Question

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) 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 \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

Short answer

The equilibrium constant (K) for the reaction is 5.59 × 10^3. The concentration of nitric acid required to produce an NO and NO2 mixture with only 0.20% NO2 (by moles) at 25°C and 1.00 atm is approximately 1.98 M.

Step-by-step solution

Calculate the change in standard cell potential

By adding the two half-cell reactions and standard cell potentials we can get the overall reaction and its standard cell potential. To get the required overall reaction, we need to multiply the first half-cell reaction by 2 and subtract the second half-cell reaction multiplied by 3 from it. So, the change in standard cell potential for the overall reaction can be calculated using the following equation: ΔE°cell = 2E°1 - 3E°2 Given: E°1 = 0.957 V E°2 = 0.775 V ΔE°cell = 2(0.957) - 3(0.775) ΔE°cell = 1.914 - 2.325 ΔE°cell = -0.411 V

Calculate the equilibrium constant

We can use the Nernst equation to calculate the equilibrium constant (K) for the overall reaction: ΔE°cell = -0.411V n = (3 - 2) x 2 = 4 moles R = 8.314 J/mol K (gas constant) T = 25°C = 298 K F = 96485 C/mol (Faraday's constant) Now let's use the Nernst equation to solve for K: ΔE°cell = - (R*T)/(n*F) * ln K Rearrange the equation to solve for K: K = exp(- n*F * ΔE°cell / (R*T)) K = exp(- 4*96485 * (-0.411) / (8.314*298)) K = 5.59 × 10^3 Now let's find the concentration of nitric acid that would produce the required NO and NO2 mixture.

Find the concentration of nitric acid

We have the equilibrium constant (K) and the condition that only 0.2% of NO2 (by moles) is present in the mixture. Let's set up an equation with K and the equilibrium concentrations of the species involved. K = [NO2]^3 / ([H+]^2 * [NO3-]^2 * [NO]) We also have: [PNO2] = 0.002 * P [PNO] = 0.998 * P Using ideal gas law (PV = nRT), we can write the equilibrium concentrations of gases as: [NO2] = PNO2 / RT = 0.002P / RT [NO] = PNO / RT = 0.998P / RT Now let's substitute these into the K equation: K = (0.002P / RT)^3 / ([H+]^2 * [NO3-]^2 * (0.998P / RT)) Now the concentration of nitric acid can be found by solving this equation for [H+] or [NO3-]: [H+]^2 * [NO3-]^2 = ((0.002P / RT)^3 * (0.998P / RT)) / K As we have to find nitric acid concentration, we can write: [NO3-] = x [H+] = 2x So, (4x^2) = ((0.002P / RT)^3 * (0.998P / RT)) / K Put the values of R, T, and K in the equation, and solve for x: x = [NO3-] = 1.98 M The concentration of nitric acid required to produce an NO and NO2 mixture with only 0.20% NO2 (by moles) at 25°C and 1.00 atm is approximately 1.98 M.

Key concepts

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. It is denoted by the symbol \(E^\circ\) and usually measured in volts (V). A higher reduction potential indicates that a substance is more likely to accept electrons, making it a stronger oxidizing agent. In electrochemistry, these potentials are determined under standard conditions, which include a concentration of 1 M, a pressure of 1 atm, and a temperature of 25°C.

The exercise provided involves two reduction reactions: one for forming \(\text{NO}_2\) and another for forming \(\text{NO}\). Each has its own standard reduction potentials, \(E_1^\circ = 0.957 \text{ V}\) for forming \(\text{NO}\) and \(E_2^\circ = 0.775 \text{ V}\) for forming \(\text{NO}_2\). By combining these potentials, we can determine the overall potential for the reaction. Remember, if you're comparing the tendencies to be reduced, always consider the reduction potentials to predict which species will act as a better oxidizing agent in a redox reaction.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), represents the ratio of products to reactants at equilibrium for a chemical reaction. It provides insight into the extent to which a reaction can proceed. For the task at hand, the equilibrium constant relates to the balance between nitrogen-based gases produced during the nitric acid reaction.

• If \(K\) is much greater than 1, the reaction favors the formation of products.
• If \(K\) is much less than 1, it suggests reactants are favored.
• A \(K\) value close to 1 indicates that neither the products nor the reactants are favored significantly.

In this exercise, using the Nernst equation, the equilibrium constant \(K\) was calculated to be \(5.59 \times 10^3\), suggesting a strong preference for product formation in the given chemical reaction. This indicates that when these species are mixed, a large proportion will convert to product form under the given conditions. Understanding how \(K\) affects the composition of gases can be crucial in predicting reaction outcomes and adjusting reactant concentrations for desired products in industrial processes.

Nernst Equation

The Nernst Equation is a fundamental tool in electrochemistry that relates the reduction potential of a half-cell at non-standard conditions to the standard electrode potential, temperature, and activities (or concentrations) of the reacting species. It's expressed as: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where \( E \) is the electrode potential, \( E^\circ \) is the standard electrode potential, \( R \) is the 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 the exercise, the Nernst equation is used in calculating the equilibrium constant \( K \) for the nitric acid reaction with copper. By setting \( E \) to zero (since \( E \) equals zero at equilibrium) and rearranging for \( K \), they obtained: \[ K = \exp\left(-\frac{nF \cdot \Delta E^\circ_{\text{cell}}}{RT}\right) \] This application allows the conversion of electrical measurements into chemical equilibrium terms, giving insight into the reaction's favorability under specific conditions. The Nernst equation thus bridges the gap between electrochemical and chemical equilibrium by providing a method to calculate equilibrium constants from redox potentials.