Chapter 10, Problem 154MP

(a)

Interpretation Introduction

Interpretation:

The sign of ΔS⁰ for the reaction should be predicted.

Concept Introduction :

Entropy is a measure of molecular disorder or randomness. Higher the randomness, higher the system’s entropy. In solids, atoms and molecules are restricted to a fixed position. In liquid phase, atoms and molecules are free to move around each other, but remain in relatively close proximity. In gas phase, atoms and molecules are free to move in much greater volume than liquid. So, entropy of substances differs as follows $S_{gas}>S_{liquid}>S_{solid}$ .

Answer to Problem 154MP

Negative.

Explanation of Solution

  $\text{Ni(s) + 4 CO(g) }𑨀{\text{ Ni(CO)}}_4\text{(g)}$

Number of moles of substances in gas phase has reduced. So, the randomness of the system decreases. So, ΔS⁰ will be negative.

(b)

Interpretation Introduction

Interpretation:

The sign of ΔS_surr for the reaction should be predicted if the spontaneity of the reaction is temperature-dependent.

Concept Introduction :

  $\Delta S_{surr}=-\frac{\Delta H}{T}$

ΔS_surr − entropy change of the surrounding

ΔH − enthalpy change

T − temperature

Answer to Problem 154MP

Positive.

Explanation of Solution

Since ΔS⁰ is negative, ΔH⁰ also should be negative for the reaction to be spontaneous. Therefore, ΔS_surr is positive.

(c)

Interpretation Introduction

Interpretation:

ΔS⁰ and ΔH⁰ for the preceding reaction should be calculated.

Concept Introduction :

  $\begin{array}{l}\Delta H^0={\displaystyle \sum \Delta H_f^0\text{(Products) }-{\displaystyle \sum \Delta H_f^0\text{(Reactants)}}}\\ \Delta S^0={\displaystyle \sum S^0\text{(Products)}-}{\displaystyle \sum S^0\text{(Reactants)}}\end{array}$

Answer to Problem 154MP

  $\Delta H^0\text{ = }-165\text{ kJ/mol}$

  $\Delta S^0\text{ = }-405\text{ J/mol}\text{.K}$

Explanation of Solution

  $\text{Ni(s) + 4 CO(g) }𑨀{\text{ Ni(CO)}}_4\text{(g)}$

  $\begin{array}{l}\Delta H^0=-607\text{ kJ/mol}-4\times \left(-110.5\text{ kJ/mol}\right)\\ \Delta H^0\text{ = }-165\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta S^0=417\text{ J/mol}\text{.K}-\left\{4\times 198\text{ J/mol}\text{.K + 30 J/mol}\text{.K}\right\}\\ \Delta S^0\text{ = }-405\text{ J/mol}\text{.K}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The temperature at which $\Delta G^0=0\text{ and K = 1}$ should be calculated.

Concept Introduction:

  $\Delta G^0=\Delta H^0-T\Delta S^0$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta {\text{H}}^0\text{ - Enthalpy change}\\ \text{T - temperature}\\ \Delta {\text{S}}^0\text{ - Entropy change}\end{array}$

Answer to Problem 154MP

  $T=407\text{ K}$

Explanation of Solution

  $\begin{array}{l}\Delta G^0=\Delta H^0-T\Delta S^0\\ 0=-165000\text{ J/mol - T }\times \left(-405\text{ J/mol}\text{.K}\right)\\ T=\frac{165000\text{ J/mol}}{405\text{ J/mol}\text{.K}}\\ T=407\text{ K}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction at 50 ⁰C should be calculated.

Concept Introduction:

  $\Delta G^0=\Delta H^0-T\Delta S^0$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta {\text{H}}^0\text{ - Enthalpy change}\\ \text{T - temperature}\\ \Delta {\text{S}}^0\text{ - Entropy change}\end{array}$

  $\Delta G^0=-RT\ln K$

  $\begin{array}{l}R\text{ - universal gas constant}\\ K\text{ - equilibrium constant}\end{array}$

Answer to Problem 154MP

  $\text{K = 3}\text{.1 }\times {\text{10}}^5$

Explanation of Solution

  $\begin{array}{l}\Delta G^0=\Delta H^0-T\Delta S^0\\ \Delta G_{323}{}^0=-165000\text{ J/mol - 323 K }\times \left(-405\text{ J/mol}\text{.K}\right)\\ \Delta G_{323}{}^0=-34000\text{ J/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -34000\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 323 K }\times \text{ ln K}\\ \text{ln K = }\frac{34000\text{ J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{ 323 K}}=12.66\\ \text{K = 3}\text{.1 }\times {\text{10}}^5\end{array}$

(f)

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction at 227 ⁰C should be calculated.

Concept Introduction:

  $\Delta G^0=\Delta H^0-T\Delta S^0$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta {\text{H}}^0\text{ - Enthalpy change}\\ \text{T - temperature}\\ \Delta {\text{S}}^0\text{ - Entropy change}\end{array}$

  $\Delta G^0=-RT\ln K$

  $\begin{array}{l}R\text{ - universal gas constant}\\ K\text{ - equilibrium constant}\end{array}$

Answer to Problem 154MP

  $\text{K = 1}\text{.1 }\times {\text{10}}^{-4}$

Explanation of Solution

  $\begin{array}{l}\Delta G^0=\Delta H^0-T\Delta S^0\\ \Delta G_{323}{}^0=-165000\text{ J/mol - 500 K }\times \left(-405\text{ J/mol}\text{.K}\right)\\ \Delta G_{323}{}^0=38000\text{ J/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ 38000\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 500 K }\times \text{ ln K}\\ \text{ln K = }-\frac{38000\text{ J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{ 500 K}}=-9.14\\ \text{K = 1}\text{.1 }\times {\text{10}}^{-4}\end{array}$

(g)

Interpretation Introduction

Interpretation:

The reason for increasing the temperature in the second step of Mond’s process should be explained.

Concept Introduction:

Le Chatelier’s principle states that if a change occurred in an equilibrium system, the equilibrium shifts in such a way that the system counteracts that change.

Answer to Problem 154MP

When temperature is increased, backward reaction is more favorable and more solid nickel will form.

Explanation of Solution

  $\text{Ni(s) + 4 CO(g) }𑨀{\text{ Ni(CO)}}_4\text{(g)}$

Enthalpy change of the forward reaction is negative. So, forward reaction is an exothermic reaction. The purpose of the second step of Mond’s process is to deposit as much nickel as possible as pure solid. So, the equilibrium should shift to left. When the temperature increases, the equilibrium shifts left in order to reduce the temperature of the system. So, rate of the backward reaction increases forming more solid of nickel.

(h)

Interpretation Introduction

Interpretation:

The maximum pressure of Ni(CO)₄ should be estimated.

Concept Introduction:

The change in Gibbs free energy is calculated as follows:

  $\Delta G^0=\Delta H^0-T\Delta S^0$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta {\text{H}}^0\text{ - Enthalpy change}\\ \text{T - temperature}\\ \Delta {\text{S}}^0\text{ - Entropy change}\end{array}$

Also,

  $\Delta G^0=-RT\ln K$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy}\\ \text{T - temperature}\\ R\text{ - universal gas constant}\\ K\text{ - equilibrium constant}\end{array}$

Answer to Problem 154MP

  $P_{{\text{Ni(CO)}}_4\text{(g)}}=17\text{ atm}$

Explanation of Solution

at 42 ⁰C (the boiling point)

  $\begin{array}{l}\Delta G^0=\Delta H^0-T\Delta S^0=0\\ \Delta H^0=T\Delta S^0\\ \Delta S^0=\frac{\Delta H^0}{T}=\frac{29000\text{ J/mol}}{315\text{ K}}=92.1\text{ J/mol}\text{.K}\end{array}$

At 152 ⁰C;

  $\begin{array}{l}\Delta G^0=\Delta H^0-T\Delta S^0\\ \Delta G_{152}{}^0=29000\text{ J/mol }-425\text{ K }\times \text{ 92}\text{.1 J/mol}\text{.K}\\ \Delta G_{152}{}^0=-10100\text{ J/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K_p\\ -10100\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{425 K }\times {\text{ ln K}}_p\\ \ln K_p=\frac{10100\text{ J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{425 K }}=2.86\\ K_p=17\end{array}$

  ${\text{Ni(CO)}}_4\text{(l) }𑨀{\text{ Ni(CO)}}_4\text{(g)}$

  ${\text{K}}_p=P_{{\text{Ni(CO)}}_4\text{(g)}}=17\text{ atm}$