Q20.52 P - Chemistry: Molecular Nature Of Matter And Change

Question

Find ${\mathbf{\Delta G}}^{\mathbf{\circ }}$  for the reactions in Problem 20.50 using  ${\mathbf{\Delta H}}_{\mathbf{f}}^{\mathbf{\circ }}$ and ${\mathbf{S}}^{\mathbf{\circ }}$ values.

Step-by-Step Solution

Answer

Reaction-A the value of standard free energy is ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=\underset{¯}{-1138\mathrm{kJ}}$.

Reaction-B the value of standard free energy is  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=\underset{¯}{-1379\mathrm{kJ}}$.

Reaction-C the value of standard free energy is  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=\underset{¯}{-226\mathrm{kJ}}$.

Step 1: Concept.

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Step 2: Find ΔG ∘ for the reactions-A

Reaction-A

Considering the given Chemical reaction:

 $\mathrm{BaO}\left(\mathrm{s}\right)+\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{BaCO}}_3\left(\mathrm{s}\right)$

The number of particles decreases as well, indicating that entropy is decreasing.

As a result, the ${\text{ΔG}}_{\mathrm{f}}^{\mathrm{o}}$ values are zero, indicating that the solid is less than the gas.

Standard enthalpy change is,

The reaction's enthalpy change is calculated as follows:

 $$\begin{gathered} \begin{array}{l}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{Products}\right)}^{\mathrm{o}}-\sum \mathrm{n}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{\mathrm{o}}\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(1\mathrm{molMgO}\right)\left({\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}} of \mathrm{MgO}\right)\right]\end{array} \\ \begin{array}{c}-\left[(2\mathrm{mol} \mathrm{Mg})\left({\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}} \mathrm{of} \mathrm{Mg}\right)+\left(1{\mathrm{molO}}_2\right)\left({\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}} \mathrm{of} O_2\right)\right]\\ =\left[\right(2\mathrm{mol} \mathrm{MgO}\left)\right(-601.2\mathrm{kJ}/\mathrm{mol}\left)\right]-\left[(2\mathrm{mol} \mathrm{Mg})(0\mathrm{kJ}/\mathrm{mol})+(1\mathrm{mol} O{)}_2\right)(0 \mathrm{kJ}/\mathrm{mol})\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-1202.4\mathrm{kJ}\end{array} \end{gathered}$$

 

The enthalpy change is negative. Hence, the enthalpy  $\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\right)$ value is  $-1202.4\mathrm{kJ}$

Entropy change  ${\mathrm{\Delta S}}_{\mathrm{system}}^{\mathrm{o}}$

The standard equation for entropy change is:

 ${\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum {\mathrm{mS}}_{\mathrm{Products}}^{\mathrm{o}}-\sum {\mathrm{nS}}_{\mathrm{reactants}}^{\mathrm{o}}$

Where, (m) and (n) are the stoichiometric co-efficient.

$\begin{array}{l}{\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[(1\mathrm{mol} \mathrm{MgO})\left({\mathrm{S}}^{\mathrm{o}}\mathrm{of} \mathrm{Mg} \mathrm{g}\right)\right]-\left[(2\mathrm{mol} \mathrm{Mg})\left({\mathrm{S}}^{\mathrm{o}} \mathrm{of} \mathrm{Mg}\right)\right.\left.+\left(1 \mathrm{mol} {\mathrm{O}}_2\right)\left({\mathrm{S}}^{\mathrm{o}} \mathrm{of} {\mathrm{O}}_2\right)\right]\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\right(1 \mathrm{mol} \mathrm{MgO}\left)\right(26.9\mathrm{J}/\mathrm{mol}\times \mathrm{K}\left)\right]-\left[\left(2 \mathrm{mol} {\mathrm{N}}_2\right)(32.69\mathrm{J}/\mathrm{mol}\times \mathrm{K})\right.\left.+\left(1 {\mathrm{mol}}_2\right)(205.0\mathrm{J}/\mathrm{mol}\times \mathrm{K})\right]\end{array}$  

Therefore, the  $\left({\mathrm{\Delta S}}_{\mathrm{rxn}}^0\right)$ of the reaction is $-216.58\mathrm{J}/\mathrm{K}$ 

Calculate the change in free energy  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}$ next.

Standard Free energy change equation is,

 ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}{=\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}{-\mathrm{T\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}$

Free energy change ${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}$

The enthalpy and entropy values calculated are

 $\begin{array}{l}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-1202.4\mathrm{kJ}\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=-216.58\mathrm{J}/\mathrm{K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

 $\begin{array}{l}{\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1202.4\mathrm{kJ}-\left[\left(298\mathrm{K}\right)(-216.58\mathrm{J}/\mathrm{K})\left(1\mathrm{kJ}/{10}^3\mathrm{J}\right)\right]\\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1137.859\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1138\mathrm{kJ}$.

step 3: Find ΔG ∘ for the reactions-B.

Reaction-B

Considering the given Chemical reaction:

 $2{\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right) \to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The number of particles decreases as well, indicating that entropy is decreasing.

As a result, the values are zero, indicating that the solid is less than the gas.

The standard enthalpy change formula is:

The reaction's enthalpy change is calculated as follows:

 $\begin{array}{c}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{Products}\right)}^{\mathrm{o}}-\sum \mathrm{n}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{\mathrm{o}}\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right) \left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}{\mathrm{ofCO}}_2\right)\right.+\left(4{\mathrm{molH}}_2\mathrm{O}\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\right.\mathrm{of}\left.\left.{\mathrm{H}}_2\mathrm{O}\right)\right]\\ -\left[\left(2{\mathrm{molCH}}_3\mathrm{OH}\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\right.\right.\mathrm{of}\left.{\mathrm{CH}}_3\mathrm{OH}\right)+\left(3{\mathrm{molO}}_2\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\right.\mathrm{of}\left.{\mathrm{O}}_2\right)]\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right)(-393.5\mathrm{kJ}/\mathrm{mol})+(4\mathrm{molH}2\mathrm{O}\left)\right(-241.826\mathrm{kJ}/\mathrm{mol})\right]\\ \left[\left(2{\mathrm{molCHH}}_3\mathrm{OH}\right)(-201.2\mathrm{kJ}/\mathrm{mol})\right.\left.+\left(3{\mathrm{molO}}_2\right)(0\mathrm{kJ}/\mathrm{mol})\right]\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-1351.904\mathrm{kJ}\end{array}$

The change in enthalpy is negative.

Hence, the enthalpy ${\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}$  value is $-1351.904\mathrm{kJ}$ 

Entropy change ${\mathrm{\Delta S}}_{\mathrm{system}}^{\mathrm{o}}$.

Standard entropy change equation is,

Where, (m) and (n) are the stoichiometric co-efficient. 

$\begin{array}{c}{\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right)\left({\mathrm{S}}^{\mathrm{o}}\right.\right.\mathrm{of}\left.{\mathrm{CO}}_2\right)+\left(4{\mathrm{molH}}_2\mathrm{O}\right)\left({\mathrm{S}}^{\mathrm{o}}\right.\mathrm{of}\left.\left.{\mathrm{H}}_2\mathrm{O}\right)\right]\\ -\left[\left(2{\mathrm{molCH}}_3\mathrm{OH}\right)\left({\mathrm{S}}^{\mathrm{o}}\right.\right.\mathrm{of}\left.{\mathrm{CH}}_3\mathrm{OH}\right)+\left(3{\mathrm{molO}}_2\right)\left({\mathrm{S}}^{\mathrm{o}}\right.\mathrm{of}\left.\left.{\mathrm{O}}_2\right)\right]\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right)(213.7\mathrm{J}/\mathrm{mol}\times \mathrm{K})+\left(4{\mathrm{molHH}}_2\mathrm{O}\right)(188.72\mathrm{J}/\mathrm{mol}\times \mathrm{K})\right]\\ \left[\left(2{\mathrm{molCH}}_3\mathrm{OH}\right)(238\mathrm{J}/\mathrm{mol}\times \mathrm{K})\right.\left.+\left(3{\mathrm{molO}}_2\right)(205.0\mathrm{J}/\mathrm{mol}\times \mathrm{K})\right]\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=-91.28\mathrm{J}/\mathrm{K}\end{array}$ 

Therefore, the $\left({\mathrm{\Delta S}}_{\mathrm{rxn}}^0\right)$ of the reaction is $-91.28\mathrm{J}/\mathrm{K}$

Next calculate the Free energy change ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}$

standard Free energy change equation is,

 ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}{=\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}{-\mathrm{T\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}$

Free energy change  ${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}$

The values of calculated enthalpy and entropy are,

 $\begin{array}{l}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-1351.904\mathrm{kJ}\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=-91.28\mathrm{J}/\mathrm{K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

 $\begin{array}{l}{\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1351.904\mathrm{kJ}-\left[\left(298\mathrm{K}\right)(-91.28\mathrm{J}/\mathrm{K})\left(1\mathrm{kJ}/{10}^3\mathrm{J}\right)\right]\\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1379.105\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=\underset{¯}{-1379\mathrm{kJ}}$.

Step 4: Find ΔG ∘ for the reactions-C.

Reaction-C

Considering the given Chemical reaction:

 $\mathrm{BaO}\left(\mathrm{s}\right)+\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{BaCO}}_3\left(\mathrm{s}\right)$

The number of particles decreases as well, indicating that entropy is decreasing.

As a result, the ${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}$ values are zero, indicating that the solid is less than the gas.

The standard enthalpy change formula is:

The reaction's enthalpy change is calculated as follows:

 $$\begin{gathered} \begin{array}{l}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{Products}\right)}^{\mathrm{o}}-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{\mathrm{o}}\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(1\mathrm{mol} {\mathrm{BaCOB}}_3\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\mathrm{of} {\mathrm{BaCO}}_3\right)\right]\end{array} \\ \begin{array}{c}-\left[\left(1\mathrm{molBaO}\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}\mathrm{ofBaO}\right)+\left(1{\mathrm{molCOCO}}_2\right)\left({\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}{\mathrm{ofCO}}_2\right)\right]\\ =\left[\right(1\mathrm{molBaCO}3\left)\right(-1219\mathrm{kJ}/\mathrm{mol}\left)\right]-\left[\right(1\mathrm{molBaO}\left)\right(-548.1\mathrm{kJ}/\mathrm{mol})\\ \left.+\left(1{\mathrm{molCO}}_2\right)(-393.5\mathrm{kJ}/\mathrm{mol})\right]\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-277.4\mathrm{kJ}\end{array} \end{gathered}$$

The enthalpy change is negative. Hence, the enthalpy  ${\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}$  value is $-277.4\mathrm{kJ}$

Entropy change ${\mathrm{\Delta S}}_{\mathrm{system}}^{\mathrm{o}}$ 

Standard entropy change equation is,

Where, (m) and (n) are the stoichiometric co-efficient.

$\begin{array}{c}{\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[\left(2{\mathrm{molBaCO}}_3\right)\left({\mathrm{S}}^{\mathrm{o}}\mathrm{ofBaCO}3\right)\right]\\ -\left[\left(1\mathrm{molBaO}\right)\left({\mathrm{S}}^{\mathrm{o}}\mathrm{ofBaO}\right)+\left(1\mathrm{molCO}2\right)\left({\mathrm{S}}^{\mathrm{o}}\mathrm{ofCO}2\right)\right]\\ =\left[\right(1\mathrm{mol}\left)\right(112\mathrm{J}/\mathrm{mol}\times \mathrm{K}\left)\right]-\left[\right(1\mathrm{mol}\left)\right(72.07\mathrm{J}/\mathrm{mol}\times \mathrm{K})\\ +\left(1\mathrm{mol}\right)(213.7\mathrm{J}/\mathrm{mol}\times \mathrm{K}\left)\right]\\ {\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}=-173.77\mathrm{J}/\mathrm{K}\end{array}$

Therefore, the ${\mathrm{\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}$ of the reaction is $-173.77\mathrm{J}/\mathrm{K}$ 

Finally calculate the Free energy change  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}$

Standard Free energy change equation is,

 ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}{=\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}{-\mathrm{T\Delta S}}_{\mathrm{rxn}}^{\mathrm{o}}$

Free energy change  ${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}$

Calculated enthalpy and entropy values are

 $\begin{array}{l}{\mathrm{DH}}_{\mathrm{rxn}}^{\mathrm{o}}=-277.4\mathrm{kJ}\\ {\mathrm{DS}}_{\mathrm{rxn}}^{\mathrm{o}}=-173.77\mathrm{J}/\mathrm{K}\end{array}$

These values are plugging above standard free energy equation,

 $\begin{array}{l}{\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-277.4\mathrm{kJ}-\left[\left(298\mathrm{K}\right)(-173.77\mathrm{J}/\mathrm{K})\left(1\mathrm{kJ}/{10}^3\mathrm{J}\right)\right]\\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-225.6265\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is  ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-225.6265 \mathrm{kJ}$.