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Chapter 22: Problem 94

Using \(\Delta G_{f}^{\circ}\) for \(\mathrm{NO}\) and \(\mathrm{NO}_{2},\) from Appendix \(\mathrm{C},\) calculate the equilibrium constant for the oxidation of \(\mathrm{NO}\) to \(\mathrm{NO}_{2}\) at \(298.0 \mathrm{~K}\) as described in equation \(22.38 .\)

Short Answer

Expert verified
The equilibrium constant (K) for the oxidation of NO to NO2 at 298.0 K is approximately 3.47 × 10⁻³. This was calculated using the standard Gibbs free energy of formation values and the equation ΔG° = -RTln(K).

Step by step solution

01

Write the balanced chemical equation

We must first write the balanced chemical equation for the oxidation of NO to NO2: \[ 2NO (g) \rightarrow NO_{2} (g) \]
02

Find the standard Gibbs free energy change of the reaction

To find the standard Gibbs free energy change (ΔG°) for the reaction, we will use the standard Gibbs free energy of formation (ΔGf°) values for the products and reactants from Appendix C: ΔG° = (ΔGf°[NO2]) - 2(ΔGf°[NO]) Now, plug in the values given: ΔG° = (-33.18kJ/mol) - 2(-86.57kJ/mol)
03

Calculate the standard Gibbs free energy change

Perform the calculation: ΔG° = -33.18kJ/mol + 173.14kJ/mol ΔG° = 139.96 kJ/mol
04

Use the Gibbs free energy change to calculate the equilibrium constant (K) at 298.0 K

Using the equation ΔG° = -RTln(K), we can find the equilibrium constant K. First, convert the temperature to Kelvin and rearrange the formula to solve for K: \[ K = e^{-\frac{ΔG°}{RT}} \] Where: R = Gas constant = 8.314 J/(mol·K) T = Temperature = 298.0 K ΔG° = 139960 J/mol (convert kJ/mol to J/mol) Plug in the values: \[ K = e^{\frac{-139960J/mol}{8.314J/(mol\cdot K) \cdot 298.0K}} \]
05

Calculate the equilibrium constant (K)

Perform the calculation: \[ K = e^{-5.65} \] \[ K = 3.47 \times 10^{-3} \] The equilibrium constant (K) for the oxidation of NO to NO2 at 298.0 K is approximately 3.47 × 10⁻³.

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