Chapter 18, Problem 90SCQ

Consider the formation of NO(g) from its elements.

N₂(g) + O₂(g) ⇄ 2 NO(g)

1. (a) Calculate K_p at 25 °C. Is the reaction product-favored at equilibrium at this temperature?
2. (b) Assuming Δ_rH° and Δ_rS° are nearly constant with temperature, calculate Δ_rG° at 700 °C. Estimate K_p from the new value of Δ_rG° at 700 °C. Is the reaction product-favored at equilibrium at 700 °C?
3. (c) Using K_p at 700 °C, calculate the equilibrium partial pressures of the three gases if you mix 1.00 bar each of N₂ and O₂.

Interpretation Introduction

Interpretation:

The equilibrium constant for formation of $\text{NO}\left(\text{g}\right)$ from its elements should be calculated and identified that whether the reaction is product favoured at equilibrium at given temperature.

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:

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

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

It is also related to entropy and entropy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{- TΔS}}^{\text{o}}$

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 90SCQ

The value of ${\text{K}}_{\text{p}}$ at $\text{25 }°\text{C }$ is $3.5\times 10^{-31}$.

The reaction is reactant-favoured at equilibrium.

Explanation of Solution

The value of ${\text{K}}_{\text{p}}$ at $\text{25 °C }$ is calculated below.

Given:

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

The given reaction is,

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

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

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ for ${\text{O}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ for ${\text{N}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

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

Substituting the respective values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{86}\text{.58 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0}\right)\end{array}\right]\end{array}\right]\\ \text{= +173}\text{.16 kJ/mol-rxn}\end{array}$

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ is positive, thus the reaction is reactant-favored at equilibrium.

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Substituting the respective values of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$, $\text{T}$ and $\text{R}$.

  $\begin{array}{c}{\text{K}}_{\text{p}}{\text{= e}}^{\text{ - }\frac{\text{+173}\text{.16 kJ/mol- rxn}}{\left(\text{0}\text{.008314 kJ/K×mol}\right)\left(\text{298}\text{.15 K}\right)}}\\ \text{= 3}{\text{.5×10}}^{\text{-31}}\end{array}$

Interpretation Introduction

Interpretation:

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ value for formation of $\text{NO}\left(\text{g}\right)$ from its elements should be calculated at ${\text{700}}^{\text{o}}\text{C}$ and ${\text{K}}_{\text{p}}$ for respective ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ at ${\text{700}}^{\text{o}}\text{C}$ should be estimated also it should be identified that the reaction is product favoured at equilibrium at this given temperature.

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:

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

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

It is also related to entropy and entropy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{- TΔS}}^{\text{o}}$

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 90SCQ

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ at $\text{700 }°\text{C }$ is $179.59kJ\text{/}mol\text{-}rxn$.

The value of ${\text{K}}_{\text{p}}$ at $\text{700 }°\text{C }$ is $4\times 10^{-9}$.

Explanation of Solution

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ and ${\text{K}}_{\text{p}}$ at $\text{700 }°\text{C }$ is calculated below.

Given:

The Appendix L referred for values of standard entropies and enthalpies.

$\begin{array}{cccccc}& {\text{N}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \rightleftarrows & \text{2NO}\left(\text{g}\right)\\ {\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{kJ/mol}\right)& \text{0}& & \text{0}& & \text{90}\text{.29}\\ {\text{S}}^{\text{o}}\left(\text{J/K×mol}\right)& \text{191}\text{.56}& & \text{205}\text{.07}& & \text{210}\text{.76}\end{array}$

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

Substitute the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\left[\begin{array}{l}\left(\text{2 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{90}\text{.29 kJ/mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 180}\text{.58 kJ/mol-rxn}\end{array}$

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

Substitute the values,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{2 mol NO}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{210}\text{.76 J/K×mol}\right)\text{-}\\ \left[\begin{array}{l}\left({\text{1 mol N}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{191}\text{.56 J/K×mol}\right)\text{+}\\ \left({\text{1 mol O}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{205}\text{.07 J/K×mol}\right)\end{array}\right]\end{array}\right]\\ \text{= 24}\text{.89 J/K×mol-rxn}\end{array}$

Now, ${\text{ΔG}}^{\text{o}}{\text{= ΔH}}^{\text{o}}{\text{- TΔS}}^{\text{o}}$

Substitute the value of ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$.

  $\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{= 180}\text{.58 kJ/mol-rxn-}\left[\left(\text{973}\text{.15K}\right)\left(\text{24}\text{.89 J/K×mol-rxn}\right)\right]\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\\ \text{= 156}\text{.36 kJ/mol-rxn}\end{array}$

The value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ is positive, thus the reaction is reactant-favored at equilibrium.

$\Delta G^{\circ }$ is related to the equilibrium constant ${\text{K}}_{\text{p}}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

Substitute the value of ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$, $\text{T}$ and $\text{R}$.

  $\begin{array}{c}{\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{\text{+156}\text{.36 kJ/mol- rxn}}{\left(\text{0}\text{.008314 kJ/K×mol}\right)\left(\text{973}\text{.15 K}\right)}}\\ \text{= 4}{\text{.0×10}}^{\text{-9}}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium partial pressures of the given three gases should be calculated if they were mixed $1bar$ of each $N_{2}and{O}_2$.

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:

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

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RTlnK}}_{\text{p}}$

The rearranged expression is,

  ${\text{K}}_{\text{p}}{\text{= e}}^{\text{- }\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}}{\text{RT}}}$

It is also related to entropy and entropy by the following expression,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{- TΔS}}^{\text{o}}$

Here, ${\text{ΔH}}^{\text{o}}$ is the change in enthalpy and ${\text{ΔS}}^{\text{o}}$ is the change in entropy.

Answer to Problem 90SCQ

The equilibrium partial pressure of ${\text{N}}_{\text{2}}\left(\text{g}\right)$ is $1bar$.

The equilibrium partial pressure of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $1bar$.

The equilibrium partial pressure of $\text{NO}\left(\text{g}\right)$ is $3\times 10^{-5}$.

Explanation of Solution

The equilibrium partial pressure of ${\text{N}}_{\text{2}}\left(\text{g}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$ and $\text{NO}\left(\text{g}\right)$ is calculated below.

Given:

The initial pressures of ${\text{N}}_{\text{2}}\left(\text{g}\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $\text{1 bar}$. The partial pressure of $\text{NO}\left(\text{g}\right)$ is supposed to be $x$.

$\begin{array}{cccccc}& {\text{N}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \rightleftarrows & \text{2NO}\left(\text{g}\right)\\ \text{Initial}& \text{1}& & \text{1}& & \text{0}\\ \text{Equilibrium}& \text{1-x}& & \text{1-x}& & \text{2x}\end{array}$

Now,

  ${\text{K}}_{\text{p}}\text{= 4}{\text{.0×10}}^{\text{-9}}$

The equilibrium constant is related to the equilibrium partial pressure by the expression,

  ${\text{K}}_{\text{p}}\text{= }\frac{{\text{p}}^{\text{2}}{}_{\text{NO}}}{\left({\text{p}}_{{\text{N}}_{\text{2}}}\right)\left({\text{p}}_{{\text{O}}_{\text{2}}}\right)}$

Substitute the values,

  $\begin{array}{c}\text{4}{\text{.0×10}}^{\text{-9}}\text{ = }\frac{{\text{4x}}^{\text{2}}}{\left(\text{1-x}\right)\left(\text{1-x}\right)}\\ {\text{x = 3×10}}^{\text{-5}}\end{array}$

Thus, the value of $\text{x}$ is very small and can be neglected in comparison to $\text{1 bar}$.