Question

Oxygen can be made in the laboratory by reacting sodium peroxide and water. $$ \begin{array}{l} 2 \mathrm{Na}_{2} \mathrm{O}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{NaOH}(s)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=-109.0 \mathrm{~kJ} ; \Delta G^{\circ}=-148.4 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{array} $$ (a) Calculate \(\Delta S^{\circ} .\) Is the sign reasonable? (b) Calculate \(S^{\circ}\) for \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (c) Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)\)

Short answer

a) The standard entropy change (ΔS°) for the reaction is approximately 132.1 J/mol·K. The positive sign is reasonable because a gas is formed in the reaction, typically increasing entropy. b) The standard molar entropy (S°) for sodium peroxide (Na2O2) is approximately 107.48 J/mol·K. c) The standard enthalpy of formation (ΔHf°) for sodium peroxide (Na2O2) is approximately 576 kJ/mol.

Step-by-step solution

Calculate ΔS° using ΔG° and ΔH°

We can calculate the standard entropy change (ΔS°) using the standard free energy change (ΔG°) and the standard enthalpy change (ΔH°) by the equation: ΔG° = ΔH° - TΔS° Where T is the temperature in Kelvin. Given that the temperature is 25°C, we convert it to Kelvin: T = 25 + 273.15 = 298.15 K Now, we can rearrange the equation to solve for ΔS°: ΔS° = (ΔH° - ΔG°) / T Substitute the given values: ΔS° = (-109000 J/mol - (-148400 J/mol)) / 298.15 K

Evaluate ΔS° and check the sign

Calculate the value of ΔS°: ΔS° = (39400 J/mol) / 298.15 K ≈ 132.1 J/mol·K The positive sign of ΔS° is reasonable because in the reaction, a solid and a liquid are converted to a solid and a gas. The formation of gas typically increases the entropy, leading to a positive ΔS°. (b) Calculate S° for Na2O2(s).

Use the standard entropy values of reactants and products

To determine the standard molar entropy (S°) for sodium peroxide (Na2O2), we can use the standard entropy values for the other reactants and products in the given reaction. The standard entropy values at 25°C for the substances in the reaction are: S°(H2O) = 69.91 J/mol·K S°(NaOH) = 64.46 J/mol·K S°(O2) = 205.14 J/mol·K

Calculate S° for Na2O2(s) using ΔS°

Recall that: ΔS° = Σ(S°_products) - Σ(S°_reactants) Rearrange the equation to solve for S°(Na2O2): S°(Na2O2) = (ΔS° + Σ(S°_reactants)) / 2 - (S°_products) Substitute the given values, remembering that there are 2 moles of Na2O2 and H2O and 4 moles of NaOH in the reaction: S°(Na2O2) = (132.1 J/mol·K + 2*69.91 J/mol·K) / 2 - 2*64.46 J/mol·K - (205.14 J/mol·K)

Evaluate S° for Na2O2(s)

Calculate the value of S°(Na2O2): S°(Na2O2) = (132.1 + 139.82) / 2 - 128.92 - 205.14 ≈ 107.48 J/mol·K (c) Calculate ΔHf° for Na2O2(s).

Use the standard enthalpies of formation for other substances

To calculate the standard enthalpy of formation (ΔHf°) for Na2O2, we can use the known ΔHf° values for the other substances in the reaction. The standard enthalpies of formation at 25°C for the substances in the reaction are: ΔHf°(H2O) = -285.83 kJ/mol ΔHf°(NaOH) = -426.7 kJ/mol

Use Hess's law to determine ΔHf° for Na2O2(s)

To calculate ΔHf°(Na2O2) using Hess's law, we have: ΔH° = Σ(n*ΔHf°_products) - Σ(n*ΔHf°_reactants) Rearrange the equation to solve for ΔHf°(Na2O2): ΔHf°(Na2O2) = (ΔH° + Σ(n*ΔHf°_reactants)) / 2 - Σ(n*ΔHf°_products) Substitute the given values, remembering that there are 2 moles of Na2O2 and H2O and 4 moles of NaOH in the reaction: ΔHf°(Na2O2) = (-109000 J/mol + 2*(-285830 J/mol)) / 2 - 4*(-426700 J/mol)

Evaluate ΔHf° for Na2O2(s)

Calculate the value of ΔHf°(Na2O2) ΔHf°(Na2O2) = (-109000 - 571660) / 2 + 1706800 ≈ 576 kJ/mol The standard enthalpy of formation for Na2O2(s) is approximately 576 kJ/mol.