Question

Using \(S^{\circ}\) values from Appendix \(\mathrm{C},\) calculate \(\Delta S^{\circ}\) values for the following reactions. In each case, account for the sign of \(\Delta S^{\circ} .\) $$ \begin{array}{l}{\text { (a) } \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} \\ {\text { (b) } \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)} \\ {\text { (c) } \mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } 2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$

Short answer

(a) ΔS° = -121.03 J/mol·K (decrease in entropy) (b) ΔS° = 175.77 J/mol·K (increase in entropy) (c) ΔS° = 155.31 J/mol·K (increase in entropy) (d) ΔS° = 33.3 J/mol·K (increase in entropy)

Step-by-step solution

(a) Calculate ΔS° for the reaction C2H4(g) + H2(g) ⟶ C2H6(g)

First, look up the standard molar entropies for C2H4(g), H2(g), and C2H6(g) from Appendix C: \(S^{\circ}_{C_{2}H_{4}(g)} = 219.52 \ J/mol\cdot K\) \(S^{\circ}_{H_{2}(g)} = 130.68 \ J/mol\cdot K\) \(S^{\circ}_{C_{2}H_{6}(g)} = 229.17 \ J/mol\cdot K\) Now, calculate the change in entropy (ΔS°): ΔS° = Total entropy of products - Total entropy of reactants ΔS° = \(S^{\circ}_{C_{2}H_{6}(g)} - (S^{\circ}_{C_{2}H_{4}(g)} + S^{\circ}_{H_{2}(g)})\) ΔS° = (229.17 - (219.52 + 130.68)) J/mol·K ΔS° = -121.03 J/mol·K Since the sign of ΔS° is negative, it means the reaction leads to a decrease in entropy.

(b) Calculate ΔS° for the reaction N2O4(g) ⟶ 2 NO2(g)

First, look up the standard molar entropies for N2O4(g) and NO2(g) from Appendix C: \(S^{\circ}_{N_{2}O_{4}(g)} = 304.29 \ J/mol\cdot K\) \(S^{\circ}_{NO_{2}(g)} = 240.03 \ J/mol\cdot K\) Now, calculate the change in entropy (ΔS°): ΔS° = Total entropy of products - Total entropy of reactants ΔS° = \(2 \times S^{\circ}_{NO_{2}(g)} - S^{\circ}_{N_{2}O_{4}(g)}\) ΔS° = (2 × 240.03 - 304.29) J/mol·K ΔS° = 175.77 J/mol·K Since the sign of ΔS° is positive, it means the reaction leads to an increase in entropy.

(c) Calculate ΔS° for the reaction Be(OH)2(s) ⟶ BeO(s) + H2O(g)

First, look up the standard molar entropies for Be(OH)2(s), BeO(s), and H2O(g) from Appendix C: \(S^{\circ}_{Be(OH)_{2}(s)} = 46.95 \ J/mol\cdot K\) \(S^{\circ}_{BeO(s)} = 13.43 \ J/mol\cdot K\) \(S^{\circ}_{H_{2}O(g)} = 188.83 \ J/mol\cdot K\) Now, calculate the change in entropy (ΔS°): ΔS° = Total entropy of products - Total entropy of reactants ΔS° = \((S^{\circ}_{BeO(s)} + S^{\circ}_{H_{2}O(g)}) - S^{\circ}_{Be(OH)_{2}(s)}\) ΔS° = (13.43 + 188.83 - 46.95) J/mol·K ΔS° = 155.31 J/mol·K Since the sign of ΔS° is positive, it means the reaction leads to an increase in entropy.

(d) Calculate ΔS° for the reaction 2 CH3OH(g) + 3 O2(g) ⟶ 2 CO2(g) + 4 H2O(g)

First, look up the standard molar entropies for CH3OH(g), O2(g), CO2(g), and H2O(g) from Appendix C: \(S^{\circ}_{CH_{3}OH(g)} = 239.82 \ J/mol\cdot K\) \(S^{\circ}_{O_{2}(g)} = 205.15 \ J/mol\cdot K\) \(S^{\circ}_{CO_{2}(g)} = 213.79 \ J/mol\cdot K\) \(S^{\circ}_{H_{2}O(g)} = 188.83 \ J/mol\cdot K\) Now, calculate the change in entropy (ΔS°): ΔS° = Total entropy of products - Total entropy of reactants ΔS° = \((2 \times S^{\circ}_{CO_{2}(g)} + 4 \times S^{\circ}_{H_{2}O(g)}) - (2 \times S^{\circ}_{CH_{3}OH(g)} + 3 \times S^{\circ}_{O_{2}(g)})\) ΔS° = (2 × 213.79 + 4 × 188.83 - 2 × 239.82 - 3 × 205.15) J/mol·K ΔS° = 33.3 J/mol·K Since the sign of ΔS° is positive, it means the reaction leads to an increase in entropy.

Key concepts

Standard Molar Entropy

Standard molar entropy, denoted as \(S^{\boxed{\textbf{∘}}}\), is a measure of the amount of disorder or randomness in a system, per mole of substance, under standard conditions (1 bar of pressure and the temperature of interest, usually 298.15 K). It is expressed in units of joules per mole kelvin (J/mol·K). This thermodynamic property helps us understand how particles are arranged in a system and how they can disperse energy in the form of heat or work.

For example, gases have higher standard molar entropies than liquids or solids due to their particles being more widely separated and having more freedom to move. In chemical reactions, calculating the change in standard molar entropy can tell us about the system's tendency towards disorder or order as reactants transform into products.

Change in Entropy

The change in entropy, denoted as \(\Delta S^{\boxed{\textbf{∘}}}\), is crucial in predicting how a chemical reaction will proceed. It is calculated by subtracting the total standard molar entropy of the reactants from that of the products. A positive value of \(\Delta S^{\boxed{\textbf{∘}}}\) signals an increase in disorder, indicating the reactants had a more ordered state than the products. Conversely, a negative value implies a decrease in disorder, suggesting the products are in a more ordered state than the reactants.

Understanding this change helps in assessing the spontaneity of reactions. In the exercise provided, reactions (b), (c), and (d) demonstrated an increase in entropy, suggesting a more disorderly system after the reactions have occurred, which is often the case when solid reactants form gaseous products or when more gaseous molecules are produced.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relationships between heat, work, temperature, and energy. The laws of thermodynamics govern how energy is transferred within a system and how these transfers can perform work or lead to changes in temperature or state of matter. These principles are vital in understanding the energy changes involved in chemical reactions and processes.

Entropy is a thermodynamic function that plays a central role in the Second Law of Thermodynamics, which states that the total entropy of an isolated system can never decrease over time. This law implies that natural processes tend to move towards a state of maximum entropy or disorder, explaining why many spontaneous reactions result in an increase in entropy.

Gibbs Free Energy

Gibbs free energy, denoted as \(G\), is a thermodynamic quantity used to predict the spontaneity of a process at constant temperature and pressure. It combines enthalpy (\(H\)), temperature (\(T\)), and entropy (\(S\)) into one value: \(G = H - TS\). When the change in Gibbs free energy, \(\Delta G\), for a process is negative, the process is spontaneous. If \(\Delta G\) is positive, the process is non-spontaneous and requires energy input to proceed.

A key aspect of applying Gibbs free energy to chemical reactions, as in the exercise, is relating \(\Delta G\) to \(\Delta H\) (change in enthalpy) and \(\Delta S\) (change in entropy). The formula \(\Delta G = \Delta H - T\Delta S\) allows us to understand how changes in heat and disorder affect the spontaneity of a reaction, ensuring a holistic view of the energy changes taking place.