Problem 53 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) \(\Delta S^{\circ}\) = \(-121\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\), indicating a decrease in entropy and a more organized reaction. (b) \(\Delta S^{\circ}\) = \(175.7\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\), indicating an increase in entropy and a more disorganized reaction. (c) For \(Be(OH)_2(s) \longrightarrow BeO(s) + H_2O(g)\), use the same process to find the \(\Delta S^{\circ}\) value and explain its sign.

Step 1: Write down the entropy values

Using Appendix C values, we have: - \(S^{\circ}_{C_2H_4(g)}\) = \(219.6\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\) - \(S^{\circ}_{H_2(g)}\) = \(130.6\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\) - \(S^{\circ}_{C_2H_6(g)}\) = \(229.2\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\)

Step 2: Calculate \(\Delta S^{\circ}\) for the reaction

Using the formula \(\Delta S^{\circ} = \sum{S^{\circ}_\text{products}} - \sum{S^{\circ}_\text{reactants}}\), we get: \(\Delta S^{\circ}\) = \((S^{\circ}_{C_2H_6(g)}) - (S^{\circ}_{C_2H_4(g)} + S^{\circ}_{H_2(g)})\) \(\Delta S^{\circ}\) = \((229.2 - (219.6 + 130.6))\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\) \(\Delta S^{\circ}\) = \(-121\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\)

Step 3: Account for the sign of \(\Delta S^{\circ}\)

Since the value of \(\Delta S^{\circ}\) is negative, this indicates a decrease in entropy and the reaction is becoming more organized as it proceeds. (b) \(N_2O_4(g) \longrightarrow 2 NO_2(g)\)

Step 1: Write down the entropy values

Using Appendix C values, we have: - \(S^{\circ}_{N_2O_4(g)}\) = \(304.3\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\) - \(S^{\circ}_{NO_2(g)}\) = \(240.0\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\)

Step 2: Calculate \(\Delta S^{\circ}\) for the reaction

Using the formula \(\Delta S^{\circ} = \sum{S^{\circ}_\text{products}} - \sum{S^{\circ}_\text{reactants}}\), we get: \(\Delta S^{\circ}\) = \((2S^{\circ}_{NO_2(g)}) - (S^{\circ}_{N_2O_4(g)})\) \(\Delta S^{\circ}\) = \((2(240.0) - 304.3)\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\) \(\Delta S^{\circ}\) = \(175.7\: \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\)

Step 3: Account for the sign of \(\Delta S^{\circ}\)

Since the value of \(\Delta S^{\circ}\) is positive, this indicates an increase in entropy and the reaction is becoming more disorganized as it proceeds. (c) \(Be(OH)_2(s) \longrightarrow BeO(s) + H_2O(g)\) Repeat the same process for the remaining reactions to obtain the \(\Delta S^{\circ}\) and explain the sign.

Key Concepts

Standard Molar Entropy

Standard molar entropy, denoted as S°, is a fundamental concept in chemical thermodynamics providing a quantitative measure of the amount of disorder or randomness in a substance at the standard state, commonly 1 bar pressure and 25°C (298 K). The units for standard molar entropy are joules per kelvin per mole \(\frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\).

Standard molar entropy values for substances are essential for calculating the entropy change \(\Delta S°\) in chemical reactions. For instance, when given a reaction, the S° values are required for both reactants and products. By summing the S° values of the products and subtracting the sum of the S° values of the reactants, we can determine \(\Delta S°\) for the entire reaction, as seen in the provided exercise. The sign of \(\Delta S°\) gives insight into whether the reaction results in an increase (- disorder or randomness) or a decrease (+ order) in the overall system entropy.

Gibbs Free Energy

Gibbs free energy, represented by the symbol \(G\), is an essential quantity in chemical thermodynamics that reveals the spontaneity of a reaction under constant pressure and temperature. It is defined by the equation \(G = H - TS\), where \(H\) is the enthalpy, \(T\) is the temperature in Kelvin, and \(S\) is the entropy.

The change in Gibbs free energy, \(\Delta G\), can be calculated with the relation \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the change in enthalpy and \(\Delta S\) is the change in entropy of the reaction. A negative \(\Delta G\) signifies a spontaneous reaction, while a positive \(\Delta G\) indicates that the reaction is non-spontaneous under those conditions. Educational resources focus on the importance of \(\Delta G\) for predicting the direction and extent of chemical reactions.

Chemical Thermodynamics

Chemical thermodynamics is the branch of physical chemistry that studies the interrelation of energy transfers and the properties of substances with the laws of thermodynamics. It provides a rigorous framework to understand how energy transformations are implicated in chemical reactions and processes.

The principles of chemical thermodynamics are employed to determine the feasibility and extent of reactions, involving concepts like enthalpy \(H\), entropy \(S\), and the aforementioned Gibbs free energy \(G\). These quantities interrelate to depict the thermodynamic landscape of a reaction. Knowing the signs and magnitudes of these quantities helps chemists to elucidate whether a chemical reaction will proceed spontaneously or if external energy input is needed.

Entropy Change of Reaction

Entropy change of reaction, symbolized as \(\Delta S_{\text{reaction}}\), quantifies the overall entropy change that occurs as a chemical reaction proceeds from reactants to products. This value can be either positive or negative, depending on whether the reaction increases or decreases the system's disorder.

A positive \(\Delta S_{\text{reaction}}\) corresponds to a gain in entropy, indicative of a more disordered state post-reaction. This is commonly seen in reactions where gases are formed from solids or liquids or when the number of gas molecules increases. A negative \(\Delta S_{\text{reaction}}\) reflects a loss in entropy and occurs when the system becomes more ordered, as in the synthesis of a complex molecule from simpler precursors. The detailed calculation of \(\Delta S_{\text{reaction}}\) involves standard molar entropy values, as highlighted in the problem's step-by-step solution.