Chapter 5, Problem 5.27E

Nitrogen dioxide, ${\text{NO}}_{\text{2}}$, dimerizes easily to form dinitrogen tetroxide, ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$:

$2{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)$

(a) Using data in Appendix 2, calculate ${\Delta }_{\text{rxn}}G°$ and $K$ for this equilibrium.

(b) Calculate $\xi$ for this equilibrium if $1.00{\text{mol NO}}_2$ were present initially and allowed to come to equilibrium with the dimer in a $20.0\text{-L}$ system.

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{\text{rxn}}G°$ and $K$ for the given equilibrium are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is calculated by subtracting the standard Gibbs free energy of formation of reactants from standard Gibbs free energy of formation of products. The formula for standard Gibbs free energy of reaction is given as,

${\Delta }_{\text{rxn}}G°={\displaystyle \sum {\Delta }_{\text{f}}G°\left(\text{products}\right)}-{\displaystyle \sum {\Delta }_{\text{f}}G°\left(\text{reactants}\right)}$

Where,

• ${\Delta }_{\text{f}}G°\left(\text{products}\right)$ represents the standard Gibbs free energy of formation of products.

• ${\Delta }_{\text{f}}G°\left(\text{reactants}\right)$ represents the standard Gibbs free energy of formation of reactants.

The relation between equilibrium constant and standard Gibbs free energy of reaction is given as,

${\Delta }_{\text{rxn}}G°=-RT\ln K$

Where,

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\cdot \text{K}$.

• $T$ represents the temperature.

• $K$ represents the equilibrium constant

Answer to Problem 5.27E

The values of ${\Delta }_{\text{rxn}}G°$ and $K$ for the given equilibrium are $-4.81\text{kJ}/\text{mol}$ and $6.96$ respectively.

Explanation of Solution

From Appendix $2$

The standard Gibbs free energy of formation of ${\text{NO}}_{\text{2}}$ is $51.30\text{kJ}/\text{mol}$.

The standard Gibbs free energy of formation of ${\text{N}}_2{\text{O}}_4$ is $97.79\text{kJ}/\text{mol}$.

Temperature at the equilibrium is $298\text{ K}$.

The given reaction is represented as,

$2{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)$

The standard Gibbs free energy of given reaction is given as,

${\Delta }_{\text{rxn}}G°={\Delta }_{\text{f}}G°\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)-2{\Delta }_{\text{f}}G°\left({\text{NO}}_2\right)$

Where,

• ${\Delta }_{\text{f}}G°\left({\text{NO}}_2\right)$ represents the standard Gibbs free energy of formation of ${\text{NO}}_2$.

• ${\Delta }_{\text{f}}G°\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ represents the standard Gibbs free energy of formation of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

Substitute the value of ${\Delta }_{\text{f}}G°\left({\text{NO}}_2\right)$ and ${\Delta }_{\text{f}}G°\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)$ in the above equation.

$\begin{array}{rll}{\Delta }_{\text{rxn}}G°& =& 97.79\text{kJ}/\text{mol}-\left(2\right)\left(51.30\text{kJ}/\text{mol}\right)\\ & =& -4.81\text{kJ}/\text{mol}\end{array}$

Therefore, the value ${\Delta }_{\text{rxn}}G°$ at equilibrium is $-4.81\text{kJ}/\text{mol}$.

The relation between equilibrium constant and standard Gibbs free energy of reaction is given as,

${\Delta }_{\text{rxn}}G°=-RT\ln K$

Where,

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\cdot \text{K}$.

• $T$ represents the temperature.

• $K$ represents the equilibrium constant

Rearrange the above equation form the value of $K$.

$K=\exp \left(-\frac{{\Delta }_{\text{rxn}}G°}{RT}\right)$

Substitute the value of $R$, $T$ and ${\Delta }_{\text{rxn}}G°$ in the above equation.

$\begin{array}{rll}K& =& \exp \left(-\frac{\left(-4.81\text{kJ}/\text{mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(298\text{K}\right)}\right)\\ & =& \exp \left(-1.6871\right)\\ & =& 6.96\end{array}$

Therefore, the value $K$ at equilibrium is $6.96$.

Conclusion

The values of ${\Delta }_{\text{rxn}}G°$ and $K$ for the given equilibrium are $-4.81\text{kJ}/\text{mol}$ and $6.97$ respectively.

Interpretation Introduction

Interpretation:

The value of $\xi$ for the given equilibrium reaction is to be calculated.

Concept introduction:

The equilibrium constant of a reaction is expressed as the ratio of partial pressure of products and reactants each raised to the power of their stoichiometric coefficients. A typical equilibrium reaction is represented as,

$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The algebraic form of equilibrium constant for the above chemical reaction is expressed as,

$K=\frac{{\left(p_{\text{C}}\right)}^c{\left(p_{\text{D}}\right)}^d}{{\left(p_{\text{A}}\right)}^a{\left(p_{\text{B}}\right)}^b}$

Where,

• $\left(p_{\text{A}}\right)$ represents the partial pressure of reactant $\text{A}$.

• $\left(p_{\text{B}}\right)$ represents the partial pressure of reactant $\text{B}$.

• $\left(p_{\text{C}}\right)$ represents the partial pressure of product $\text{C}$.

• $\left(p_{\text{D}}\right)$ represents the partial pressure of product $\text{D}$.

• $a$ represents the stoichiometric coefficient of reactant $\text{A}$.

• $b$ represents the stoichiometric coefficient of reactant $\text{B}$.

• $c$ represents the stoichiometric coefficient of product $\text{C}$.

• $d$ represents the stoichiometric coefficient of product $\text{D}$.

Answer to Problem 5.27E

The value of $\xi$ for the given equilibrium reaction is $\text{0}\text{.4 mol}$.

Explanation of Solution

The initial number of moles of ${\text{NO}}_{\text{2}}$ is $1.00\text{mol}$.

The volume of the reaction system is $20.0\text{ L}$.

The temperature of the reaction system is $298\text{K}$.

The value $K$ at equilibrium is $6.96$.

The ideal gas equation is given as,

$PV=nRT$ (1)

Where,

• $V$ represents the volume occupied by the ideal gas.

• $P$ represents the pressure of the ideal gas.

• $n$ represents the number of moles of the ideal gas.

• $T$ represents the temperature of the ideal gas.

• $R$ represents the ideal gas constant with value $0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}$.

Rearrange the above equation for the value of $P$.

$P=\frac{nRT}{V}$

Substitute the value of $V$, $T$, $n$ and $R$ in the above equation.

$\begin{array}{rll}P& =& \frac{\left(1.00\text{mol}\right)\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(298\text{K}\right)}{20.0\text{ L}}\\ & =& 1.2227\text{atm}\end{array}$

The table for initial and equilibrium amounts of the substances involved in the reaction is represented as,

$\begin{array}{l}2{\text{NO}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)\\ \begin{array}{cccc}\text{Pressure}\left(\text{atm}\right)& 2{\text{NO}}_{\text{2}}\left(g\right)& \rightleftharpoons & {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(g\right)\\ \text{Initial}& 1.2227\text{atm}& & 0\text{atm}\\ \text{equilibrium}& \left(1.2227-2x\right)\text{atm}& & x\text{atm}\end{array}\end{array}$

The expression forequilibrium constant for the given equilibrium reaction is represented as,

$K=\frac{\left(p_{{\text{N}}_2{\text{O}}_4}\right)}{{\left(p_{{\text{NO}}_{\text{2}}}\right)}^2}$

Substitute the value of $p_{{\text{N}}_2{\text{O}}_4}$, $p_{{\text{NO}}_{\text{2}}}$ and $K$ in the above expression.

$6.96=\frac{\left(x\text{atm}\right)}{{\left(\left(1.2227-2x\right)\text{atm}\right)}^2}$

Rearrange the equation to form a quadratic equation.

$27.84x^2-18.02x+10.4052=0$

Solve the quadratic equation form the value of $x=0.481\text{ atm}$.

Rearrange equation (1) form the value of $n$.

$n=\frac{PV}{RT}$

Substitute the value of $V$, $T$, $P$ and $R$ in the above equation.

$\begin{array}{rll}n& =& \frac{\left(0.481\text{atm}\right)\left(20.0\text{ L}\right)}{\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(298\text{K}\right)}\\ & =& 0.3934\text{ mol}\end{array}$

The number of mole of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ at equilibrium is $0.3934\text{ mol}$.

The initial number of mole of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is $0\text{ mol}$.

The expression for extent of reaction for ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is given as,

$\xi =\frac{n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{f}}-n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{i}}}{v_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}}$

Where,

• $n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{f}}$ represent the number of moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ at equilibrium.

• $n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{f}}$ represent the initial number of moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

• $v_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ represent the stoichiometric coefficient of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

Substitute the value of $n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{f}}$, $n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{f}}$ and $v_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ in the above expression.

$\begin{array}{rll}\xi & =& \frac{0.3934\text{ mol}-0\text{ mol}}{1}\\ & =& 0.3934\text{ mol}\\ & \approx & \text{0}\text{.4 mol}\end{array}$

Therefore, the extent of the reaction is $\text{0}\text{.4 mol}$.

Conclusion

The extent of the reaction is $\text{0}\text{.4 mol}$.