Chapter 7, Problem 79E

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

$\begin{array}{ll}4{\text{NH}}_3\left(g\right)+5{\text{O}}_2\hfill & \to 4\text{NO}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)\hfill \\ 2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\hfill & \to 2{\text{NO}}_2\left(g\right)\hfill \\ 3{\text{NO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\hfill & \to 2{\text{HNO}}_3\left(aq\right)+\text{NO}\left(g\right)\hfill \end{array}$

a. Use the values of $\Delta H_{\text{f}}^{°}$ in Appendix 4 to calculate the value of ∆Hº for each of the preceding reactions.

b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

Interpretation Introduction

Interpretation: The enthalpy change $\text{ΔH}$ of given reactions has to be calculated and type of overall reaction should be explained.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
• Formula:

               ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• In equation the enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.
• The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
• Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

              Formula:

                              $\text{ΔH}\left(\text{total}\right)\text{= }{\displaystyle \sum \text{ΔH}}\left(\text{number of steps}\right)\mathrm{......}\text{(2)}$

Rules:

• The sign of ΔH reversed when the reaction is reversed.
• The quantities of reactants and products is directly proportional to the magnitude of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.
• The standard enthalpy of formation of homo atomic molecules is zero.

Answer to Problem 79E

The overall reaction is ${\text{12NH}}_{\text{3(g)}}{\text{+21O}}_{\text{2(g)}}\to {\text{8HNO}}_{\text{3(aq)}}{\text{+4NO}}_{\text{(g)}}{\text{+14H}}_{\text{2}}{\text{O}}_{\text{(g)}}$ and it is a exothermic reaction.

Explanation of Solution

Given:

Ostwald process for the commercial production of nitric acid from ammonia and oxygen

                              $\begin{array}{l}{\text{4NH}}_{\text{3(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4NO}}_{\text{(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\mathrm{......}(1)\\ {\text{2NO}}_{\text{(g)}}{\text{+O}}_{\text{2}}{}_{\text{(g)}}\to {\text{2NO}}_{\text{2(g)}}\mathrm{......}(2)\\ {\text{3NO}}_{\text{2(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{2HNO}}_{\text{3(aq)}}{\text{+NO}}_{\text{(g)}}\mathrm{......}(3)\end{array}$

To calculate: Enthalpy change of the reactions.

                                  ${\text{4NH}}_{\text{3(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4NO}}_{\text{(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\mathrm{......}(1)$

                           $\begin{array}{l}\text{ΔH°}\text{=}\left[\text{4mol}\left(\frac{\text{90}\text{KJ}}{\text{mol}}\right)\text{+6mol}\left(\frac{\text{-242KJ}}{\text{mol}}\right)\right]\text{-}\left[\text{4mol}\left(\frac{\text{-46KJ}}{\text{mol}}\right)\right]\\ \text{=}\text{-908KJ}\end{array}$

                                             ${\text{2NO}}_{\text{(g)}}{\text{+O}}_{\text{2}}{}_{\text{(g)}}\to {\text{2NO}}_{\text{2(g)}}\mathrm{......}(2)$

                           $\begin{array}{l}\text{ΔH°=}\left[\text{2mol}\left(\frac{\text{34KJ}}{\text{mol}}\right)\right]\text{-}\left[\text{2mol}\left(\frac{\text{90KJ}}{\text{mol}}\right)\right]\\ \text{=}\text{-112KJ}\end{array}$

                                     ${\text{3NO}}_{\text{2(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{2HNO}}_{\text{3(aq)}}{\text{+NO}}_{\text{(g)}}\mathrm{......}(3)$

                       $\begin{array}{l}\text{ΔH°}\text{=}\left[\text{2mol}\left(\frac{\text{-207KJ}}{\text{mol}}\right)\text{+1mol}\left(\frac{\text{90KJ}}{\text{mol}}\right)\right]\text{-}\left[\text{3mol}\left(\frac{\text{34KJ}}{\text{mol}}\right)\text{+1mol}\left(\frac{\text{-286KJ}}{\text{mol}}\right)\text{+}\right]\\ \text{=}\text{-140KJ}\end{array}$

Standard enthalpy of formation of reactant and product are multiplied by its magnitude and subtract the value of reactant from product to get enthalpy of reaction (1) is $\text{-908KJ}$, reaction (2) is $\text{-112KJ}$ and reaction (3) is $\text{-140KJ}$.

Interpretation Introduction

Interpretation: The enthalpy change $\text{ΔH}$ of given reactions has to be calculated and type of overall reaction should be explained.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
• Formula:

               ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• In equation the enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.
• The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
• Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

              Formula:

  $\text{ΔH}\left(\text{total}\right)\text{= }{\displaystyle \sum \text{ΔH}}\left(\text{number of steps}\right)\mathrm{......}\text{(2)}$

Rules:

• The sign of ΔH reversed when the reaction is reversed.
• The quantities of reactants and products is directly proportional to the magnitude of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.
• The standard enthalpy of formation of homo atomic molecules is zero.

Answer to Problem 79E

The overall reaction is ${\text{12NH}}_{\text{3(g)}}{\text{+21O}}_{\text{2(g)}}\to {\text{8HNO}}_{\text{3(aq)}}{\text{+4NO}}_{\text{(g)}}{\text{+14H}}_{\text{2}}{\text{O}}_{\text{(g)}}$ and it is an exothermic reaction.

Explanation of Solution

Given:

Ostwald process for the commercial production of nitric acid from ammonia and oxygen

                              $\begin{array}{l}{\text{4NH}}_{\text{3(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4NO}}_{\text{(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\mathrm{......}(1)\\ {\text{2NO}}_{\text{(g)}}{\text{+O}}_{\text{2}}{}_{\text{(g)}}\to {\text{2NO}}_{\text{2(g)}}\mathrm{......}(2)\\ {\text{3NO}}_{\text{2(g)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{2HNO}}_{\text{3(aq)}}{\text{+NO}}_{\text{(g)}}\mathrm{......}(3)\end{array}$

Ostwald process for the commercial production of nitric acid from ammonia and oxygen involving in steps are recorded as shown above.

To write: The overall reaction.

                           $\begin{array}{l}{\text{12NH}}_{\text{3(g)}}{\text{+15O}}_{\text{2(g)}}\to {\text{12NO}}_{\text{(g)}}{\text{+18H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ {\text{12NO}}_{\text{(g)}}{\text{+6O}}_{\text{2}}{}_{\text{(g)}}\to {\text{12NO}}_{\text{2(g)}}\\ {\text{12NO}}_{\text{2(g)}}{\text{+4H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{8HNO}}_{\text{3(aq)}}{\text{+4NO}}_{\text{(g)}}\\ {\text{4H}}_{\text{2}}{\text{O}}_{\text{(g)}}\to {\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{The}\text{overall}\text{reaction}\text{is:}\\ \\ {\text{12NH}}_{\text{3(g)}}{\text{+21O}}_{\text{2(g)}}\to {\text{8HNO}}_{\text{3(aq)}}{\text{+4NO}}_{\text{(g)}}{\text{+14H}}_{\text{2}}{\text{O}}_{\text{(g)}}\end{array}$

The all the reactions are having a negative sign of $\text{ΔΗ}$ so the overall reaction is exothermic reaction.

To write the balanced reaction and multiplied to the integer by calculate the mole ration of products and reactants.