Chapter 10, Problem 10.88PAE

Nickel metal reacts with carbon monoxide to form tetra-carbonyl nickel, Ni(CO)₄:

   $\text{Ni}\left(\text{s}\right)+\text{4CO}\left(\text{g}\right)\to \text{Ni}{\left(\text{CO}\right)}_{\text{4}}\left(\text{g}\right)$

This reaction is exploited in the Mond process in order to separate pure nickel from other metals. The reaction above separates nickel from impurities by dissolving it into the gas phase. Conditions are then changed so that the reaction runs in the opposite direction to recover the purified metal.

(a) Predict the sign of $\Delta \text{S}°$ for the reaction as written above.

(b) Use tabulated thermodynamic data to calculate $\Delta \text{H}°$, $\Delta \text{S}°$ , and $\Delta \text{G}°$ for the reaction.

(c) Find the range of temperatures at which this reaction is spontaneous in the forward direction.

Interpretation Introduction

Concept Introduction:

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}$

Where $\Delta S$ is the total change in the entropy of the system and $\Delta H$ is the total change in the enthalpy of the system. For a spontaneous process $\Delta G$ must be negative.

If a process is exothermic and entropy of the process decreases, that is $\Delta H<0$ and $\Delta S<0$ Gibb’s free energy depends on temperature. At higher temperatures the term $T\Delta S$ dominates, so the process will be spontaneous at lower temperatures where $\Delta H$ dominates. Such process is enthalpy driven.

If a process is endothermic and entropy of the process increases, that is $\Delta H>0$ and $\Delta S>0$ Gibb’s free energy again depends on temperature. At higher temperatures the term $T\Delta S$ dominates, so the process will be spontaneous at higher temperatures where $\Delta S$ dominates. Such process is entropy driven.

Enthalpy change for a process is determined as:

$\Delta H^{\text{o}}={\displaystyle \sum _i{v}_i}\Delta H_{\text{f}}{}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}\Delta H_{\text{f}}{}^{\text{o}}{\left(\text{reactant}\right)}_j$

Where v is for stoichiometric coefficients.

Similarly entropy change for a process is determined as:

$\Delta S^{\text{o}}={\displaystyle \sum _i{v}_i}{S}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}{S}^{\text{o}}{\left(\text{reactant}\right)}_j$

Where v is for stoichiometric coefficients and $S^{\text{o}}$ is for absolute entropies of reactants and products..

Similar to enthalpy and entropy, standard Gibb’s free energy change can be calculated as:

$\Delta G^{\text{o}}={\displaystyle \sum _i{v}_i}\Delta G_{\text{f}}{}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}\Delta G_{\text{f}}{}^{\text{o}}{\left(\text{reactant}\right)}_j$

Also $\Delta H_{\text{f}}{}^{\text{o}}$ and $\Delta G_{\text{f}}{}^{\text{o}}$ for elements in their standard states is zero.

Answer to Problem 10.88PAE

Solution:

a) The sign of $\Delta S^{\text{o}}$ for the reaction is negative.

b)

$\Delta H^{\text{o}}$, $\Delta S^{\text{o}}$ and $\Delta G^{\text{o}}$ for the reaction are $-160.9{\text{ kJ mol}}^{-1}$, $-410.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and $-38.5{\text{ kJ mol}}^{-1}$ respectively.

c)

The reaction is spontaneous in the forward direction at temperatures below 392.3 K.

a)

Explanation of Solution

Entropy is defined as the measure of randomness or disorder in a system.

More number of particles would mean more random arrangements of particles of a system. And so entropy of any system increases if it moves towards more random distribution of particles constituting the system.

One way to increase entropy of a system is to increase the number of particles present. So a chemical reaction that would increase the number of moles of gas in the system would increase entropy.

Given reaction is:

$\text{Ni}\left(\text{s}\right)+4\text{ CO}\left(\text{g}\right)\rightleftarrows \text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)$

The reactants have 4 moles of gas while the product has just one. Hence entropy of products is less than the reactants. Thus the sign of $\Delta S^{\text{o}}$ for the reaction is negative.

b)

Given reaction is:

$\text{Ni}\left(\text{s}\right)+4\text{ CO}\left(\text{g}\right)\rightleftarrows \text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)$

The thermodynamic enthalpies of formation for the compounds as tabulated are $\Delta H_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]=-110.5{\text{ kJ mol}}^{-1}$, $\Delta H_{\text{f}}{}^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]=0{\text{ kJ mol}}^{-1}$ and $\Delta H_{\text{f}}{}^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]=-602.9{\text{ kJ mol}}^{-1}$.

Calculate enthalpy change the above reaction that is $\Delta H^{\text{o}}$, enthalpy of formation for Nickel would be zero.

$\begin{array}{c}\Delta H^{\text{o}}=\Delta H_{\text{f}}{}^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]-4\Delta H_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-\Delta H_{\text{f}}{}^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]\\ =-602.9{\text{ kJ mol}}^{-1}-4\left(-110.5{\text{ kJ mol}}^{-1}\right)-0\\ =-160.9{\text{ kJ mol}}^{-1}\end{array}$

The thermodynamic Gibb’s free emergies of formation for the compounds as tabulated are $\Delta G_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]=-137.2{\text{ kJ mol}}^{-1}$, $\Delta G_{\text{f}}{}^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]=0{\text{ kJ mol}}^{-1}$ and $\Delta G_{\text{f}}{}^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]=-587.3{\text{ kJ mol}}^{-1}$.

Calculate enthalpy change the above reaction that is $\Delta G^{\text{o}}$, enthalpy of formation for Nickel would be zero.

$\begin{array}{c}\Delta G^{\text{o}}=\Delta G_{\text{f}}{}^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]-4\Delta G_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-\Delta G_{\text{f}}{}^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]\\ =-587.3{\text{ kJ mol}}^{-1}-4\left(-137.2{\text{ kJ mol}}^{-1}\right)-0\\ =-38.5{\text{ kJ mol}}^{-1}\end{array}$

Similarly, thermodynamic entropies for the compounds as tabulated are $S^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]=197.6{\text{ J K}}^{-1}{\text{ mol}}^{-1}$, $S^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]=30.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and $S^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]=410.4{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

Calculate enthalpy change the above reaction that is $\Delta G^{\text{o}}$, enthalpy of formation for Nickel would be zero.

$\begin{array}{c}\Delta S^{\text{o}}=S^{\text{o}}\left[\text{Ni}{\left(\text{CO}\right)}_4\left(\text{g}\right)\right]-4S^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-S^{\text{o}}\left[\text{Ni}\left(\text{s}\right)\right]\\ =410.4{\text{ J K}}^{-1}{\text{ mol}}^{-1}-4\left(\text{197}{\text{.6 J K}}^{-1}{\text{ mol}}^{-1}\right)-30.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ =-410.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}\end{array}$

c)

If a process is exothermic and entropy of the process decreases, that is $\Delta H<0$ and $\Delta S<0$ Gibb’s free energy depends on temperature. At higher temperatures the term $T\Delta S$ dominates, so the process will be spontaneous at lower temperatures where $\Delta H$ dominates.

Calculate the temperature at which $\Delta G^{\text{o}}=0$.

$\begin{array}{c}\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ 0=-160.9\times {10}^3{\text{ J mol}}^{-1}-T\left(-410.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)\\ T=\frac{-160.9\times {10}^3{\text{ J mol}}^{-1}}{-410.1{\text{ J K}}^{-1}{\text{ mol}}^{-1}}\\ =392.3\text{ K}\end{array}$

The reaction is spontaneous in the forward direction at temperatures below 392.3 K.