Chapter 10, Problem 10.102PAE

10.102 Ammonia can react with oxygen gas to form nitrogen dioxide and water. (a) Write a balanced chemical equation for this reaction. (b) Use tabulated data to determine the free energy change for the reaction and comment on its spontaneity. (c) Use tabulated data to calculate the enthalpy change of the reaction, (d) Determine how much heat flows and in what direction when 11.4 g of ammonia gas is burned in excess oxygen.

Interpretation Introduction

Interpretation:

Ammonia reacts with Oxygen gas to form Nitrogen dioxide and water. The feasibility of this reaction can be identified by using free energy change. The free energy change, enthalpy change and entropy change can be calculated by using standard data.

Concept introduction: The free energy change of the reaction $\Delta G^0$ is calculated by using $\Delta H^0$ and $\Delta S^0$ at the given temperature. The formula that can be used is as follows.

$\Delta G^0=\Delta H^0-T\Delta S^0$

$\Delta H^0,\Delta S^0$ is calculated by using the following mathematical expressions.

$\Delta H^0={\displaystyle \sum \Delta H^0\left(\text{products}\right)-}{\displaystyle \sum \Delta H^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta S^0={\displaystyle \sum \Delta S^0\left(\text{products}\right)-}{\displaystyle \sum \Delta S^0\left(\text{reactants}\right)}\\ \Delta G^0={\displaystyle \sum \Delta G^0\left(\text{products}\right)-}{\displaystyle \sum \Delta G^0\left(\text{reactants}\right)}\end{array}$

If the sign of the free energy change is negative, it indicates that the reaction is spontaneous.

Answer to Problem 10.102PAE

Solution: a) The balanced equation is $4{\text{NH}}_3\left(g\right)+7{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }4{\text{NO}}_2\left(g\right)+\text{6}{\text{H}}_2\text{O}\left(l\right)$

b) $\Delta G_f^0=\boxed{-1152\text{kJ}}$

c) $\Delta H_f^0=\boxed{-1397.56\text{kJ}}$

d) The amount of heat produced by 11.4 g of NH₃ = 233.88 kJ

Given:

From the standard data −

$\begin{array}{l}\Delta G_f^0\left[{\text{NH}}_3\left(g\right)\right]=-16.5\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{NO}}_{\text{2}}\left(g\right)\right]=51.3\text{kJ/mol}\\ \Delta G_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]=-237.2\text{kJ/mol,}\Delta {\text{G}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\begin{array}{l}\Delta H_f^0\left[{\text{NH}}_3\left(g\right)\right]=-46.11\text{kJ/mol}\text{,}\Delta {\text{H}}_f^0\left[{\text{NO}}_2\left(g\right)\right]=33.2\text{kJ/mol}\\ \Delta H_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{kJ/mol,}\Delta {\text{H}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

Explanation of Solution

a)

The balanced chemical equation of the given reaction is as follows.

$4{\text{NH}}_3\left(g\right)+7{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }4{\text{NO}}_2\left(g\right)+\text{6}{\text{H}}_2\text{O}\left(l\right)$

b) From the standard data −

$\begin{array}{l}\Delta G_f^0\left[{\text{NH}}_3\left(g\right)\right]=-16.5\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{NO}}_{\text{2}}\left(g\right)\right]=51.3\text{kJ/mol}\\ \Delta G_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]=-237.2\text{kJ/mol,}\Delta {\text{G}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\Delta G_{rxn}^0={\displaystyle \sum \Delta G_f^0\left(\text{products}\right)-}{\displaystyle \sum \Delta G_f^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta G_{rxn}^0=\left[4\times \Delta G_f^0\left[{\text{NO}}_2\left(\text{g}\right)\right]+6\times \Delta {\text{G}}_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]\right]-\left[4\times \Delta G_f^0\left[{\text{NH}}_3\left(g\right)\right]-7\times \Delta {\text{G}}_f^0\left[{\text{O}}_2\left(g\right)\right]\right]\\ =\left[4\left(51.3\right)+6\left(-237.2\right)-4\left(-16.5\right)-0\right]\text{kJ}\\ =\boxed{-1152\text{kJ}}\end{array}$

As $\Delta G_f^0$ is in negative sign, we conclude that the reaction is spontaneous.

c)

From the standard data −

$\begin{array}{l}\Delta H_f^0\left[{\text{NH}}_3\left(g\right)\right]=-46.11\text{kJ/mol}\text{,}\Delta {\text{H}}_f^0\left[{\text{NO}}_2\left(g\right)\right]=33.2\text{kJ/mol}\\ \Delta H_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{kJ/mol,}\Delta {\text{H}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\Delta H_{rxn}^0={\displaystyle \sum \Delta H_f^0\left(\text{products}\right)-}{\displaystyle \sum \Delta H_f^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta H_{rxn}^0=\left[4\times \Delta H_f^0\left[{\text{NO}}_2\left(\text{g}\right)\right]+6\times \Delta {\text{H}}_f^0\left[{\text{H}}_2\text{O}\left(l\right)\right]\right]-\left[4\times \Delta H_f^0\left[{\text{NH}}_3\left(g\right)\right]+7\times \Delta {\text{H}}_f^0\left[{\text{O}}_2\left(g\right)\right]\right]\\ =\left[4\left(33.2\right)+6\left(-285.8\right)-4\left(-46.11\right)-0\right]\\ =\boxed{-1397.56\text{kJ}}\end{array}$

d)

From the balanced equation, 4 mols $\left(4\text{mol×17}\text{g/mol}\text{=}68.12\text{g}\right)$ ammonia $\left({\text{NH}}_3\right)$ produce 1397.56 kJ of heat.

Therefore, the amount of heat produced by 11.4 g of ammonia $\text{=}\left(\frac{\text{-1397}\text{.56×11}\text{.4}}{\text{68}\text{.12}}\right)\text{kJ}$

$=\boxed{-233.88\text{kJ}}$

As the above value is in negative, we conclude that the reaction is exothermic. Hence heat transfers to surroundings.

Conclusion

The given reaction is spontaneous as free energy change of the reaction is negative. The given reaction is an exothermic reaction as enthalpy change is negative and it produces 1397.56 kJ of energy.