Chapter 10, Problem 10.61PAE

Estimate the temperature range over which each of the following reactions is spontaneous.

(a) $\text{2Al}\left(\text{s}\right)+{\text{3Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{2AlCl}}_{\text{3}}\left(\text{s}\right)$

(b) $\text{2NOCl}\left(\text{g}\right)\to \text{2NO}\left(\text{g}\right)+C{l}_2(g)$

(c) $\text{4NO}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{4NH}}_{\text{3}}\left(\text{g}\right)+5O_2(g)$

(d) ${\text{2PH}}_{\text{3}}\left(\text{g}\right)\to {\text{3H}}_{\text{2}}\left(\text{g}\right)+\text{2P}\left(\text{g}\right)$

Interpretation Introduction

Interpretation:

The temperature range over which the reaction $\text{2NOCl}\left(\text{g}\right)\to \text{2NO}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)$ is spontaneous should be estimated.

Concept introduction:

A process which happens without any outward intervention is recognized as the spontaneous process. As per the 2^nd law of thermodynamics, that states the universe’s entropy rises for any spontaneous process.

A process which happens when pressure and temperature both are constant, the determination of spontaneity can be done using Gibbs free energy:

Here, a sign of $\text{ΔG}$ is dependent on changes in entropy $\text{ΔS}$ and enthalpy $\text{ΔH}$.

Answer to Problem 10.61PAE

Solution:

The reaction will be spontaneous for temperature higher than 647 K.

Explanation of Solution

In the given reaction due to the presence of more number of moles of gas in the product, the change in entropy is estimated as positive, and due to the presence of more number of bonds in the product compared to the reactant, the enthalpy is also positive because in products bonds are formed more than the reactants.

$\text{2NOCl(g) }\to {\text{ 2NO + Cl}}_2\text{(g) }$

First, calculate the change in enthalpy by subtracting all of the product enthalpies from the reactant enthalpies:

$\begin{array}{l}\Delta {\text{H}}_{\text{rxn}}^0\text{ = }{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{reactants}}^0}\\ \text{ = 2}\Delta {\text{H}}_{\text{f NO(g)}}^0\text{ + }\Delta {\text{H}}_{{\text{f Cl}}_2(\text{g)}}^0-\text{ 2 }\Delta {\text{H}}_{\text{f NOCl(g)}}^0\\ \text{ = 2}\times \text{90}\text{.25 }\frac{\text{kJ}}{\text{mol}}\text{ + 1}\times \text{0 }\frac{\text{kJ}}{\text{mol}}-2\times 52.29\frac{\text{kJ}}{\text{mol}}\\ \text{ = 75}\text{.32 kJ/mol}\end{array}$

In next step, find the change in entropy after subtracting all the product entropies from the reactant entropies:

$\begin{array}{l}\Delta {\text{S}}_{\text{rxn}}^0\text{ = }{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{reactants}}^0}\\ \text{ = 2}\Delta {\text{S}}_{\text{NO(g)}}^0\text{ + }\Delta {\text{S}}_{{\text{Cl}}_2}^0-\text{ 2 }\Delta {\text{S}}_{\text{NOCl(g)}}^0\\ \text{ = 2}\times \text{210}\text{.7}\frac{\text{J}}{\text{mol}}\text{ + }223.0\frac{\text{kJ}}{\text{mol}}-2\times 264\frac{\text{J}}{\text{mol}}\\ \text{ = 116}\text{.4 kJ/mol}\end{array}$

In the final step, using Gibbs free energy form the first section to find the temperature of spontaneity. Spontaneous reactions occur when the change in Gibb’s free energy is less than zero which means that energy is released from the system.

$\Delta \text{G = }75.32\frac{\text{kJ}}{\text{mol}}-\text{T}\times 116.4\frac{\text{J}}{\text{mol}\text{.K}}$

To find out the range of the temperature which can cause the negative change in $\Delta \text{G }$ the set $\Delta \text{G }$ to zero and solve for T

$\begin{array}{l}75.32\frac{\text{kJ}}{\text{mol}}\times \frac{{10}^3\text{J}}{1\text{ kJ}}=\text{T}\times 116.4\frac{\text{J}}{\text{mol}\text{.K}}\\ \text{T}>\text{647}\text{.079 K}\end{array}$

This temperature is the cutoff for temperatures for spontaneity. All temperatures higher than this temperature will result in spontaneous reaction because of the larger contribution from entropy which has a positive sign. Therefore, the reaction will be spontaneous for temperature higher than $\boxed{>\text{647 K}}$.

Interpretation Introduction

Interpretation:

The temperature range over which the reaction $\text{4NO}\left(\text{g}\right){\text{+6H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{4NH}}_{\text{3}}\left(\text{g}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right)$ is spontaneous should be estimated.

Concept introduction:

A process which happens without any outward intervention is recognized as the spontaneous process. As per the 2^nd law of thermodynamics, that states the universe’s entropy rises for any spontaneous process.

A process which happens when pressure and temperature both are constant, the determination of spontaneity can be done using Gibbs free energy:

Here, a sign of $\text{ΔG}$ is dependent on changes in entropy $\text{ΔS}$ and enthalpy $\text{ΔH}$.

Answer to Problem 10.61PAE

Solution:

There are no temperatures that this reaction will be spontaneous.

Explanation of Solution

The first change in enthalpy will be calculated after subtracting product enthalpies from the reactant enthalpies.

$\begin{array}{l}\Delta {\text{H}}_{\text{rxn}}^0\text{ = }{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{reactants}}^0}\\ \text{ = 4}\Delta {\text{H}}_{{\text{f NH}}_3\text{(g)}}^0\text{ + }\Delta {\text{H}}_{{\text{f O}}_2\text{(g)}}^0-\text{ 4 }\Delta {\text{H}}_{\text{f NO(g)}}^0-\text{ 6 }\Delta {\text{H}}_{{\text{f H}}_2\text{O(g)}}^0\\ \text{ = 4}\times -\text{46}\text{.11}\frac{\text{kJ}}{\text{mol}}\text{ + 0}\frac{\text{kJ}}{\text{mol}}-4\times 90.25\frac{\text{kJ}}{\text{mol}}+6\times 241.8\frac{\text{kJ}}{\text{mol}}\\ \text{ = 905}\text{.4 kJ/mol}\end{array}$

In next step find the entropy change by subtracting product entropies from the reactant entropies.

$\begin{array}{l}\Delta {\text{S}}_{\text{rxn}}^0\text{ = }{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{reactants}}^0}\\ \text{ = 4}\Delta {\text{S}}_{{\text{NH}}_3\text{(g)}}^0\text{ + 5}\Delta {\text{S}}_{{\text{O}}_2}^0-\text{ 6}\Delta {\text{S}}_{{\text{H}}_2\text{O(g)}}^0\\ \text{ = 4}\times \text{192}\text{.3}\frac{\text{J}}{\text{mol}\cdot \text{K}}\text{ + 5}\times 205.0\frac{\text{J}}{\text{mol}\cdot \text{K}}-4\times 210.7\frac{\text{J}}{\text{mol}\cdot \text{K}}\\ \text{ = }-\text{180}\text{.8 kJ/mol}\end{array}$

In the final step using the expression for Gibb’s free energy from the first section to find the temperature of spontaneity. Spontaneous reactions occur when the change in Gibb’s free energy is less than zero, meaning that energy is released from the system

$\Delta \text{G = }905.4\frac{\text{kJ}}{\text{mol}}\text{ + T}\times \text{ 180}\text{.8}\frac{\text{J}}{\text{mol}\cdot \text{K}}$

The reaction will be never being spontaneous because all temperatures are positive, meaning that $\Delta \text{G }$ will always be positive. There are no temperatures that this reaction will be spontaneous.

Interpretation Introduction

Interpretation:

The temperature range over which the reaction ${\text{2 PH}}_{\text{3}}\left(\text{g}\right)\to {\text{3 H}}_{\text{2}}\left(\text{g}\right)\text{+ 2P}\left(\text{g}\right)$ is spontaneous should be estimated.

Concept introduction:

A process which happens without any outward intervention is recognized as the spontaneous process. As per the 2^nd law of thermodynamics, that states the universe’s entropy rises for any spontaneous process.

A process which happens when pressure and temperature both are constant, the determination of spontaneity can be done using Gibbs free energy:

Here, a sign of $\text{ΔG}$ is dependent on changes in entropy $\text{ΔS}$ and enthalpy $\text{ΔH}$.

Answer to Problem 10.61PAE

Solution:

The temperature must be greater than 201.20 K for the reaction to be spontaneous

Explanation of Solution

This reaction is the opposite of a formation reaction because a compound is split into its elemental states. The following equation is the balanced overall reaction:

${\text{2PH}}_3\text{(g) }\to {\text{ 3H}}_2\text{ + 2P(g) }$

According, the change in entropy for the formation of phosphine can be found in a table of common values:

$\Delta {\text{H}}_{\text{f}}^0\text{ = 5}\text{.4}\frac{\text{kJ}}{\text{mol}}$

$\begin{array}{l}\Delta {\text{H}}_{\text{rxn}}^0\text{ = }{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{H}}_{\text{reactants}}^0}\\ \text{ = 3}\Delta {\text{H}}_{{\text{f H}}_2\text{(g)}}^0\text{ + 2}\Delta {\text{H}}_{\text{f P(g)}}^0-\text{ 2 }\Delta {\text{H}}_{{\text{f PH}}_3\text{(g)}}^0\\ \text{ = 4}\times 0\frac{\text{kJ}}{\text{mol}}\text{ + }2\times 58.9\frac{\text{kJ}}{\text{mol}}-2\times 5.4\frac{\text{kJ}}{\text{mol}}\\ \text{ = 107 kJ/mol}\end{array}$

Next, calculate the change in entropy by subtracting all the product entropies from the reactant entropies:

$\Delta \text{S = }{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{products}}^0}-{\displaystyle \sum \text{n}\Delta {\text{S}}_{\text{reactants}}^0}$

Plugging the values for the reactants and products as found in the table of common thermodynamic values. Multiply each product or reactant through its coefficient listed in the overall balanced reaction.

$\begin{array}{l}\Delta {\text{S}}_{\text{rxn}}^0\text{ = 3}\times \text{130}\text{.6}\frac{\text{J}}{\text{mol}\cdot \text{K}}\text{ + 2}\times 280\frac{\text{J}}{\text{mol}\cdot \text{K}}-2\times 210.1\frac{\text{J}}{\text{mol}\cdot \text{K}}\\ \text{ = }\boxed{\text{531}\text{.8 J/mol}}\end{array}$

Then the standard Gibbs free energy of reaction is: -

$\text{ΔG = ΔH}-\text{TΔS}$

For the reaction to be spontaneous, $\Delta \text{G}$< 0

$\begin{array}{l}\Rightarrow \text{ΔH}-\text{TΔS < 0}\\ \text{T >}\frac{\text{ΔH}}{\text{ΔS}}\\ \text{T > }\frac{\text{107}\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{1\text{ KJ}}}{\text{531}\text{.8}\frac{\text{J}}{\text{mol}\cdot \text{K}}}\\ \text{T > 201}\text{.20 K}\end{array}$

So, the temperature must be greater than 201.20 K for the reaction to be spontaneous.