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

Short Answer

Expert verified
Question: Calculate (a) the standard entropy change for the reaction between sodium peroxide and water, (b) the standard entropy for sodium peroxide, and (c) the standard enthalpy of formation for sodium peroxide, given the standard enthalpy and free energy changes, and the standard entropies for NaOH, H鈧侽, and O鈧.

Step by step solution

01

a) Calculating the standard entropy change (螖S掳)

Using the Gibbs-Helmholtz equation, which relates the standard Gibbs free energy change (螖G掳) with the standard enthalpy change (螖H掳) and the standard entropy change (螖S掳) at a specific temperature (T), $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $$ We are given 螖H掳 = -109.0 kJ and 螖G掳 = -148.4 kJ at T = 25掳C (298 K). We can now solve for 螖S掳: $$ \Delta S^{\circ} = \frac{\Delta H^{\circ}-\Delta G^{\circ}}{T} $$ Plug in the values and calculate 螖S掳.
02

Discussing the reasonability of the 螖S掳 sign

We will look at the products and reactants of the given reaction: 2 Na鈧侽鈧(s) + 2 H鈧侽(l) 鈫 4 NaOH(s) + O鈧(g) We can see that there's an increase in the number of gas molecules in the products compared to the reactants. This results in an increase in disorder (entropy). So, having a positive 螖S掳 would be reasonable.
03

b) Calculating the standard entropy (S掳) for Na鈧侽鈧(s)

Using the standard entropies provided below (all values in J mol鈦宦 K鈦宦): NaOH(s): 64.5 H鈧侽(l): 70.0 O鈧(g): 205.2 We will calculate the standard entropy of Na鈧侽鈧(s) by rearranging the entropy-change equation for the reaction: $$ \Delta S^{\circ}=\sum S^{\circ}_{\text{products}}-\sum S^{\circ}_{\text{reactants}} $$ Plug in the known entropy values and solve for the standard entropy of Na鈧侽鈧(s).
04

c) Calculating the standard enthalpy of formation (螖Hf掳) for Na鈧侽鈧(s)

We will use the standard enthalpy-of-formation-change equation for the reaction as follows: $$ \Delta H^{\circ}=\sum \Delta H_{f}^{\circ}(\text{products})-\sum \Delta H_{f}^{\circ}(\text{reactants}) $$ Using the given standard enthalpies of formation (all values in kJ mol鈦宦): H鈧侽(l): -285.8 NaOH(s): -425.6 and rearranging the equation above, we can solve for the standard enthalpy of formation of Na鈧侽鈧(s): $$ \Delta H_{f}^{\circ}(\text{Na}_{2}\text{O}_{2}\text{(s)}) = \frac{\Delta H^{\circ} - \sum \Delta H_{f}^{\circ}(\text{products}) + \sum \Delta H_{f}^{\circ}(\text{reactants})}{2} $$ Plug in the values and calculate 螖Hf掳 for Na鈧侽鈧(s).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy, represented as \( G \), plays a pivotal role in thermodynamics within chemistry; primarily, it helps predict whether a chemical reaction will occur spontaneously under constant pressure and temperature. A reaction is considered spontaneous if it leads to a decrease in the system's free energy (\(\Delta G^\circ < 0\)). The concept revolves around energy conservation and conversions between different forms, such as heat and work.When discussing Gibbs free energy in the context of chemical reactions, the Gibbs-Helmholtz equation is often utilized:\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\]This equation links the change in Gibbs free energy (\(\Delta G^\circ\)) to the change in enthalpy (\(\Delta H^\circ\)) and change in entropy (\(\Delta S^\circ\)) at a constant temperature (\(T\)). In essence, it's a balance between the energy given off or absorbed (in the form of heat) and the degree of freedom at the molecular level鈥攁n interplay of enthalpy and entropy. The temperature, measured in Kelvin, scales the impact of entropy change on the free energy change.In our exercise, the given reaction is spontaneous as the calculated Gibbs free energy change is negative (\(\Delta G^\circ = -148.4 \text{kJ}\)), indicating the process can proceed without external energy input. This forms the groundwork for many predictions and designs in chemical processes, such as reaction feasibility and the direction of equilibria.
Standard Enthalpy Change
Standard enthalpy change, represented as \( \Delta H^\circ \), signifies the heat absorbed or released during a chemical reaction at standard conditions (1 atm pressure and, usually, 25掳C or 298K). It is intrinsic to the types of bonds formed and broken in the reaction, with endothermic reactions absorbing heat (\(\Delta H^\circ > 0\)) and exothermic reactions releasing heat (\(\Delta H^\circ < 0\)).

Chemists use this concept to assess the thermal aspect of reactions and how much energy is involved in breaking and forming chemical bonds. It鈥檚 also handy in calculating how heat changes affect the surroundings and the overall energy efficiency of chemical processes.

In the exercise, the reaction exhibits a negative standard enthalpy change (\(\Delta H^\circ = -109.0 \text{kJ}\)), classifying it as exothermic. Thus, it releases heat to its surroundings. The standard enthalpy change is also pivotal in calculating other thermodynamic properties, such as Gibbs free energy.
Standard Entropy Change
Standard entropy change, indicated as \( \Delta S^\circ \), quantifies the change in disorder or randomness in a system during a chemical process. Entropy is a fundamental concept in thermodynamics embodying the distribution of energy within a system, and by extension, the possible microstates a system can occupy. Increases in entropy (\(\Delta S^\circ > 0\)) suggest a transition towards more disorder, while decreases imply the opposite.

In chemical reactions, entropy considers molecular complexity, phase changes, and changes in the number of particles. Reactions that produce gases typically show a significant increase in entropy because gases have more random movement and higher disorder than solids or liquids.

Our exercise explores the entropy change in a reaction forming a gas, oxygen, from solid and liquid reactants. The expectation is a positive entropy change, supported by the computation using the Gibbs-Helmholtz equation. Analyzing the entropy change offers insight into the 'driving force' behind reaction spontaneity; it's not just about the energy exchanges but also how the system鈥檚 order is affected.

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Most popular questions from this chapter

In your own words, explain why (a) \(\Delta S^{\circ}\) is negative for a reaction in which the number of moles of gas decreases. (b) we take \(\Delta S^{\circ}\) to be independent of \(T\), even though entropy increases with \(T\). (c) a solid has lower entropy than its corresponding liquid.

Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}\) for \(\mathrm{HF}(a q)\) is \(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\), find \(K_{\mathrm{a}}\) for \(\mathrm{HF}\) at \(25^{\circ} \mathrm{C}\).

Given the following standard free energies at \(25^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}(g)+\frac{3}{2} \mathrm{O}_{2}(g) & \Delta G^{\circ}=-59.2 \mathrm{~kJ} \\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & \Delta G^{\circ}=-35.6 \mathrm{~kJ} \end{array} $$ calculate \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{NO}_{2}(g)+{ }_{2}^{1} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$

Fill in the blanks: (a) \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) become equal at _____ \(K\). (b) \(\Delta G^{\circ}\) and \(\Delta G\) are equal when \(Q=\) _____. (c) \(S^{\circ}\) for steam is _____ than \(S^{\circ}\) for water.

Sulfur has about 20 different allotropes. The most common are rhombic sulfur (the stable form at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) ) and monoclinic sulfur. They differ in their crystal structures. Given $$ \mathrm{S}(s, \text { monoclinic }): \Delta H_{\mathrm{f}}^{\circ}=0.30 \mathrm{~kJ} / \mathrm{mol}, S^{\circ}=0.0326 \mathrm{~kJ} / \mathrm{mol} \cdot \mathrm{K} $$ at what temperature are the two forms in equilibrium?

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