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 \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ 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.

Short answer

The equilibrium constant for the given reaction is approximately \(1.06 * 10^{12}\). To produce a gas mixture containing 0.20% NO2 at 1.00 atm and 25°C, a nitric acid concentration of \(8.6 * 10^{-4}\; mol/L\) is required.

Step-by-step solution

a. Calculate the equilibrium constant for the given reaction.

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 \(1.06 * 10^{12}\).

b. Find the concentration of nitric acid required to produce the desired NO and NO2 mixture.

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.

Key concepts

Standard Reduction Potentials

Standard reduction potentials are essential in understanding the directionality of a chemical reaction. They are given in volts and signify the tendency of a compound to gain electrons; hence its ability to be reduced. The higher the standard reduction potential, the stronger the substance is as an oxidizing agent.
In the given problem, we have two reactions each with their own standard reduction potentials: one forming NO with a potential of 0.957 V and another forming NO2 with 0.775 V.

• The reaction with the higher potential (0.957 V) indicates that forming NO is more favorable than NO2.
• These potentials are used to determine the electrochemical feasibility of the reactions and hence guide the calculations for the equilibrium constant using the Nernst Equation.

By subtracting multiple times the standard reduction potential of the second reaction from the first, we establish the potential for the entire reaction of interest.
The formula for this overall potential is crucial in predicting and calculating whether the reaction will proceed in the forward or backward direction.

Nernst Equation

The Nernst Equation plays a pivotal role in connecting standard reduction potentials to equilibrium conditions. It allows us to calculate the equilibrium constant, \( K \), of a reaction by incorporating the reaction’s standard electrode potential.
The equation is formulated as:\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]
where \( E \) is the cell potential at non-standard conditions, \( E^{\circ} \) is the standard cell potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient.

• In equilibrium, since \( E \) becomes zero, the equation simplifies to solve for \( K \) as seen in the exercise.
• We use the computed value of the reaction’s standard potential (\( E_{ ext{overall}}^{\circ} = -0.368 \, V \)) to find the \( K \), which typically indicates the position of the equilibrium.

This equation thus elegantly relates electrochemistry to equilibrium thermodynamics.

Equilibrium Constant Calculation

Calculating the equilibrium constant, \( K \), provides insights into the extent of a reaction’s completion. In the given chemical equation, the calculated \( K \) value of approximately \(1.06 \times 10^{12}\) is indicative of a highly favorable formation of NO2 from NO under standard conditions.
The significance of \( K \) is as follows:

• A large \( K \) value suggests that products are heavily favored at equilibrium. For this reaction, this high \( K \) means that NO2 is extensively formed.
• Calculation of \( K \) involves the Nernst equation and detailed stoichiometric balancing of the involved half-reactions, determining how potentials add up or cancel each other.

This calculation is critical, as it further assists in understanding the conditions required to achieve particular product ratios or concentrations in chemical reactions, like the nitric acid concentration required in our example.

Stoichiometry in Chemical Reactions

Understanding stoichiometry is essential for balancing chemical equations and determining quantities of reactants and products in a reaction. In the provided equilibrium reaction of nitric acid with copper, stoichiometry tells us how the amounts of NO and NO2 are related.
Key aspects of stoichiometry include:

• The coefficients in the chemical equation (2 for H\(^+\), 2 for NO\(_3^-\), 1 for NO, 3 for NO2, etc.) represent the molar ratio of reactants and products.
• This ratio is used to determine how changes in concentrations of reactants (like H\(^+\) and NO\(_3^-\) in this case) affect the amount of products formed, and vice versa.

Applying stoichiometry in concert with equilibrium concepts allows chemists to predict how different concentrations and conditions affect the outcome of a reaction. In our exercise, it helped in establishing the required concentration of nitric acid to achieve a specific product gas composition, emphasizing the integral role of balancing and reaction ratios.