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Chapter 16: Problem 107

Using data from Appendix \(4,\) calculate \(\Delta H^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{km}\) above the surface of the earth, the temperature is about \(230 . . \mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

Short Answer

Expert verified
We can calculate ΔH°, ΔG°, and the equilibrium constant K at 298 K for the given reaction using thermodynamic data: ΔH° = 285.4 kJ, ΔG° = 326.4 kJ, and K ≈ 1.3 × 10^(-56) Using the K-value, we estimate the partial pressure of ozone (pO3) at 30 km above the earth's surface to be approximately 1.16 × 10^(-19) atm. The extremely low partial pressure for ozone suggests that the equilibrium state is not maintained under these conditions. Due to the low concentration of ozone, the reaction between 3 oxygen molecules is extremely rare, so it is not reasonable to assume that the equilibrium between oxygen and ozone is maintained.

Step by step solution

01

Calculate standard enthalpy change (ΔH°)

Using the thermodynamic data from Appendix 4, we find the standard enthalpies of the formation of O2 and O3: ΔHf° (O2)(g) = 0 kJ/mol (by definition) ΔHf° (O3)(g) = 142.7 kJ/mol Using these values, we can calculate ΔH° for the reaction: ΔH° = [(2 mol O3) × (142.7 kJ/mol)] - [(3 mol O2) × (0 kJ/mol)] = 285.4 kJ
02

Calculate standard Gibbs free energy change (ΔG°)

Now, let's calculate ΔG° for the reaction using the standard Gibbs free energies of formation found in Appendix 4: ΔGf° (O2)(g) = 0 kJ/mol (by definition) ΔGf° (O3)(g) = 163.2 kJ/mol ΔG° = [(2 mol O3) × (163.2 kJ/mol)] - [(3 mol O2) × (0 kJ/mol)] = 326.4 kJ
03

Calculate the equilibrium constant (K) at 298 K

We can calculate the equilibrium constant K at 298 K using the relationship: K = exp(-ΔG°/(RT)) where R = 8.314 J/(mol K) and T = 298 K K = exp(-326400 J/(8.314 J/(mol K) × 298 K)) ≈ 1.3 × 10^(-56) Step 2: Estimate partial pressure of ozone at 30 km above the earth's surface.
04

Calculate the partial pressure of ozone (pO3)

We have calculated K-value, and the partial pressure of O2, pO2, is given as 1.0 × 10^(-3) atm. From the given reaction, we can write down the equilibrium expression: K = [pO3]^2 / [pO2]^3 Now, let's solve for the partial pressure of ozone pO3: pO3 = sqrt(K * [pO2]^3) pO3 = sqrt(1.3 × 10^(-56) * (1.0 × 10^(-3) atm)^3) ≈ 1.16 × 10^(-19) atm Step 3: Assess if equilibrium assumption is reasonable
05

Determine if equilibrium assumption is reasonable

The equilibrium between oxygen and ozone results in an extremely low partial pressure for ozone, which suggests that the equilibrium state is not maintained under these conditions. Due to the low concentration of ozone, the reaction between 3 oxygen molecules is extremely rare, so it is not reasonable to assume that the equilibrium between oxygen and ozone is maintained.

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Most popular questions from this chapter

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}.\) a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(S_{\text {rhombic }}(s) \longrightarrow S_{\text {monoclinic }}(s).\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Consider the following reaction at \(800 . \mathrm{K}:\) $$\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{atm}, P_{\mathrm{NF}_{3}}=0.48\) atm. Calculate \(\Delta G^{\circ}\) for the reaction at \(800 .\) K.

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q,\) specifically for aqueous reactions. With this in mind, consider the reaction $$HF(a q) \rightleftharpoons H^{+}(a q)+F^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C}.\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 M\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} M\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S ?\) \(\Delta S_{\text {univ }}\) ? Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of AB is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.
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