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) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Step-by-Step Solution

Answer

The calculated \(\Delta S^{\circ}\) values for the reactions are: (a) \(\Delta S^{\circ}_{1} = S^{\circ}(C_2H_6) - [S^{\circ}(C_2H_4) + S^{\circ}(H_2)]\) (b) \(\Delta S^{\circ}_{2} = 2 \cdot S^{\circ}(NO_2) - S^{\circ}(N_2O_4)\) (c) \(\Delta S^{\circ}_{3} = [S^{\circ}(BeO) + S^{\circ}(H_2O_{(g)})] - S^{\circ}(Be(OH)_2_{(s)})\) (d) \(\Delta S^{\circ}_{4} = [2 \cdot S^{\circ}(CO_2) + 4 \cdot S^{\circ}(H_2O_{(g)})] - [2 \cdot S^{\circ}(CH_3OH_{(g)}) + 3 \cdot S^{\circ}(O_2)]\) Substitute the standard entropy values from Appendix C for each species and calculate the sum of entropies for reactants and products. The sign of \(\Delta S^{\circ}\) depends on whether the sum of the products' entropy is greater or lesser than the sum of the reactants' entropy.

Step 1: (a) Calculate \(\Delta S^{\circ}\) for the first reaction.

For the first reaction, we have: C2H4(g) + H2(g) → C2H6(g) To calculate the \(\Delta S^{\circ}\), substitute the standard entropy values of each species from Appendix C into the formula, then calculate the sum of entropies for reactants and products, and find the difference. \(\Delta S^{\circ}_{1} = S^{\circ}(C_2H_6) - [S^{\circ}(C_2H_4) + S^{\circ}(H_2)]\)

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

For the second reaction, we have: N2O4(g) → 2NO2(g) To calculate the \(\Delta S^{\circ}\), substitute the standard entropy values of each species from Appendix C into the formula, then calculate the sum of entropies for reactants and products, and find the difference. \(\Delta S^{\circ}_{2} = 2 \cdot S^{\circ}(NO_2) - S^{\circ}(N_2O_4)\)

Step 3: (c) Calculate \(\Delta S^{\circ}\) for the third reaction.

For the third reaction, we have: Be(OH)2(s) → BeO(s) + H2O(g) To calculate the \(\Delta S^{\circ}\), substitute the standard entropy values of each species from Appendix C into the formula, then calculate the sum of entropies for reactants and products, and find the difference. \(\Delta S^{\circ}_{3} = [S^{\circ}(BeO) + S^{\circ}(H_2O_{(g)})] - S^{\circ}(Be(OH)_2_{(s)})\)

Step 4: (d) Calculate \(\Delta S^{\circ}\) for the fourth reaction.

For the fourth reaction, we have: 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g) To calculate the \(\Delta S^{\circ}\), substitute the standard entropy values of each species from Appendix C into the formula, then calculate the sum of entropies for reactants and products, and find the difference. \(\Delta S^{\circ}_{4} = [2 \cdot S^{\circ}(CO_2) + 4 \cdot S^{\circ}(H_2O_{(g)})] - [2 \cdot S^{\circ}(CH_3OH_{(g)}) + 3 \cdot S^{\circ}(O_2)]\) Now, refer to Appendix C to substitute the \(S^{\circ}\) values and then calculate \(\Delta S^{\circ}\) for each reaction. The sign of \(\Delta S^{\circ}\) will be determined by whether the entropy of the products is greater or lesser than the entropy of the reactants.

Key Concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemical reactions, thermodynamics helps us understand how energy changes during reactions. A crucial aspect we look at is the entropy change, which refers to the disorder or randomness associated with the chemical substances involved.

Entropy, denoted by the symbol \(S\), is a measure of the number of specific ways a thermodynamic system can be arranged. The second law of thermodynamics states that the entropy of an isolated system will increase over time or remain constant in ideal cases where the system is at equilibrium.

• In simpler terms, natural processes tend to move towards a state of maximum disorder or randomness.
• This is why understanding entropy helps predict the direction in which a chemical reaction will proceed.

In our exercise, thermodynamics guides us through using standard entropy values to calculate changes in entropy for reactions. This provides insights into whether the reactions are likely to occur spontaneously.

Chemical reactions

Chemical reactions involve the reorganization of atoms, leading to the transformation of substances. The molecules or compounds present before the reaction are called reactants, and the new substances formed are the products.

Chemical reactions can be influenced by various factors, including temperature, pressure, and the presence of catalysts, but a core aspect is the change in entropy associated with these reactions.

• Entropy change helps determine the feasibility and spontaneity of a reaction.
• In our exercise, each chemical equation represents a different reaction where the entropy change is calculated based on the standard entropy values of the reactants and products.

Understanding these reactions and the related entropy changes enables us to predict whether a reaction will result in increased or decreased disorder.

Entropy change

Entropy change, represented as \(\Delta S^\circ\), is the difference in entropy between the products and reactants of a chemical reaction. Each reaction has a specific standard entropy change, which helps in evaluating the reaction's spontaneity.

To compute the entropy change for a reaction, subtract the sum of the standard entropies of the reactants from the sum of the standard entropies of the products:

• \(\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}\)

In our exercise, the chemical reactions provided guide us to use given standard entropy values to determine \(\Delta S^\circ\):

• For positive \(\Delta S^\circ\), the products are more disordered than the reactants, suggesting a likely spontaneous reaction.
• For negative \(\Delta S^\circ\), the reactants are more disordered than the products, generally implying non-spontaneity under standard conditions.

This calculation helps to emphasize the fundamental aspect that all physical and chemical changes aim towards increasing entropy, aligning with the second law of thermodynamics.