In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

Step by step solution

01

Write the chemical equation

To write the chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\), we need to combine the provided reactions in such a way that the intermediate compounds cancel out. This could be done by: \[\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) +\frac{3}{2} \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\] and \[\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\] Using the subtraction rule, overdraft the reaction that leads to \(\mathrm{NH}_{3}(\mathrm{g})\) oxidation to \(\mathrm{NO}_{2}(\mathrm{g})\). This gives: \[\mathrm{NH}_{3}(\mathrm{g})+\frac{3}{4} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) +\frac{3}{2} \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\]

02

Determine the composite \(K_{\mathrm{p}}\)

In order to find \(K_{\mathrm{p}}\) for the new reaction, we multiply the \(K_{\mathrm{p}}\) of the original reactions, taking into account that the reverse reaction has an inverse \(K_{\mathrm{p}}\). From the given data, \(K_{\mathrm{p1}}=2.11 \times 10^{19}\) and \(K_{\mathrm{p2}}=0.524\). The \(K_{\mathrm{p2}}\) corresponds to the reverse reaction for the reaction needed, so we take its inverse. The new \(K_{\mathrm{p}}\) would be: \[K_{\mathrm{p(new)}} = K_{\mathrm{p1}} / K_{\mathrm{p2}}\] Substituting the given equilibrium constants we get: \[K_{\mathrm{p(new)}} = 2.11 \times 10^{19} / 0.524 = 4.03 \times 10^{19}\]

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