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 \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) 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.

Step by step solution

01

Write provided equations in reduction form

\[ 2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \] We can convert this equation into a reduction form by reversing it, and doubling it: \[ 2 \mathrm{NO}(g) \longrightarrow 4\mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) + 3 \mathrm{NO}_{2}(g)-2\mathrm{H}_{2} \mathrm{O}(l) \] Now let's find the cell potential (E) for this reaction using the given reduction potentials:

02

Determine the cell potential

The Nernst Equation can be used to determine the difference in cell potential for the two desired reactions. However, first we need to multiply the provided equations to obtain the desired balanced equation: \[ 2(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)) \] \[ 6\mathrm{e}^{-} + 8\mathrm{H}^{+}(a q) + 2\mathrm{NO}_{3}^{-}(a q) \longrightarrow 2\mathrm{NO}(g)+ 4 \mathrm{H}_{2} \mathrm{O}(l) \] Using the provided reduction potentials, we can determine the cell potential difference (E) for the reaction: \( \mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{1} -\mathscr{E}^{\circ}_{2} = (2 \times 0.957) - 0.775 = 1.139\ \mathrm{V} \) Now, let's use the Nernst Equation to determine the equilibrium constant (K) for the given reaction.

03

Calculate the equilibrium constant

The Nernst Equation can be written as: \[ \mathscr{E}^{\circ}_{cell} = \dfrac{RT}{nF} \ln{K} \] Where R is the gas constant (8.314 J/(mol K)), T is the temperature (25˚C or 298.15 K), n is the number of electrons transferred in the reaction (6), and F is Faraday's constant (96485 C/mol). We can rearrange the equation to solve for K: \[ K = e^{\frac{\mathscr{E}^{\circ}_{cell} \times nF}{RT}} \] Now, let's calculate K: \[ K = e^{\frac{1.139 \times 6 \times 96485}{8.314 \times 298.15}} = 4.72 \times 10^{19} \] a. The equilibrium constant for this reaction is approximately \(4.72 \times 10^{19}\). For part b, we are asked to find the concentration of nitric acid that will produce a NO and NO2 mixture with only 0.20% NO2 (by moles) at 25˚C and 1.00 atm.

04

Apply the equilibrium constant and Ideal Gas Law

We can use the equilibrium constant and the ideal gas law to relate the pressure of NO and NO2 to their concentrations in the reaction. Let x be the pressure of NO and (1-x) be the pressure of NO2 at equilibrium. Then, the moles fraction of NO2 is given by: \[ \frac{1-x}{1} = 0.002 \] Now, we can write the expression for K using molar concentrations: \[ K = \dfrac{[\mathrm{H}^{+}]^4[\mathrm{NO}_{3}^{-}]^2{x}^2}{(1-x)^3} \] Hence we can solve for the concentration of nitric acid (\([\mathrm{H}^{+}]\), \([\mathrm{NO}_{3}^{-}]\)): \[ [\mathrm{H}^{+}]^4[\mathrm{NO}_{3}^{-}]^2=\frac{K(1-x)^3}{x^2}= \frac{(4.72 \times 10^{19})(0.998)^3}{(0.002)^2} = 1.17 \times 10^{24} \] Considering nitric acid to be a strong monoprotic acid under these conditions, we can assume that \([\mathrm{H}^{+}]=[\mathrm{NO}_{3}^{-}]\). Thus: \[ ([\mathrm{H}^{+}])^6=(1.17 \times 10^{24}) \] Now we can solve for \([\mathrm{H}^{+}]\): \[ [\mathrm{H}^{+}]=(1.17 \times 10^{24})^{\frac{1}{6}} = 2.77 \ \mathrm{M} \] b. The concentration of nitric acid will be around 2.77 M to produce a NO and NO2 mixture with only 0.20% NO2 (by moles) at 25˚C and 1.00 atm.

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