Q22.62CP - Chemistry: Molecular Nature Of Matter And Change

Question

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

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

(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   ${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.

Step-by-Step Solution

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  $- 1236 kJ/mol$.

Step 1: 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 in  1902. The Ostwald process is a cornerstone of contemporary chemical manufacturing, and it provides the primary raw material for the most prevalent type of fertiliser.

Step 2: Nature of Change

(a)

Step   includes 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.

Step 3: 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.

Step 4: Calculation for Δ H r x n o

(c)

The formula is –

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

Substitute the values and calculate –

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

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