanation of Solution

Given reaction: ${\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}\text{3}{\text{H}}_{\text{2}}\left(\text{g}\right)\to \text{2}{\text{NH}}_{\text{3}}\left(\text{g}\right)$ ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{=}\text{-92}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}$

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}\text{-199}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The change in entropy of the given reaction is ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}\text{-199}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The standard entropy of the surrounding is calculated as follows,

$\begin{array}{l}{\text{ΔS}}^{\circ }{}_{\text{surr}}\text{=}-\frac{{\Delta }_r{\text{H}}^{\circ }}{\text{T}}\\ \\ =-\frac{\left(\text{-92}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\right)}{\left({25}^{\circ }C+273\right)}\\ \\ =+309J{K}^{-1}mo{l}^{-1}\end{array}$

The negative sign of standard enthalpy is because the energy is released (exothermic) to the surrounding and the sign is negative. The entropy is increased in the surrounding results in positive value.

The standard entropy of the surrounding is $+309J{K}^{-1}mo{l}^{-1}$.

(c)

Interpretation Introduction

Interpretation:

The standard Gibbs energy of the reaction has to be calculated.

Concept Information:

Standard reaction entropy: Difference in molar entropy between the products and the reactants in their standard states is called the standard reaction entropy,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}{\displaystyle \sum _{\text{products}}{\text{νS}}_{\text{m}}{}^{\text{o}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{νS}}_{\text{m}}{}^{\text{o}}}$

Where,

$\text{ν}$ is the stoichiometric coefficients in the chemical equation

Gibbs free energy: A thermodynamic quantity equal to the enthalpy (of a system or process) minus the product of the entropy and the absolute temperature.

${\text{ΔG}}_{\text{r}}^{\text{o}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{-}{\text{TΔ}}_{\text{r}}{\text{S}}^{\text{o}}$

Where,

${\text{ΔG}}_{\text{r}}^{\text{o}}$ is the Gibbs free energy,

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change of the reaction,

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}$ is the entropy change of the reaction.

Explanation of Solution

Given reaction: ${\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}\text{3}{\text{H}}_{\text{2}}\left(\text{g}\right)\to \text{2}{\text{NH}}_{\text{3}}\left(\text{g}\right)$ ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{=}\text{-92}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}$

${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}\text{-199}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$; $\text{T}\text{=}\text{298}\text{K}$

The change in entropy of the given reaction is ${\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{=}\text{-199}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Standard Gibb’s free energy is calculated as follows,

$\begin{array}{l}{\text{ΔG}}_{\text{r}}^{\text{o}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{-}{\text{TΔ}}_{\text{r}}{\text{S}}^{\text{o}}\\ \\ =\left(\text{-92}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\right)-\left(298K\right)\left(\text{-199}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \\ =\left(\text{-92}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\right)+\left(59.302\text{kJ}{\text{.mol}}^{\text{-1}}\right)\\ \\ =-32.9kJmo{l}^{-1}\end{array}$

The negative sign of standard Gibbs energy because the energy is released (exothermic) to the surrounding and the sign is negative.

The standard Gibb’s free energy is $-32.9kJmo{l}^{-1}$.