Question

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.

Short answer

a. The equilibrium constant for this reaction is approximately \(4.72 \times 10^{19}\). 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.

Step-by-step solution

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:

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.

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.

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.

Key concepts

Reduction Potentials

In electrochemistry, reduction potential refers to the tendency of a chemical species to gain electrons and thereby be reduced. It is usually measured in volts (V) under standard conditions.

Reduction potentials are crucial because they help us predict the direction of electron flow in a redox reaction. When comparing reduction potentials, the species with the higher (more positive) reduction potential is more likely to gain electrons.

In the exercise, the standard reduction potential for the conversion of nitrate to nitrogen monoxide (\( \mathrm{NO} \)) and nitrogen dioxide (\( \mathrm{NO}_{2} \)) are given as 0.957 V and 0.775 V, respectively. This indicates that nitrate more readily gains electrons to form \( \mathrm{NO} \) over \( \mathrm{NO}_{2} \) under standard conditions. Hence, the equilibrium between these gases depends on these potential differences.

Equilibrium Constants

The equilibrium constant (\( K \)) is a vital concept in chemistry that provides insight into the balance between products and reactants in a chemical reaction at equilibrium. It is the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation.

Large \( K \) values (much greater than 1) suggest that the formation of products is favored at equilibrium. Conversely, small \( K \) values indicate that reactants are favored. The calculated \( K \) for the given chemical reaction is approximately \( 4.72 \times 10^{19} \), implying that the reaction heavily favors product formation under standard conditions.

This value provides a quantitative basis to determine how an equilibrium mixture will behave when conditions change.

Nernst Equation

The Nernst Equation is a fundamental tool in electrochemistry that relates the reduction potential of a cell to the standard electrode potential, temperature, and activities (or concentrations) of chemical species involved.

The equation is expressed as:\[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q\]where:

• \( \mathscr{E} \) is the cell potential under non-standard conditions.
• \( \mathscr{E}^{\circ} \) is the standard cell potential.
• \( R \) is the universal gas constant (8.314 J/(mol K)).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant (96485 C/mol).
• \( Q \) is the reaction quotient.

In our problem, the Nernst Equation was essential in connecting the cell potential to the equilibrium constant \( K \) of the reaction, elucidating the energetics and equilibria of redox reactions.

Gas Law

The Ideal Gas Law is an equation of state for an ideal gas that relates pressure, volume, temperature, and number of moles of a gas. It is usually expressed as:\[PV = nRT\]where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume of the gas.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant (0.0821 L atm/mol K).
• \( T \) is the temperature in Kelvin.

In this exercise, we applied the ideal gas law to relate the pressure of \( \mathrm{NO} \) and \( \mathrm{NO}_{2} \) to their mole fractions in the reacted mixture, given the overall conditions of temperature and pressure. This relationship helps us determine how varying concentrations of reactants change the equilibrium and final product distribution.