Chapter 18.7, Problem 6CYU

(a)

Interpretation Introduction

Interpretation:

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for the given reaction ${\text{N}}_{\text{2}}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows \text{2NO}\left(\text{g}\right)$ should be calculated using standard free energies of reactant and product formation and should be identified that whether reaction is reactant or product favoured.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{reactants}\right)$

${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ gets related to the reaction quotient $\text{Q}$ by the expression ${\text{Δ}}_{\text{r}}{\text{G = Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{+ RT lnQ}$

For a general reaction, $\text{aA + bB}\to \text{cC + dD}$

  $\text{Q = }\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Answer to Problem 6CYU

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for the given reaction is $173.2kJ/mol-rxn$. The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ value is positive, thus the reaction is reactant favoured.

Explanation of Solution

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for the given reaction is calculated below.

Given: ${\text{N}}_{\text{2}}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows \text{2NO}\left(\text{g}\right)$

The Appendix L was referred for the values of standard free energy.

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for $\text{NO}\left(\text{g}\right)$ is $\text{86}\text{.58 kJ/mol}$

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for ${\text{N}}_{\text{2}}\left(\text{g}\right)$ is $\text{0 kJ/mol}$

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $\text{0 kJ/mol}$

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{reactants}\right)$

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{2 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{G}}^{\text{°}}\left[\text{NO}\left(\text{g}\right)\right]\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{G}}^{\text{°}}\left[{\text{N}}_{\text{2}}\left(\text{g}\right)\right]\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{G}}^{\text{°}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]$

Substituting the free energy values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{86}\text{.58 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0}\right)\end{array}\right]\end{array}\right]\\ \text{= 173}\text{.2 kJ/mol- rxn}\end{array}$

Therefore, the reaction is reactant favoured since the value is positive.

(b)

Interpretation Introduction

Interpretation:

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}$ for the reaction when $\text{0}\text{.10 atm}{\text{N}}_{\text{2}}\text{,}\text{0}\text{.10}{\text{atm O}}_{\text{2}}\text{and}\text{0}\text{.010}\text{atm NO}$ are mixed should be calculated and should be identified that whether this reaction is spontaneous or not under such given conditions.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{reactants}\right)$

${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ gets related to the reaction quotient $\text{Q}$ by the expression ${\text{Δ}}_{\text{r}}{\text{G = Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{+ RT lnQ}$

For a general reaction, $\text{aA + bB}\to \text{cC + dD}$

  $\text{Q = }\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Answer to Problem 6CYU

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ for the given reaction of formation of nitrogen monoxide is $161.7kJ/mol-rxn$. The reaction will proceed not spontaneously since the value is positive.

Explanation of Solution

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ for the given reaction is calculated below,

Given: $\text{0}\text{.10 atm}{\text{N}}_{\text{2}}\text{,}\text{0}\text{.10}{\text{atm O}}_{\text{2}}\text{and}\text{0}\text{.010}\text{atm NO}$ are mixed.

${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ is related to the reaction quotient $\text{Q}$ by the expression,

  ${\text{Δ}}_{\text{r}}{\text{G = Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{+RT lnQ}$

  ${\text{Δ}}_{\text{r}}{\text{G = Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{+RT ln}\left(\frac{{\text{P}}_{{\text{N}}_{\text{2}}}{\text{P}}_{{\text{O}}_{\text{2}}}}{{\text{P}}_{\text{NO}}{}^{\text{2}}}\right)$

Substituting the respective values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}\text{G = 173}\text{.2 kJ/mol- rxn +}\left(\text{0}\text{.08314 kJ/K}\right)\left(\text{298}\text{.15}\right)\text{ln}\left(\frac{\left(\text{0}\text{.10}\right)\left(\text{0}\text{.10}\right)}{{\left(\text{0}\text{.010}\right)}^{\text{2}}}\right)\\ \text{=161}\text{.7 kJ/mol- rxn}\end{array}$

Thus the reaction will proceed not spontaneously since the value is positive