Question

The Ostwald process for the production of ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$ is

role="math" localid="1663397733473" $$\begin{gathered} {\mathbf{\text{(1)4NH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + 5O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\stackrel{\mathbf{\text{Pt/Rh Catalyst}}}{\mathbf{\to }}{\mathbf{\text{4NO(g) + 6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}} \\ {\mathbf{\text{(2)2NO(g) +O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{2NO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}} \\ {\mathbf{\text{(3)3NO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) +H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}\mathbf{\to }{\mathbf{\text{2HNO}}}_{\mathbf{\text{3}}}\mathbf{\text{(aq) +NO(g)}} \end{gathered}$$

(a) Describe the nature of the change that occurs in step .

(b) Write an overall equation that includes ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$and ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$as the only nitrogen-containing species.

(c) Calculate $\mathit{\Delta }{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{o}}$ (in kJ/mol atoms) for this reaction at role="math" localid="1663397880389" ${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.

Short answer

(a) The nature of the change that occurs in step three is disproportionation.

(b) The overall reaction is $4N{H}_3\left(g\right)+8O_2\left(g\right)+3{\text{H}}_2\text{O}\left(\text{l}\right)\to 6{\text{H}}_2\text{O}\left(\text{g}\right)+4{\text{HNO}}_3(\text{aq})$.

(c) The value for $\Delta H_{rxn}^o$ at ${\text{25}}^{\text{o}}\text{C}$ is $-1236kJ/mol$.

Step-by-step solution

Concept Introduction

The Ostwald process is a chemical reaction that produces nitric acid (${\text{HNO}}_{\text{3}}$ ). Wilhelm Ostwald invented the method, which he patented in1902. The Ostwald process is a cornerstone of contemporary chemical manufacturing, and it provides the primary raw material for the most prevalent type of fertiliser.

Nature of Change

(a)

Stepincludes the disproportionation of nitrogen from oxidation state$\left(IV\right)$to$V$and$II$.

Disproportionation is a redox process in which one compound converts to two different compounds out of which one has a higher oxidation state, and the other one a lower oxidation state than the starting compound.

Therefore, it is disproportionation.

Overall Reaction

(b)

Combine the given three reaction into one reaction –

$$\begin{gathered} 4{\text{NH}}_3\left(g\right)+5{\text{O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right) \\ 2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{NO}}_2\left(g\right) \\ 3{\text{NO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\to 2{\text{HNO}}_3\left(aq\right)+\text{NO}\left(g\right) \\ \overline{4{\text{NH}}_3\left(g\right)+8{\text{O}}_2\left(g\right)+6\text{NO}\left(g\right)+6{\text{NO}}_2\left(g\right)+3{\text{H}}_2\text{O}\left(\text{l}\right)\to 6\text{NO}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)+6{\text{NO}}_2\left(g\right)+4{\text{HNO}}_3\left(aq\right)} \\ 4N{H}_3\left(g\right)+8O_2\left(g\right)+3{\text{H}}_2\text{O}\left(\text{l}\right)\to 6{\text{H}}_2\text{O}\left(\text{g}\right)+4{\text{HNO}}_3(\text{aq}) \end{gathered}$$

Therefore, the overall reaction is obtained.

Calculation for  ΔHrxno

(c)

The formula is –

$$\begin{gathered} \Delta H_{rxn}^o=\Delta H_{products}^o-\Delta H_{reac\tan ts}^o\Delta H_{rxn}^o \\ =\left[4mol·\Delta H_{HN{O}_3,aq}^o+6mol·\Delta H_{H_{2}O,g}^o\right]-\left[4mol·\Delta H_{N{H}_3,g}^o+8mol·\Delta H_{O_{2,g}}^o+3mol·\Delta H_{\left.H_{2}O,l\right]}^o\right] \end{gathered}$$

Substitute the values and calculate –

$$\begin{gathered} \Delta H_{rxn}^o=[4\times (-206.57kJ/mol)+6mol\times (-241.826kJ/mol\left)\right] \\ -[4mol\times (-45.9kJ/mol)+8·0kJ/mol+3\times (-285.840kJ/mol\left)\right] \\ \Delta H_{rxn}^o=-1236kJ/mol \end{gathered}$$

Therefore, the value obtained as $-1236kJ/mol$.