A sample of air with a mole ratio of \(\mathrm{N}_{2}\) to \(\mathrm{O}_{2}\) of 79: 21 is heated to 2500 K. When equilibrium is established in a closed container with air initially at 1.00 atm, the mole percent of \(\mathrm{NO}\) is found to be \(1.8 \% .\) Calculate \(K_{\mathrm{p}}\) for the reaction. $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$$

The mole fractions of nitrogen and oxygen given by the ratio 79: 21 sum up to 100. Thus the mole fractions are \(0.79\) for \(\mathrm{N}_{2}\) and \(0.21\) for \(\mathrm{O}_{2}\). Remember that the mole fraction is defined as the amount of a specific component divided by the total amount.

The total pressure at equilibrium is the sum of the partial pressures of each gas. Since the initial pressure is 1.00 atm and no gas was added or removed (only the proportions changed), the total pressure at equilibrium is still 1.00 atm. Use the mole percent of nitric oxide at equilibrium to find its partial pressure. A \(1.8 \%\) mole percent corresponds to a partial pressure of \(0.018 \times 1.00 \, \mathrm{atm} = 0.018 \, \mathrm{atm}\). The reaction consumes equal moles of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to produce \(\mathrm{NO}\), so the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) decrease by the same amount, \(0.018 \, \mathrm{atm}\). The final partial pressures are therefore \(0.79 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.772 \, \mathrm{atm}\) for \(\mathrm{N}_{2}\) and \(0.21 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.192 \, \mathrm{atm}\) for \(\mathrm{O}_{2}\).

The equilibrium constant, \(K_{\mathrm{p}}\), for the reaction is defined as \[K_{\mathrm{p}}= \frac{\left(\mathrm{P}_{\mathrm{NO}}\right)^2}{\mathrm{P}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{O}_{2}}}\] where \(\mathrm{P}_{\mathrm{NO}}\), \(\mathrm{P}_{\mathrm{N}_{2}}\), and \(\mathrm{P}_{\mathrm{O}_{2}}\) are the partial pressures of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) at equilibrium. Substituting the calculated pressures, we get \[K_{\mathrm{p}}= \frac{(0.018 \, \mathrm{atm})^2}{0.772 \, \mathrm{atm} \cdot 0.192 \, \mathrm{atm}}\]

Finally, perform the multiplication and division to obtain the value of \(K_{\mathrm{p}}\).