When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at $25^{\circ} \mathrm{C} :$ $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

We have been given two half-reactions with their standard cell potentials \(\mathscr{E}^{\circ}\): Reaction 1 : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \\ \( \mathscr{E}_{1}^{\circ}=0.957 \mathrm{V}\) Reaction 2: \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \\ \( \mathscr{E}_{2}^{\circ}=0.775 \mathrm{V}\) First, we need to multiply Reaction 2 by 3, and then subtract it from Reaction 1 to get the overall reaction: Overall Reaction: \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Now, subtract the standard cell potentials: \\ \(\mathscr{E}_{overall}^{\circ} = \mathscr{E}_{1}^{\circ} - 3 {\mathscr{E}_{2}^{\circ}} \\ \mathscr{E}_{overall}^{\circ} = 0.957 - 3(0.775) = -0.368V\) Apply the Nernst equation for the equilibrium constant K: \( K = e^{\frac{-n \mathscr{E}_{overall}^{\circ} F}{RT}}\) Here, n is the number of electrons exchanged, R is the gas constant, T is the temperature in Kelvin, and F is the Faraday's constant. For the overall reaction, the number of electrons exchanged is n = 3. R = 8.314 J K-1 mol-1 T = 25 + 273.15 = 298.15K F = 96485.34 C mol-1 Now plug in the values and calculate K: \( K = e^{\frac{-3(-0.368)(96485.34)}{(8.314)(298.15)}}\) \( K \approx 1.06 * 10^{12}\) The equilibrium constant for the given reaction is approximately \(1.06 * 10^{12}\).

We are given that the molar percentage of NO2 in the gas mixture is 0.20% and asked to determine the concentration of nitric acid required. Let's denote the initial concentration of H+ and NO3- ions as x. The assumption that the change in acid concentration can be neglected means that the concentrations at equilibrium will still be x. Let the change in NO concentration be y. According to the stoichiometry of the reaction: At equilibrium, the NO concentration is y; NO2 concentration is 0.002y; H2O has no effect on the equilibrium constant as it is a liquid. So, the equilibrium expression for K will be: \(1.06 * 10^{12} = \frac{(0.002y)^3}{(x^2)(y)}\) We also have the ideal gas law: \(PV = nRT\) The combined molar volume of NO and NO2 at 1.00 atm and 25°C is: \(\frac{n}{V} = \frac{1.002y}{RT}\) Now we can substitute the expression for the ideal gas law into the equilibrium expression: \(1.06 * 10^{12} = \frac{(0.002 * \frac{1.002x}{RT})^3}{(x^2)(\frac{1.002x}{RT})}\) Solve this equation for x to find the nitric acid concentration: \(x \approx 8.6 * 10^{-4}\; mol/L\) Therefore, a nitric acid concentration of \(8.6 * 10^{-4}\; mol/L\) is required to produce the desired mixture of NO and NO2 with 0.20% of NO2 at 1.00 atm and 25°C.