Chapter 17, Problem 93AE

Acrylonitrile is the starting material used in tire manufacture of acrylic fibers (U.S. annual production capacity is more than two million pounds). Three industrial processes for the production of acrylonitrile are given below. Using data from Appendix 4, calculate ΔS°, ΔH°, and ΔG° for each process. For part a, assume that T = 25°C; for part b. T = 70.°C: and for part c. T = 700.°C. Assume that ΔH° and ΔH° do not depend on temperature.

1. a.
2. b. $HC\equiv CH(\text{g})+HCN(\text{g})\underset{70°C-90°C}{\overset{Ca{C}_2\cdot Hcl}{\to }}C{H}_2=CHCN(\text{g})$
3. c. $4C{H}_2=CHC{H}_3(\text{g})+6N{O}_2(\text{g})\underset{\text{Ag}}{\overset{\text{700°C}}{\to }}4C{H}_2=CHCN(\text{g})+6H{O}_2(\text{g})+N_2(\text{g})$

Interpretation Introduction

Interpretation:

The value of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ for the given reaction by the use of data given in appendix $4$ should be determined.

Concept Introduction:

• Enthalpy $\text{H}$: it is the total amount of heat in a particular system.
• Entropy $\text{S}$ : it is used to describe the disorder. It is the amount of arrangements possible in a system at a particular state.
• Free energy change ${\text{ΔG}}^0$: change in the free energy takes place while reactants converts to product where both are in standard state.

Answer to Problem 93AE

$\begin{array}{cccc}& {\text{ΔH}}^{\text{0}}\left(\text{kJ}\right)& {\text{ΔS}}^{\text{0}}\left(\text{J/K}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ}\right)\\ \text{(a)}& -183.1& -100& -153.6\\ \text{(b)}& -177.1& -129& -138.6\\ \text{(c)}& -1229.6& -917.6& -1022.4\end{array}$

Explanation of Solution

To determine: The value of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ for the given reaction by the use of data given in appendix $4$

The given reaction is,

$\text{Ethylene}\text{epoxide}\text{+}\text{HCN}\stackrel{}{\to }\text{H2C=CHCN}\text{+}\text{water}$

Given,

The given values of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ are as follows

$\begin{array}{cccc}\text{Compound}& {\text{ΔH}}^{\text{0}}\left(\text{kJ/mol}\right)& {\text{ΔS}}^{\text{0}}\left(\text{kJ}\text{.mol}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ/mol}\right)\\ \begin{array}{l}\text{ethylene}\\ \text{oxide}\end{array}& -53& 242& -13\\ \text{HCN}& 135.1& 202& 125\\ \text{acrylonitrile}& 185& 274& 195.4\\ {\text{H}}_{\text{2}}\text{O}& -286& 70& -237\end{array}$

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard enthalpy of formation for the reactants}\\ {\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard enthalpy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔH}}^{\text{ο}}{\text{=ΔH}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\begin{array}{l}\text{ethylene}\\ \text{oxide}\left(\text{g}\right)\end{array}}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(185\right)\text{+}\text{1}\text{mol}\left(-\text{286}\right)-\text{1}\text{mol}\left(-53\right)-\text{1}\text{mol}\left(135.1\right)\\ =-183.1\text{kJ}\end{array}$

Therefore,

${\text{ΔH}}^{\text{ο}}=-183.1\text{kJ}$

The value of standard entropy change ${\text{ΔS}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard entropy of formation for the reactants}\\ {\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard entropy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔS}}^{\text{ο}}{\text{=ΔS}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\begin{array}{l}\text{ethylene}\\ \text{oxide}\left(\text{g}\right)\end{array}}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(274\right)\text{+}\text{1}\text{mol}\left(70\right)-\text{1}\text{mol}\left(242\right)-\text{1}\text{mol}\left(202\right)\\ =-100\text{J/K}\end{array}$

Therefore,

${\text{ΔS}}^{\text{ο}}=-100\text{J/K}$

The value of standard free energy change ${\text{ΔG}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard free energy of formation for the reactants}\\ {\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard free energy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔG}}^{\text{ο}}{\text{=ΔG}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\begin{array}{l}\text{ethylene}\\ \text{oxide}\left(\text{g}\right)\end{array}}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(195.4\right)\text{+}\text{1}\text{mol}\left(-237\right)-\text{1}\text{mol}\left(-13\right)-\text{1}\text{mol}\left(125\right)\\ =-153.6\text{kJ}\end{array}$

Therefore,

${\text{ΔG}}^{\text{ο}}=-153.6\text{kJ}$

Interpretation Introduction

Interpretation:

The value of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ for the given reaction by the use of data given in appendix $4$ should be determined.

Concept Introduction:

• Enthalpy $\text{H}$: it is the total amount of heat in a particular system.
• Entropy $\text{S}$ : it is used to describe the disorder. It is the amount of arrangements possible in a system at a particular state.
• Free energy change ${\text{ΔG}}^0$: change in the free energy takes place while reactants converts to product where both are in standard state.

Answer to Problem 93AE

$\begin{array}{cccc}& {\text{ΔH}}^{\text{0}}\left(\text{kJ}\right)& {\text{ΔS}}^{\text{0}}\left(\text{J/K}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ}\right)\\ \text{(a)}& -183.1& -100& -153.6\\ \text{(b)}& -177.1& -129& -138.6\\ \text{(c)}& -1229.6& -917.6& -1022.4\end{array}$

Explanation of Solution

The given reaction is,

$\text{HC}\equiv \text{CH}\text{+}\text{HCN}\stackrel{}{\to }\text{H2C=CHCNH}$

Given,

The given values of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ are as follows

$\begin{array}{cccc}\text{Compound}& {\text{ΔH}}^{\text{0}}\left(\text{kJ/mol}\right)& {\text{ΔS}}^{\text{0}}\left(\text{kJ}\text{.mol}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ/mol}\right)\\ \text{HCCH}& 227& 201& 209\\ \text{HCN}& 135.1& 202& 125\\ \text{acrylonitrile}& 185& 274& 195.4\end{array}$

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard enthalpy of formation for the reactants}\\ {\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard enthalpy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔH}}^{\text{ο}}{\text{=ΔH}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\text{HCCH}\left(\text{g}\right)}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(185\right)-\text{1}\text{mol}\left(227\right)-\text{1}\text{mol}\left(135.1\right)\\ =-177.1\text{kJ}\end{array}$

Therefore,

${\text{ΔH}}^{\text{ο}}=-177.1\text{kJ}$

The value of standard entropy change ${\text{ΔS}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard entropy of formation for the reactants}\\ {\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard entropy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔS}}^{\text{ο}}{\text{=ΔS}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\text{HCCH}\left(\text{g}\right)}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(274\right)-\text{1}\text{mol}\left(201\right)-\text{1}\text{mol}\left(202\right)\\ =-129\text{kJ}\end{array}$

Therefore,

${\text{ΔS}}^{\text{ο}}=-129\text{J/K}$

The value of standard free energy change ${\text{ΔG}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard free energy of formation for the reactants}\\ {\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard free energy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔG}}^{\text{ο}}{\text{=ΔG}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\text{HCCH}\left(\text{g}\right)}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\text{HCN}\left(\text{g}\right)}\\ \text{=}\text{1mol}\left(195.4\right)-\text{1}\text{mol}\left(209\right)-\text{1}\text{mol}\left(125\right)\\ =-138.6\text{kJ}\end{array}$

Therefore,

${\text{ΔG}}^{\text{ο}}=-138.6\text{kJ}$

Interpretation Introduction

Interpretation:

The value of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ for the given reaction by the use of data given in appendix $4$ should be determined.

Concept Introduction:

• Enthalpy $\text{H}$: it is the total amount of heat in a particular system.
• Entropy $\text{S}$ : it is used to describe the disorder. It is the amount of arrangements possible in a system at a particular state.
• Free energy change ${\text{ΔG}}^0$: change in the free energy takes place while reactants converts to product where both are in standard state.

Answer to Problem 93AE

$\begin{array}{cccc}& {\text{ΔH}}^{\text{0}}\left(\text{kJ}\right)& {\text{ΔS}}^{\text{0}}\left(\text{J/K}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ}\right)\\ \text{(a)}& -183.1& -100& -153.6\\ \text{(b)}& -177.1& -129& -138.6\\ \text{(c)}& -1229.6& -917.6& -1022.4\end{array}$

Explanation of Solution

The given reaction is,

$\text{4HC=CHCH}\text{+}\text{NO}\stackrel{}{\to }4{\text{H}}_{\text{2}}\text{C=CHCN}{\text{+6H}}_{\text{2}}\text{O}{\text{+N}}_{\text{2}}$

Given,

The given values of ${\text{ΔH}}^{\text{ο}}\text{,}{\text{ΔS}}^{\text{ο}}$ and ${\text{ΔG}}^{\text{ο}}$ are as follows

$\begin{array}{cccc}\text{Compound}& {\text{ΔH}}^{\text{0}}\left(\text{kJ/mol}\right)& {\text{ΔS}}^{\text{0}}\left(\text{kJ}\text{.mol}\right)& {\text{ΔG}}^{\text{0}}\left(\text{kJ/mol}\right)\\ {\text{CH}}_{\text{2}}{\text{CHCH}}_{\text{3}}& 20.9& 266.9& 62.7\\ \text{NO}& 90& 211& 87\\ \text{acrylonitrile}& 185& 274& 195.4\\ {\text{H}}_{\text{2}}\text{O}& -286& 70& -237\\ {\text{N}}_{\text{2}}& 0& 192& 0\end{array}$

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard enthalpy of formation for the reactants}\\ {\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard enthalpy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔH}}^{\text{ο}}{\text{=ΔH}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}\text{+}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{{\text{N}}_2\left(\text{g}\right)}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{2}}{\text{CHCH}}_{\text{3}}\left(\text{g}\right)}-{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(\text{g}\right)}\\ \text{=}6\text{mol}\left(185\right)\text{+}6\text{mol}\left(-\text{286}\right)+1\text{mol}\left(0\right)-4\text{mol}\left(20.9\right)-6\text{mol}\left(90\right)\\ =1110-1716-83.6-540\\ =-1229.6\text{kJ}\end{array}$

Therefore,

${\text{ΔH}}^{\text{ο}}=-1229.6\text{kJ}$

The value of standard entropy change ${\text{ΔS}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard entropy of formation for the reactants}\\ {\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard entropy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔS}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔS}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔS}}^{\text{ο}}{\text{=ΔS}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}\text{+}{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{{\text{N}}_2\left(\text{g}\right)}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{2}}{\text{CHCH}}_{\text{3}}\left(\text{g}\right)}-{\text{ΔS}}_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(\text{g}\right)}\\ \text{=}6\text{mol}\left(274\right)\text{+}6\text{mol}\left(70\right)+1\text{mol}\left(192\right)-4\text{mol}\left(266.9\right)-6\text{mol}\left(211\right)\\ =1644-420+192-1067.6-1266\\ =-917.6\text{J/K}\end{array}$

Therefore,

${\text{ΔS}}^{\text{ο}}=-917.6\text{J/K}$

The value of standard free energy change ${\text{ΔG}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard free energy of formation for the reactants}\\ {\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard free energy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Substitute the values of standard enthalpy of formations,

$\begin{array}{c}{\text{ΔG}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ {\text{ΔG}}^{\text{ο}}{\text{=ΔG}}_{{\text{fCH}}_{\text{2}}\text{CHCN}\left(\text{g}\right)}^{\text{ο}}\text{+}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(\text{l}\right)}\text{+}{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{{\text{N}}_2\left(\text{g}\right)}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{{\text{CH}}_{\text{2}}{\text{CHCH}}_{\text{3}}\left(\text{g}\right)}-{\text{ΔG}}_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(\text{g}\right)}\\ \text{=}6\text{mol}\left(195.4\right)\text{+}6\text{mol}\left(-\text{237}\right)+1\text{mol}\left(0\right)-4\text{mol}\left(62.7\right)-6\text{mol}\left(87\right)\\ =1172.4-1422-250.8-522\\ =-1022.4\text{kJ}\end{array}$

Therefore,

${\text{ΔG}}^{\text{ο}}=-1022.4\text{kJ}$