Chapter 5, Problem 5.17E

(a)

Interpretation Introduction

Interpretation:

The expression for K for the given equilibrium is to be stated.

Concept introduction:

When any reaction is at equilibrium, a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K$. The equilibrium constant is independent of the initial amount of the reactant and product.

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\text{ΔG}}^{°}$ whereas $\text{ΔG}$ depends on the reaction condition and extent of reaction.

Answer to Problem 5.17E

The expression for K for the given equilibrium is,

$K_{\text{C}}=\frac{{\left[{\text{NO}}_{\text{2}}\right]}^2}{\left[\text{NO}\right]\left[{\text{O}}_{\text{2}}\right]}$

Explanation of Solution

The given reaction is,

$\text{2NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{2NO}}_2\left(g\right)$

The expression for K for the given equilibrium is,

$K_{\text{C}}=\frac{{\left[{\text{NO}}_{\text{2}}\right]}^2}{{\left[\text{NO}\right]}^2\left[{\text{O}}_{\text{2}}\right]}$

Where,

• ${\left[{\text{NO}}_{\text{2}}\right]}^2$ is the concentration of ${\text{NO}}_{\text{2}}$.

• ${\left[\text{NO}\right]}^2$ is the concentration of $\text{NO}$.

• $\left[{\text{O}}_{\text{2}}\right]$ is the concentration of ${\text{O}}_{\text{2}}$.

Conclusion

The expression for K for the given equilibrium is,

$K_{\text{C}}=\frac{{\left[{\text{NO}}_{\text{2}}\right]}^2}{\left[\text{NO}\right]\left[{\text{O}}_{\text{2}}\right]}$

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta \text{G}°$ for the given equilibrium by using ${\Delta }_{\text{f}}G°$ values given in Appendix 2 is to be calculated.

Concept introduction:

When any reaction is at equilibrium, a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K$. The equilibrium constant is independent of the initial amount of the reactant and product.

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\text{ΔG}}^{°}$ whereas $\text{ΔG}$ depends on the reaction condition and extent of reaction.

Answer to Problem 5.17E

The value of $\Delta \text{G}°$ for the given equilibrium by using ${\Delta }_{\text{f}}G°$ values given in Appendix 2 is $-70.6\text{kJ/mol}$.

Explanation of Solution

The value of ${\Delta }_{\text{f}}G°$ for $\text{NO}$ that is given in Appendix 2 is $86.60\text{kJ/mol}$.

The value of ${\Delta }_{\text{f}}G°$ for ${\text{NO}}_2$ that is given in Appendix 2 is $51.30\text{kJ/mol}$.

The value of ${\Delta }_{\text{f}}G°$ for ${\text{O}}_2$ that is given in Appendix 2 is $0.00\text{kJ/mol}$.

The value of Gibbs free energy of the complete reaction is calculated by the expression,

${\text{ΔG}}_{\text{reaction}}^{°}=2\times {\text{Δ}}_{\text{f}}{G}_{{\text{NO}}_2}^{°}-\left(\text{2}\times {\text{Δ}}_{\text{f}}{G}_{\text{NO}}^{°}+{\text{Δ}}_{\text{f}}{G}_{{\text{O}}_2}^{°}\right)$

Substitute the respective values of Gibbs free energy of the product and the reactant in the above expression.

$\begin{array}{rll}{\text{ΔG}}_{\text{reaction}}^{°}& =& 2\times 51.30\text{kJ/mol}-\left(\text{2}\times 86.60\text{kJ/mol}+0.00\text{kJ/mol}\right)\\ & =& -70.6\text{kJ/mol}\end{array}$

Conclusion

The value of $\Delta \text{G}°$ for the given equilibrium by using ${\Delta }_{\text{f}}G°$ values given in Appendix 2 is $-70.6\text{kJ/mol}$.

(c)

Interpretation Introduction

Interpretation:

The value of K for the given equilibrium is to be calculated.

Concept introduction:

When any reaction is at equilibrium, a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K$. The equilibrium constant is independent of the initial amount of the reactant and product.

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\text{ΔG}}^{°}$ whereas $\text{ΔG}$ depends on the reaction condition and extent of reaction.

Answer to Problem 5.17E

The value of K for the given equilibrium is $2.37\times {10}^{12}$.

Explanation of Solution

The calculated value of $\Delta \text{G}°$ is $-70.6\text{kJ/mol}$.

The conversion of kilojoule to joule is done as,

$1\text{kJ}=1000\text{J}$

Thus, the Gibbs free energy of the reaction becomes $-70600\text{J/mol}$.

The given temperature in appendix 2 at which all the ${\Delta }_{\text{f}}G°$ values are taken is $298\text{K}$.

The value of gas constant, R is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The Gibbs expression is also expressed as,

${\text{Δ}}_{\text{f}}{\text{G}}_{\text{reaction}}^{°}=-\text{R}T\ln K_{\text{C}}$

Substitute the respective values of Gibbs energy, gas constant and temperature in the above expression.

$\begin{array}{rll}-70600\text{J/mol}& =& -\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\times 298\text{K}\times \ln K_{\text{C}}\\ \ln K_{\text{C}}& =& \frac{-70600\text{J/mol}}{-\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\times 298\text{K}}\\ K_{\text{C}}& =& 2.37\times {10}^{12}\end{array}$

Conclusion

The value of the value of K for the given equilibrium is $2.37\times {10}^{12}$.

(d)

Interpretation Introduction

Interpretation:

The direction in which the reaction will move if $1.00\text{bar}$ of $\text{NO}$, $1.00\text{bar}$ of ${\text{O}}_2$, and $0.250\text{bar}$ of ${\text{NO}}_2$ were placed in a chamber is to be predicted.

Concept introduction:

When any reaction is at equilibrium, a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K$. The equilibrium constant is independent of the initial amount of the reactant and product.

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\text{ΔG}}^{°}$ whereas $\text{ΔG}$ depends on the reaction condition and extent of reaction.

Answer to Problem 5.17E

If $1.00\text{bar}$ of $\text{NO}$, $1.00\text{bar}$ of ${\text{O}}_2$, and $0.250\text{bar}$ of ${\text{NO}}_2$ were placed in a chamber then the reaction will move in the forward direction.

Explanation of Solution

The given pressure for $\text{NO}$ is, $p_{\text{NO}}=1.00\text{bar}$.

The given pressure for ${\text{O}}_2$ is, $p_{{\text{O}}_2}=1.00\text{bar}$.

The given pressure for ${\text{NO}}_2$ is, $p_{{\text{NO}}_2}=0.250\text{bar}$.

The calculated value of equilibrium constant, $K_{\text{p}}$ is $1.29\times {10}^{-8}$.

The expression for the equilibrium constant can be shown as,

$K_{\text{p}}=\frac{{\left(p_{{\text{NO}}_2}\right)}^2}{{\left(p_{\text{NO}}\right)}^2\left(p_{{\text{O}}_2}\right)}$

Substitute the respective values of pressure of hydrogen cyanide and equilibrium constant in the above expression.

$\begin{array}{rll}{K}_{\text{p}}& =& \frac{{\left(0.250\text{bar}\right)}^2}{{\left(1.00\text{bar}\right)}^2\left(1.00\text{bar}\right)}\\ & =& 0.0625\end{array}$

The given temperature in appendix 2 is $298\text{K}$.

The value of gas constant, R is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The equilibrium constant with respect to concentration of the reaction is calculated by the expression,

$Q_{\text{C}}=K_{\text{p}}\text{R}T$

Where,

• $Q_{\text{C}}$ is the equilibrium constant for concentration.

• $K_{\text{p}}$ is the equilibrium with respect to pressure.

• $\text{R}$ is the gas constant.

• $T$ is the temperature.

Substitute all the respective values in the above expression.

$\begin{array}{rll}{Q}_{\text{C}}& =& 0.0625\times 8.314\times 298\\ & =& 154.8\end{array}$

Thus, the value of $Q_{\text{C}}$ is greater than $K_{\text{p}}$ which states that the reaction will move in the forward direction.

Conclusion

If $1.00\text{bar}$ of $\text{NO}$, $1.00\text{bar}$ of ${\text{O}}_2$, and $0.250\text{bar}$ of ${\text{NO}}_2$ were placed in a chamber then the reaction will move in the forward direction.