Question

Consider the following reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \qquad K_{298}=0.090\) For \(\mathrm{Cl}_{2} \mathrm{O}(g)\) $$\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{kJ} / \mathrm{mol}$$ $$\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{kJ} / \mathrm{mol}$$ $$S^{\circ}=266.1 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}$$ a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=-R T \ln (K)\) b. Use bond energy values (Table 8.5\()\) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and b to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\) e. Estimate the value of \(K\) at \(500 . \mathrm{K}\) . f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=\) 2.0 torr, and \(P_{\mathrm{HOC}}=0.10\) torr.

Short answer

a. Using the given equilibrium constant and the formula ΔG° = -RT ln(K), we find ΔG° = 4.60 kJ/mol. b. Using bond energy values, we calculated ΔH° for the reaction to be -6 kJ/mol. c. From the results of part a and b, we estimated ΔS° for the reaction to be 35.6 J/molK. d. We estimated the ΔHf° and S° for HOCl(g) to be 37.15 kJ/mol and 245.25 J/molK, respectively. e. Using the van't Hoff equation, we estimated the value of K at 500 K to be approximately 0.587. f. Given the pressures of the gases involved, we calculated ΔG at 25°C to be 4.38 kJ/mol.

Step-by-step solution

Plug the values into the equation

Using the given equilibrium constant, K, at 298 K, we can find ΔG° with the formula ΔG° = -RT ln(K): ΔG° = -(8.314 J/mol·K)(298 K) ln(0.090)

Calculate ΔG°

Rearrange the equation and compute the result: ΔG° = -(8.314)(298) ln(0.090) = 4597.5 J/mol = 4.60 kJ/mol b. Use bond energy values to estimate ΔH° for the reaction.

Calculate bond energies for reactants and products

Refer to Table 8.5 to find the bond energy values for the reactants and products: Reactants: H₂O(g) - Bond energies: H-O (467 kJ), H-O (467 kJ) - Total: 934 kJ Cl₂O(g) - Bond energies: Cl-O (249 kJ), Cl-Cl (243 kJ) - Total: 492 kJ Products: 2 HOCl(g) - Bond energies: 2(H-O) (467 kJ) + 2(Cl-O) (249 kJ) - Total: 1432 kJ

Calculate ΔH° for the reaction

ΔH° = [Σ(Bond Energies of Reactants) - Σ(Bond Energies of Products)]: ΔH° = (934 + 492) - 1432 = 1426 - 1432 = -6 kJ/mol c. Use the results from parts a and b to estimate ΔS° for the reaction.

Use ΔG° = ΔH° - TΔS° to find ΔS°

Rearrange the equation: ΔS° = (ΔG° - ΔH°) / T ΔS° = (4.60 kJ/mol - (-6 kJ/mol)) / 298 K

Calculate ΔS°

Compute the result: ΔS° = (10.6 kJ/mol) / 298 K = 0.0356 kJ/molK = 35.6 J/molK d. Estimate ΔHf° and S° for HOCl(g)

Calculate ΔHf° for HOCl(g)

Using the overall reaction (ΔH° = -6 kJ/mol) and the information given (ΔHf° for Cl2O(g) = 80.3 kJ/mol), calculate ΔHf° for HOCl(g): ΔHf°(HOCl) = (2 * ΔHf°(HOCl) - ΔHf°(Cl2O) = ΔH°) ΔHf°(HOCl) = (ΔH° + ΔHf°(Cl2O)) / 2

Calculate ΔHf°

Compute the result: ΔHf°(HOCl) = (-6 kJ/mol + 80.3 kJ/mol) / 2 = 37.15 kJ/mol

Estimate S° for HOCl(g)

To estimate S° for HOCl(g), use the overall reaction (ΔS° = 35.6 J/molK) and the information given (S° for Cl2O(g) = 266.1 J/molK) to form the equation: ΔS°(HOCl) = (2 * S°(HOCl) - S°(H2O) - S°(Cl2O)) Assume S° for water is approximately 188.8 J/molK (since entropies are similar for gases and it is given in the question): ΔS°(HOCl) = (2 * S°(HOCl) - 188.8 J/molK - 266.1 J/molK)

Calculate S°

Rearrange the equation and compute the result: S°(HOCl) = (35.6 + 188.8 + 266.1) / 2 = 245.25 J/molK e. Estimate the value of K at 500 K

Use the van't Hoff equation

The van't Hoff equation is used relating the change in equilibrium constant with temperature: \( \ln{\frac{K_{2}}{K_{1}}} = \frac{-ΔH°}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}}) \) Given information: K1 = 0.090 at T1 = 298 K T2 = 500 K ΔH° = -6 kJ/mol

Calculate K2

Rearrange the equation and compute the result: \( \ln{\frac{K_{2}}{0.090}} = \frac{-(-6) (1000)}{8.314} (\frac{1}{500} - \frac{1}{298}) \) Find K2: K2 ≈ 0.587 f. Calculate ΔG at 25°C when PH2O = 18 torr, PCl2O = 2.0 torr, and PHOC = 0.10 torr.

Convert pressures to atm

Convert the given pressures from torr to atm: PH2O = 18 torr / 760 = 0.0237 atm PCl2O = 2.0 torr / 760 = 0.00263 atm PHOC = 0.10 torr / 760 = 0.0001316 atm

Calculate Reaction Quotient (Q)

Calculate the Reaction Quotient (Q) using the given pressures: Q = (PHOC^2) / (PH2O * PCl2O) Q = (0.0001316^2) / (0.0237 * 0.00263)

Calculate ΔG at 25°C

Next, find the ΔG given the temperature and reaction quotient from Step 2: ΔG = ΔG° + R * T * ln(Q) ΔG = 4600 + (8.314 J/mol·K)(298 K) * ln(Q) Compute the result: ΔG ≈ 4376.3 J/mol = 4.38 kJ/mol

Key concepts

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a crucial concept in thermodynamics in chemistry that helps us predict whether a chemical reaction will occur spontaneously. When you calculate \( \Delta G \), you determine the energy available to do work during a constant temperature and pressure process. This makes it invaluable for chemists trying to understand reaction feasibility.
The formula \( \Delta G = \Delta H - T \Delta S \) illustrates how free energy relies on both enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) changes. Here, \( \Delta H \) represents heat change, and \( \Delta S \) relates to disorder change. At a given temperature \( T \), if \( \Delta G \) is negative, the process is spontaneous.

• If \( \Delta G = 0 \), the system is at equilibrium.
• If \( \Delta G > 0 \), the reaction is non-spontaneous.

For the specific reaction in our problem, we calculated \( \Delta G^{\circ} \) using the relationship \( \Delta G^{\circ} = -RT \ln{K} \) with known equilibrium constants and temperature conditions. Understanding these relationships enhances our ability to predict reaction behavior.

Chemical Equilibrium

Chemical equilibrium occurs in reversible reactions where the rate of the forward reaction equals the rate of the reverse reaction. It is a dynamic state where concentrations remain constant over time. At equilibrium, the ratio of product to reactant concentrations is described by the equilibrium constant (\( K \)).
In our example, the equilibrium constant \( K \) is provided at 298 K, facilitating the calculation of \( \Delta G^{\circ} \). Knowing \( K \) and using the van't Hoff equation, we estimate how \( K \) changes with temperature, enabling predictions at 500 K.
Understanding chemical equilibrium and its constant allows us to:

• Evaluate reaction extent and direction.
• Predict changes with temperature adjustments.
• Calculate other thermodynamic quantities efficiently.

These insights are essential for chemists managing reaction conditions and process optimization.

Entropy

Entropy (\( S \)) is a measure of disorder or randomness in a system. It underlines the tendency of systems towards states with maximum probability, often equated with increasing disorder. In this context, a reaction which leads to greater disorder is favored if the entropy change (\( \Delta S \)) is positive.
For our reaction, knowing \( \Delta G^{\circ} \) and \( \Delta H^{\circ} \), we calculated \( \Delta S^{\circ} \) using the formula \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \).
This calculation involved rearranging the equation to solve for entropy change, revealing how entropy changes contribute to overall free energy.

• Higher \( \Delta S \) suggests increased disorder favored by the reaction.
• Entropy change balances the energy equation, affecting spontaneity.

Understanding entropy provides deeper insights into why certain reactions proceed naturally and others do not.

Enthalpy

Enthalpy (\( \Delta H \)) represents the total heat content of a system under constant pressure. It's an indicator of energy absorbed or released when a chemical reaction occurs.
For the exercise provided, we estimated the \( \Delta H \) using bond energy values from Table 8.5. Bond energy gives the average amount of energy required to break one mole of a bond in a gaseous substance. The calculated \( \Delta H \) value indicated whether the reaction absorbed or released heat.\( \Delta H^{\circ} \) values:

• Negative \( \Delta H \): Exothermic reactions release energy.
• Positive \( \Delta H \): Endothermic reactions absorb energy.

In our case, the reaction was mildly exothermic. Understanding enthalpy changes allows chemists to assess reaction heat changes, crucial for manipulating energy output or ensuring safety in chemical processes.