At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}} \approx 1 \times 10^{-31}\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ a. Calculate the concentration of \(\mathrm{NO}\), in molecules \(/ \mathrm{cm}^{3}\), that can exist in equilibrium in air at \(25^{\circ} \mathrm{C}\). In air, \(P_{\mathrm{N}_{2}}=0.8 \mathrm{~atm}\) and \(P_{\mathrm{o},}=0.2\) atm. b. Typical concentrations of NO in relatively pristine environments range from \(10^{8}\) to \(10^{10}\) molecules \(/ \mathrm{cm}^{3}\). Why is there a discrepancy between these values and your answer to part a?

We are given the reaction: \[ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \] The equilibrium expression for this reaction is given by: \[ K_p = \frac{P_{NO}^2}{P_{N2} \cdot P_{O2}} \]

We are given the partial pressures as: \(P_{N2} = 0.8\) atm \(P_{O2} = 0.2\) atm

We know the value of \(K_p\): \[ K_p = 1 \times 10^{-31} \] Now we can substitute the values of \(K_p\), \(P_{N2}\), and \(P_{O2}\) into the equilibrium expression: \[ 1 \times 10^{-31} = \frac{P_{NO}^2}{0.8 \cdot 0.2} \] Solve for \(P_{NO}\): \[ P_{NO}^2 = 10^{-31} \times 0.8 \times 0.2 \] \[ P_{NO} = \sqrt{1.6 \times 10^{-32}} \] \[ P_{NO} \approx 4 \times 10^{-17} \mathrm{~atm} \]

Now we need to convert the partial pressure of NO into concentration using the ideal gas law equation in terms of concentration: \[ P = n \frac{RT}{V} \] where \(n\) = number of moles \(R\) = gas constant = \(8.314 \times 10^{-3} \frac{L \cdot atm}{K \cdot mol}\) (in the units we need) \(T\) = temperature in Kelvin (\(273.15 + 25 = 298.15 K\)) \(V\) = volume in L Since we need concentration in molecules/cm³, let's convert the units: \[ 1 \mathrm{mol} = 6.022 \times 10^{23} \mathrm{molecules} \] \[ 1 \mathrm{L} = 1000 \mathrm{cm}^3 \] Now we can multiply the pressure by the conversion factor: \[ \frac{1 \mathrm{mol}}{R \cdot T} \times \frac{6.022 \times 10^{23} \mathrm{molecules}}{\mathrm{mol}} \times \frac{1000\mathrm{ ~cm}^{3}}{\mathrm{L}} \] Plug in the values and solve for concentration: \[ C_{NO} = P_{NO} \times \frac{1}{8.314 \times 10^{-3} \frac{L \cdot atm}{K \cdot mol}} \times \frac{1}{298.15K} \times 6.022 \times 10^{23} \mathrm{molecules/mol} \times \frac{1000\mathrm{ ~cm}^{3}}{\mathrm{L}} \] \[ C_{NO} \approx 8.1 \times 10^{7} \mathrm{molecules/cm^3} \]

From our calculation, we found that the concentration of NO at equilibrium is about \(8.1 \times 10^7\) molecules/cm³. However, typical concentrations of NO in relatively pristine environments range from \(10^8\) to \(10^{10}\) molecules/cm³. The discrepancy between the calculated value and the observed value might be due to various factors: incomplete reactions, local variations of the concentrations of reactants, and the simplifications made when considering equilibrium. Equilibrium predictions assume that the system has reached a stable state and does not account for any transitory chemical reactions that might cause deviations in the observed values. Additionally, our calculation assumes ideal conditions as per the ideal gas law, which might not apply to real, non-ideal environmental systems.