Chapter 5, Problem 5.51E

The decomposition of ${\text{NaHCO}}_{\text{3}}$, used in kitchens to extinguish fires, is

$2{\text{NaHCO}}_3\left(\text{s}\right)\rightleftharpoons {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

(a) Using the data in Appendix $2$, calculate ${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}$ and ${\Delta }_{\text{rxn}}{H}_{{}_{}}^{\text{°}}$.

(b) Calculate $K$ at $298\text{K}$.

(c) What are $p_{{\text{CO}}_{\text{2}}}$ and $p_{{\text{H}}_2\text{O}}$ at equilibrium at $298\text{K}$?

(d) In a grease fire, the temperature can be around $1150°\text{C}$. What are $p_{{\text{CO}}_{\text{2}}}$ and $p_{{\text{H}}_2\text{O}}$ at equilibrium at $1150°\text{C}$?

Interpretation Introduction

(a)

Interpretation:

The Gibbs free energy of reaction and standard enthalpy of reaction are to be calculated.

Concept introduction:

The standard Gibbs free energy of reaction gives the feasibility of reaction. It is related to equilibrium constant by the equation given below.

${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}=-\text{R}T\ln K$

In the above reaction, $\text{R}$ is the gas constant, ${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}$ is the Gibbs free energy of reaction and $T$ is the temperature. Also, $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ can be calculated for the any given reaction as follows.

${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}=\Delta G°\left(\text{reactants}\right)-\Delta G°\left(\text{products}\right)$.

Answer to Problem 5.51E

The Gibbs free energy of reaction $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ and standard enthalpy of reaction $\Delta H_{{}_{\text{rxn}}}^{\text{°}}$ are $31.03\text{kJ}$ and $135.54\text{kJ}$ respectively.

Explanation of Solution

The reaction given is as written below.

$2{\text{NaHCO}}_3\left(\text{s}\right)\rightleftharpoons {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

For calculation of ${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}$, the above equation can be written as given below.

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

On substitution of the values in above equation, $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ obtained is given below.

$\begin{array}{rll}{\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}& =& \Delta G°\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\right)+\Delta G°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)+\Delta G°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)-2\Delta G°\left({\text{NaHCO}}_3\left(\text{s}\right)\right)\\ & =& \left(-1048.01\right)+\left(-394.35\right)+\left(-228.61\right)-2\left(-851.0\right)\\ & =& 31.03\text{kJ}\end{array}$

Similarly, for calculation of $\Delta H_{{}_{\text{rxn}}}^{\text{°}}$, the equation can be written as given below.

$\Delta H_{{}_{\text{rxn}}}^{\text{°}}=\Delta H°\left(\text{products}\right)-\Delta H°\left(\text{reactants}\right)$

On substitution of the values in above equation, $\Delta H_{{}_{\text{rxn}}}^{\text{°}}$ obtained is given below.

$\begin{array}{rll}\Delta H_{{}_{\text{rxn}}}^{\text{°}}& =& \Delta H°\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\right)+\Delta H°\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)+\Delta H°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)-2\Delta H°\left({\text{NaHCO}}_3\left(\text{s}\right)\right)\\ & =& \left(-1130.77\right)+\left(-393.51\right)+\left(-241.8\right)-2\left(-950.81\right)\\ & =& 135.54\text{kJ}\end{array}$

Conclusion

The Gibbs free energy of reaction and standard enthalpy of reaction are calculated as $31.03\text{kJ}$ and $135.54\text{kJ}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The value of equilibrium constant $K$ at $298\text{K}$ for the given reaction is to be determined.

Concept introduction:

The standard Gibbs free energy of reaction gives the feasibility of reaction. It is related to equilibrium constant by the equation given below.

${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}=-\text{R}T\ln K$

In the above reaction, $\text{R}$ is the gas constant, ${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}$ is the Gibbs free energy of reaction and $T$ is the temperature. Also, $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ can be calculated for the any given reaction as follows.

${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}=\Delta G°\left(\text{reactants}\right)-\Delta G°\left(\text{products}\right)$.

Answer to Problem 5.51E

The value of equilibrium constant $K$ at $298\text{K}$ for the given reaction is $3.6\times {10}^{-6}$.

Explanation of Solution

For calculation of equilibrium constant, the above equation can be written as given below.

${\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}=-\text{R}T\ln K$

Rearranged equation for calculation of equilibrium constant can be written as follows.

$K=e^{\left(-{\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}/\text{R}T\right)}$

On substitution of the values in above equation, equilibrium constant obtained is given below.

$\begin{array}{rll}K& =& e^{\left(-{\Delta }_{\text{rxn}}{G}_{{}_{}}^{\text{°}}/\text{R}T\right)}\\ & =& e^{\left(-31.03\times {10}^3/8.314\times 298\right)}\\ & =& 3.6\times {10}^{-6}\end{array}$

Conclusion

The value of equilibrium constant $K$ at $298\text{K}$ for the given reaction is calculated as $3.6\times {10}^{-6}$.

Interpretation Introduction

(c)

Interpretation:

The partial pressure at $298\text{K}$ for the given reaction is to be determined.

Concept introduction:

The standard Gibbs free energy of reaction gives the feasibility of reaction. It is related to equilibrium constant by the equation given below.

$\Delta G_{{}_{\text{rxn}}}^{\text{°}}=-\text{R}T\ln K$

In the above reaction, $\text{R}$ is the gas constant, $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ is the Gibbs free energy of reaction and $T$ is the temperature.

The equilibrium constant can be written in terms of partial pressure of gases as shown below.

$K=\frac{p_{\text{products}}}{p_{\text{reactants}}}$

Answer to Problem 5.51E

The value of partial pressure at $298\text{K}$ for the given reaction is $1.9\times {10}^{-3}\text{ atm}$.

Explanation of Solution

For calculation of partial pressure, the above equation can be written as given below.

$K=\frac{p_{\text{products}}}{p_{\text{reactants}}}$

Rearranged equation for calculation of partial pressure can be written as follows.

$K=p_{{\text{H}}_2\text{O}\left(\text{g}\right)}{p}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}$

Also, the partial pressures in the given equation have the relation of $p_{{\text{H}}_2\text{O}\left(\text{g}\right)}=p_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}$.

On substitution of the values in above equation, partial pressure obtained is given below.

$\begin{array}{rll}K& =& p_{{\text{H}}_2\text{O}\left(\text{g}\right)}{p}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\\ 3.6\times {10}^{-6}& =& {\left(p_{{\text{H}}_2\text{O}\left(\text{g}\right)}\right)}^2\\ p_{{\text{H}}_2\text{O}\left(\text{g}\right)}& =& 1.9\times {10}^{-3}\text{ atm}\end{array}$

Conclusion

The value of partial pressure at $298\text{K}$ for the given reaction is calculated as $1.9\times {10}^{-3}\text{ atm}$.

Interpretation Introduction

(d)

Interpretation:

The value of partial pressure at $\text{1150}°\text{C}$ for the given reaction is to be determined.

Concept introduction:

The standard Gibbs free energy of reaction gives the feasibility of reaction. It is related to equilibrium constant by the equation given below.

$\Delta G_{{}_{\text{rxn}}}^{\text{°}}=-\text{R}T\ln K$

In the above reaction, $\text{R}$ is the gas constant, $\Delta G_{{}_{\text{rxn}}}^{\text{°}}$ is the Gibbs free energy of reaction and $T$ is the temperature.

The equilibrium constant can be written in terms of partial pressure of gases as shown below.

$K=\frac{p_{\text{products}}}{p_{\text{reactants}}}$

Answer to Problem 5.51E

The value of partial pressure at $\text{1150}°\text{C}$ for the given reaction is $0.27\text{ atm}$.

Explanation of Solution

For calculation of equilibrium constant at $\text{1150}°\text{C}\left(\text{1423K}\right)$, the above equation can be written as given below.

$\Delta G_{{}_{\text{rxn}}}^{\text{°}}=-\text{R}T\ln K$

Rearranged equation for calculation of equilibrium constant can be written as follows.

$K=e^{\left(-\Delta G_{{}_{\text{rxn}}}^{\text{°}}/\text{R}T\right)}$

On substitution of the values in above equation, equilibrium constant obtained is given below.

$\begin{array}{rll}K& =& e^{\left(-\Delta G_{{}_{\text{rxn}}}^{\text{°}}/\text{R}T\right)}\\ & =& e^{\left(-31.03\times {10}^3/8.314\times 1423\right)}\\ & =& 0.074\end{array}$

For calculation of partial pressure, the above equation can be written as given below.

$K=\frac{p_{\text{products}}}{p_{\text{reactants}}}$

Rearranged equation for calculation of partial pressure can be written as follows.

$K=p_{{\text{H}}_2\text{O}\left(\text{g}\right)}{p}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}$

Also, the partial pressures in the given equation have the relation of $p_{{\text{H}}_2\text{O}\left(\text{g}\right)}=p_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}$.

On substitution of the values in above equation, partial pressure obtained is given below.

$\begin{array}{rll}K& =& p_{{\text{H}}_2\text{O}\left(\text{g}\right)}{p}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}\\ 0.074& =& {\left(p_{{\text{H}}_2\text{O}\left(\text{g}\right)}\right)}^2\\ p_{{\text{H}}_2\text{O}\left(\text{g}\right)}& =& 0.27\text{ atm}\end{array}$

Conclusion

The value of partial pressure at $\text{1150}°\text{C}$ for the given reaction is calculated as $0.27\text{ atm}$.